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If we apply the transformation \( \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right\rbrack \) to the unit square, then rotate the result by 45 degrees counterclockwise using the matrix \( \left\lbrack \begin{array}{rr} \cos \pi /4 & - \sin \pi /4 \\ \sin \pi /4 & \cos \pi /4 \end{array}\right\rbrack \) , ... | \[ \left\lbrack \begin{array}{rr} \cos \pi /4 & - \sin \pi /4 \\ \sin \pi /4 & \cos \pi /4 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} 2\cos \pi /4 & - \sin \pi /4 \\ 2\sin \pi /4 & \cos \pi /4 \end{array}\right\rbrack .\;\bigtriangleu... | Yes |
Example 1.3.12 (Associativity of matrix multiplication). For the food plant of Example 1.3.1, one might make a line matrix \( D, n \) wide, giving the price, per pound or liter, of each of the \( n \) ingredients. Since \( A \) is an \( n \times 3 \) matrix, the product \( {DA} \) is a line matrix 3 wide; \( A \) tells... | \( \bigtriangleup \) | No |
Proposition 1.3.13. A linear transformation \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) is invertible if and only if the \( m \times n \) matrix \( \left\lbrack T\right\rbrack \) is invertible. If it is invertible, then\n\n\[ \left\lbrack {T}^{-1}\right\rbrack = {\left\lbrack T\right\rbrack }^{-1} \] | Proof. Suppose that \( \left\lbrack T\right\rbrack \) is invertible, and let \( S : {\mathbb{R}}^{m} \rightarrow {\mathbb{R}}^{n} \) be the linear transformation such that \( \left\lbrack S\right\rbrack = {\left\lbrack T\right\rbrack }^{-1} \) . Then for any \( \overrightarrow{\mathbf{y}} \in {\mathbb{R}}^{m} \), we ha... | Yes |
Let \( \overrightarrow{\mathbf{x}},\overrightarrow{\mathbf{y}} \) be vectors in \( {\mathbb{R}}^{2} \) or in \( {\mathbb{R}}^{3} \), and let \( \alpha \) be the angle between them. Then\n\n\[ \overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{y}} = \left| \overrightarrow{\mathbf{x}}\right| \left| \overrightarrow... | Proof. This is an application of the cosine law from trigonometry, which says that if you know all of a triangle's sides, or two sides and the angle between them, or two angles and a side, then you can determine the others. Let a triangle have sides of length \( a, b, c \), and let \( \gamma \) be the angle opposite th... | Yes |
Theorem 1.4.5 (Schwarz’s inequality). For any vectors \( \overrightarrow{\mathbf{v}},\overrightarrow{\mathbf{w}} \in {\mathbb{R}}^{n} \), \[ \left| {\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}}\right| \leq \left| \overrightarrow{\mathbf{v}}\right| \left| \overrightarrow{\mathbf{w}}\right| \] | Proof. If either \( \overrightarrow{\mathbf{v}} \) or \( \overrightarrow{\mathbf{w}} \) is \( \overrightarrow{\mathbf{0}} \), the statement is obvious, so suppose both are nonzero. Consider the function \( {\left| \overrightarrow{\mathbf{v}} + t\overrightarrow{\mathbf{w}}\right| }^{2} \) as a function of \( t \). It is... | Yes |
What is the angle between the diagonal of a cube and any edge? | Let us assume that our cube is the unit cube \( 0 \leq x, y, z \leq 1 \), so that the standard basis vectors \( {\overrightarrow{\mathbf{e}}}_{1},{\overrightarrow{\mathbf{e}}}_{2},{\overrightarrow{\mathbf{e}}}_{3} \) are edges, and the vector \( \overrightarrow{\mathbf{d}} = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \e... | Yes |
Theorem 1.4.9 (The triangle inequality). For vectors \( \overrightarrow{\mathrm{x}},\overrightarrow{\mathrm{y}} \in {\mathbb{R}}^{n} \), \[ \left| {\overrightarrow{\mathbf{x}} + \overrightarrow{\mathbf{y}}}\right| \leq \left| \overrightarrow{\mathbf{x}}\right| + \left| \overrightarrow{\mathbf{y}}\right| \] | Proof. This inequality is proved by the following computation: \[ {\left| \overrightarrow{x} + \overrightarrow{y}\right| }^{2} = {\left| \overrightarrow{x}\right| }^{2} + 2\overrightarrow{x} \cdot \overrightarrow{y} + {\left| \overrightarrow{y}\right| }^{2}\underset{\text{ Schwarz }}{ \leq }{\left| \overrightarrow{x}\r... | Yes |
Example 1.4.12. Let \( A = \left\lbrack \begin{array}{lll} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 1 & 2 & 1 \end{array}\right\rbrack ,\overrightarrow{\mathbf{b}} = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack, B = \left\lbrack \begin{array}{lll} 1 & 0 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 0 \end{array}\right\rbrack \) .... | Proof. In inequality 1.4.27, note that if \( A \) consists of a single row (i.e., if \( A = {\overline{\mathbf{a}}}^{\top } \) is the transpose of a vector \( \overrightarrow{\mathbf{a}} \) ), the theorem is Schwarz’s inequality: \[ \underset{\left| {\overrightarrow{\mathbf{a}}}^{\top }\overrightarrow{\mathbf{b}}\right... | Yes |
Proposition 1.4.14 (Geometric interpretation of the determinant in \( {\mathbb{R}}^{2} \) ).\n\n1. The area of the parallelogram spanned by the vectors \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \) is \( \left| {\det \left\lbrack {\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}}\right\... | Proof. 1. The area of the parallelogram is height times base. We will choose as base \( \left| \overrightarrow{\mathbf{b}}\right| = \sqrt{{b}_{1}^{2} + {b}_{2}^{2}} \) . If \( \theta \) is the angle between \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \), the height \( h \) of the parallelogram ... | Yes |
Example 1.4.16 (Determinant of a \( 3 \times 3 \) matrix). | \[ \det \left\lbrack \begin{array}{rrr} 3 & 1 & - 2 \\ 1 & 2 & 4 \\ 2 & 0 & 1 \end{array}\right\rbrack = 3\det \left\lbrack \begin{array}{ll} 2 & 4 \\ 0 & 1 \end{array}\right\rbrack - 1\det \left\lbrack \begin{array}{rr} 1 & - 2 \\ 0 & 1 \end{array}\right\rbrack + 2\det \left\lbrack \begin{array}{rr} 1 & - 2 \\ 2 & 4 \... | Yes |
Example 1.4.18 (Cross product of two vectors in \( {\mathbb{R}}^{3} \) ). | \[ \left\lbrack \begin{array}{l} 3 \\ 0 \\ 1 \end{array}\right\rbrack \times \left\lbrack \begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right\rbrack = \left\lbrack \begin{array}{r} \det \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 4 \end{array}\right\rbrack \\ - \det \left\lbrack \begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\... | Yes |
Proposition 1.4.19. The cross product and the determinant satisfy\n\n\[ \n\det \left\lbrack {\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}}\right\rbrack = \overrightarrow{\mathbf{a}} \cdot \left( {\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}}\right) \n\] | Proof. This follows immediately from Definitions 1.4.15 and 1.4.17:\n\n\[ \n\overset{\overrightarrow{\mathbf{a}}}{\overbrace{\left\lbrack \begin{matrix} {a}_{1} \\ {a}_{2} \\ {a}_{3} \end{matrix}\right\rbrack }} \cdot \overset{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}}{\overbrace{\left\lbrack \begi... | Yes |
Proposition 1.4.20 (Geometric interpretation of cross product).\n\nThe cross product \( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \) is the vector satisfying three properties:\n\n1. It is orthogonal to the plane spanned by \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \) :\n\... | Proof. 1. To check that the cross product \( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \) is orthogonal to \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \), we check that the dot product in each case is zero (Corollary 1.4.8). For instance, \( \overrightarrow{\mathbf{a}} \tim... | Yes |
Is the natural domain of the formula\n\n\[ f\left( \begin{array}{l} x \\ y \end{array}\right) = \sqrt{\frac{x}{y}}\;\text{ open or closed? } \] | If the argument of the square root is nonnegative, the square root can be evaluated, so the first and the third quadrants are in the natural domain. The \( x \) -axis is not (since \( y = 0 \) there), but the \( y \) -axis with the origin removed is in the natural domain, since \( x/y \) is zero there. So the natural d... | Yes |
Proposition 1.5.13 (Convergence in terms of coordinates). \( A \) sequence \( m \mapsto {\mathbf{a}}_{m} \) with \( {\mathbf{a}}_{m} \in {\mathbb{R}}^{n} \) converges to \( \mathbf{a} \) if and only if each coordinate converges; i.e., if for all \( j \) with \( 1 \leq j \leq n \) the \( j \) th coordinate \( {\left( {a... | Proof. Let us first see the easy direction: that \( m \mapsto {\mathbf{a}}_{m} \) converges to a implies that for each \( j = 1,\ldots, n \), the \( j \) th coordinate of \( {\mathbf{a}}_{m} \) converges to \( {a}_{j} \) .\n\nThe challenger hands you an epsilon. Write \( {\mathbf{a}}_{m} \) as \( \left( \begin{matrix} ... | Yes |
Proposition 1.5.15 (Limit of sequence is unique). If the sequence \( i \mapsto {\mathbf{a}}_{i} \) of points in \( {\mathbb{R}}^{n} \) converges to \( \mathbf{a} \) and to \( \mathbf{b} \), then \( \mathbf{a} = \mathbf{b} \) . | Proof. This could be reduced to the one-dimensional case, but we will use it as an opportunity to play the \( \left( {\epsilon, M}\right) \) -game in more sober fashion. Suppose \( \mathbf{a} \neq \mathbf{b} \), and set \( {\epsilon }_{0} = \left( \left| {\mathbf{a} - \mathbf{b}}\right| \right) /4 \) ; our assumption \... | Yes |
1. Let \( i \mapsto {\mathbf{x}}_{i} \) be a sequence in a closed set \( C \subset {\mathbb{R}}^{n} \) converging to \( {\mathbf{x}}_{0} \in {\mathbb{R}}^{n} \) . Then \( {\mathbf{x}}_{0} \in C.\n\n2. Conversely, if every convergent sequence in a set \( C \subset {\mathbb{R}}^{n} \) converges to a point in \( C \), the... | 1. If \( {\mathbf{x}}_{0} \notin C \), then \( {\mathbf{x}}_{0} \in \left( {{\mathbb{R}}^{n} - C}\right) \), which is open, so there exists \( r > 0 \) such that \( {B}_{r}\left( {\mathbf{x}}_{0}\right) \subset \left( {{\mathbb{R}}^{n} - C}\right) \) . Then for all \( m \) we have \( \left| {{\mathbf{x}}_{m} - {\mathbf... | No |
Theorem 1.5.22 (Limit of a composition). Let \( U \subset {\mathbb{R}}^{n}, V \subset {\mathbb{R}}^{m} \) be subsets, and \( \mathbf{f} : U \rightarrow V \) and \( \mathbf{g} : V \rightarrow {\mathbb{R}}^{k} \) be mappings, so that \( \mathbf{g} \circ \mathbf{f} \) is defined in \( U \) . If \( {\mathbf{x}}_{0} \) is a... | Proof. For all \( \epsilon > 0 \) there exists \( {\delta }_{1} > 0 \) such that if \( \left| {\mathbf{y} - {\mathbf{y}}_{0}}\right| < {\delta }_{1} \), then \( \left| {\mathbf{g}\left( \mathbf{y}\right) - {\mathbf{z}}_{0}}\right| < \epsilon \) . Next, there exists \( \delta > 0 \) such that if \( \left| {\mathbf{x} - ... | Yes |
Consider the functions \( f, g : \mathbb{R} \rightarrow \mathbb{R} \)\n\n\[ f\left( x\right) = \left\{ {\begin{array}{ll} x\sin \frac{1}{x} & \text{ if }x \neq 0 \\ 0 & \text{ if }x = 0 \end{array}\;\text{ and }\;g\left( y\right) = \left\{ \begin{array}{ll} 1 & \text{ if }y \neq 0 \\ 0 & \text{ if }y = 0 \end{array}\ri... | but \( \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {g \circ f}\right) \left( x\right) \) does not exist. With Definition 1.5.20, \( \mathop{\lim }\limits_{{y \rightarrow 0}}g\left( y\right) \) doesn't exist, but Theorem 1.5.22 is true (vacuously: the hypotheses aren't satisfied). | No |
Does \( \mathop{\lim }\limits_{{\left( \begin{array}{l} x \\ y \end{array}\right) \rightarrow \left( \begin{array}{l} 0 \\ 0 \end{array}\right) }}f\left( \begin{array}{l} x \\ y \end{array}\right) \) exist? | A first idea is to approach the origin along straight lines. Set \( y = {mx} \) . When \( m \neq 0 \), the limit becomes\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left| \frac{m}{x}\right| {e}^{-\left| \frac{m}{x}\right| }\n\]\n\nthis limit exists and is always 0, for all values of \( m \) . Indeed,\n\n\[ \mathop{... | No |
If \( \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\mathbf{f}\left( \mathbf{x}\right) \) and \( \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\mathbf{g}\left( \mathbf{x}\right) \) exist, then \( \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\left( {\mathbf{f} + ... | \[ \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\left( {\mathbf{f} + \mathbf{g}}\right) \left( \mathbf{x}\right) = \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\mathbf{f}\left( \mathbf{x}\right) + \mathop{\lim }\limits_{{\mathbf{x} \rightarrow {\mathbf{x}}_{0}}}\mathbf{g}\left( \... | Yes |
Proposition 1.5.28 (Criterion for continuity). Let \( X \subset {\mathbb{R}}^{n} \) . A function \( \mathbf{f} : X \rightarrow {\mathbb{R}}^{m} \) is continuous at \( {\mathbf{x}}_{0} \in X \) if and only if for every sequence \( {\mathbf{x}}_{i} \in X \) converging to \( {\mathbf{x}}_{0} \) , \[ \mathop{\lim }\limits_... | Proof. Suppose the \( \epsilon ,\delta \) condition is satisfied, and let \( {\mathbf{x}}_{i}, i = 1,2,\ldots \) be a sequence in \( X \) that converges to \( {\mathbf{x}}_{0} \in X \) . We must show that the sequence \( i \mapsto \mathbf{f}\left( {\mathbf{x}}_{i}\right) \) converges to \( \mathbf{f}\left( {\mathbf{x}}... | Yes |
Theorem 1.5.33 (Linear functions are uniformly continuous). Every linear transformation \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) is uniformly continuous. | Proof. We have to show that for all \( \epsilon > 0 \) there exists \( \delta > 0 \) such that for any \( {\overrightarrow{\mathbf{x}}}_{0},{\overrightarrow{\mathbf{x}}}_{1} \in {\mathbb{R}}^{n} \) with \( \left| {{\overrightarrow{\mathbf{x}}}_{1} - {\overrightarrow{\mathbf{x}}}_{0}}\right| < \delta \), then \( \left| ... | Yes |
Proposition 1.5.35 (Absolute convergence implies convergence).\n\n\\[ \n\\text{If}\\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\left| {\\overrightarrow{\\mathbf{a}}}_{i}\\right| \\text{converges, then}\\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }{\\overrightarrow{\\mathbf{a}}}_{i}\\text{converges.} \n\\] | Proof. Set \\( {\\overrightarrow{\\mathbf{a}}}_{i} = \\left\\lbrack \\begin{matrix} {a}_{1, i} \\\\ \\vdots \\\\ {a}_{n, i} \\end{matrix}\\right\\rbrack \\) . Then \\( \\left| {a}_{k, i}\\right| \\leq \\left| {\\overrightarrow{\\mathbf{a}}}_{i}\\right| \\), so \\( \\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }\\left| {a... | Yes |
Proposition 1.5.36 (Complex exponentials). For any complex number \( z \), the series\n\n\[ \n{e}^{z}\overset{\text{ def }}{ = }1 + z + \frac{{z}^{2}}{2!} + \cdots = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{z}^{k}}{k!}\;\text{ converges. }\n\] | Proof. In one-variable calculus, you learn (using the ratio test, for instance) that the series \( \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{x}^{k}}{k!} \) converges to \( {e}^{x} \) for all \( x \in \mathbb{R} \) . Since\n\n\[ \n\left| \frac{{z}^{k}}{k!}\right| = \frac{{\left| z\right| }^{k}}{k!}\n\]\n\nthe seri... | Yes |
Theorem 1.5.37. For any real number \( t \) we have \( {e}^{it} = \cos t + i\sin t \) | Proof. In terms of power series we have\n\n\[ \sin t = t - \frac{{t}^{3}}{3!} + \frac{{t}^{5}}{5!} - \frac{{t}^{7}}{7!} + \cdots \]\n\n1.5.58\n\n\[ \cos t = 1 - \frac{{t}^{2}}{2!} + \frac{{t}^{4}}{4!} - \frac{{t}^{6}}{6!} + \cdots \]\n\n1.5.59\n\n\[ {e}^{it} = 1 + \left( {it}\right) + \frac{{\left( it\right) }^{2}}{2!}... | No |
Proposition 1.5.38. Let \( A \) be a square matrix. If \( \left| A\right| < 1 \), the series\n\n\[ S\overset{\text{ def }}{ = }I + A + {A}^{2} + \cdots \;\text{ converges to }\;{\left( I - A\right) }^{-1}. \] | Proof. We use the same trick used in the scalar case of Example 0.5.6:\n\n\[ {S}_{k}\overset{\text{ def }}{ = }I + A + {A}^{2} + \cdots + {A}^{k} \]\n\n\[ {S}_{k}A = \;A + {A}^{2} + \cdots + {A}^{k} + {A}^{k + 1} \]\n\n\[ {S}_{k}\left( {I - A}\right) = {S}_{k} - {S}_{k}A = I \]\n\nWe know (Proposition 1.4.11) that\n\n\... | Yes |
Corollary 1.5.40. The set of invertible \( n \times n \) matrices is open. | Proof. Suppose \( B \) is invertible, and \( \left| H\right| < 1/\left| {B}^{-1}\right| \) . Then \( \left| {-{B}^{-1}H}\right| < 1 \) , so \( I + {B}^{-1}H \) is invertible (by Corollary 1.5.39), and\n\n\[ \n{\left( I + {B}^{-1}H\right) }^{-1}{B}^{-1}\underset{\text{ Prop. 1.2.15 }}{ = }{\left( B\left( I + {B}^{-1}H\r... | Yes |
Theorem 1.6.3 (Convergent subsequence in a compact set). If a compact set \( C \subset {\mathbb{R}}^{n} \) contains a sequence \( i \mapsto {\mathbf{x}}_{i} \), then that sequence has a convergent subsequence \( j \mapsto {\mathbf{x}}_{i\left( j\right) } \) whose limit is in \( C \) . | Proof. The set \( C \) is contained in a box \( - {10}^{N} \leq {x}_{i} < {10}^{N} \) for some \( N \) . Decompose this box into boxes of sidelength 1 in the obvious way. Then at least one of these boxes, which we’ll call \( {B}_{0} \), must contain infinitely many terms of the sequence, since the sequence is infinite ... | Yes |
Consider the sequence \[ m \mapsto {x}_{m}\overset{\text{ def }}{ = }\sin {10}^{m} \] This is a sequence in the compact set \( C = \left\lbrack {-1,1}\right\rbrack \), so it contains a convergent subsequence. But how do you find it? | The first step of the construction is to divide the interval \( \left\lbrack {-1,1}\right\rbrack \) into three subintervals (our \ | No |
Theorem 1.6.9 (Existence of minima and maxima). Let \( C \subset {\mathbb{R}}^{n} \) be a compact subset, and let \( f : C \rightarrow \mathbb{R} \) be a continuous function. Then there exists a point \( \mathbf{a} \in C \) such that \( f\left( \mathbf{a}\right) \geq f\left( \mathbf{x}\right) \) for all \( \mathbf{x} \... | Proof of Theorem 1.6.9. We will prove the statement for the maximum. The proof is by contradiction. Assume \( f \) is unbounded. Then for any integer \( N \), no matter how large, there exists a point \( {\mathbf{x}}_{N} \in C \) such that \( \left| {f\left( {\mathbf{x}}_{N}\right) }\right| > N \) . By Theorem 1.6.3, t... | Yes |
Proposition 1.6.12 (Derivative 0 at maximum or minimum). Let \( g : \left( {a, b}\right) \rightarrow \mathbb{R} \) be a differentiable function on the open interval \( \left( {a, b}\right) \) . If \( c \in \left( {a, b}\right) \) is a maximum or minimum of \( g \), then \( {g}^{\prime }\left( c\right) = 0 \) . | Proof. We will prove it only for the maximum. If \( c \) is a maximum, then \( g\left( c\right) - g\left( {c + h}\right) \geq 0 \), so\n\n\[ \frac{g\left( c\right) - g\left( {c + h}\right) }{h}\;\left\{ {\begin{array}{ll} \geq 0 & \text{ if }h > 0 \\ \leq 0 & \text{ if }h < 0; \end{array}\;\text{ i.e.,}\;{g}^{\prime }\... | Yes |
Theorem 1.6.13 (Mean value theorem). If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous, and \( f \) is differentiable on \( \left( {a, b}\right) \), then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[{f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right)... | Proof. Think of \( f \) as representing distance traveled (by a car or, as in Figure 1.6.4, by a hare). The distance the hare travels in the time interval \( b - a \) is \( f\left( b\right) - f\left( a\right) \), so its average speed is\n\n\[m = \frac{f\left( b\right) - f\left( a\right) }{b - a}.\n\nThe function \( g \... | Yes |
Theorem 1.6.14 (Fundamental theorem of algebra). Let\n\n\\[ \np\\left( z\\right) = {z}^{k} + {a}_{k - 1}{z}^{k - 1} + \\cdots + {a}_{0} \n\\]\n\nbe a polynomial of degree \\( k > 0 \\) with complex coefficients. Then \\( p \\) has a root: there exists a complex number \\( {z}_{0} \\) such that \\( p\\left( {z}_{0}\\rig... | Proof of Theorem 1.6.14. We want to show that there exists a number \\( z \\) such that \\( p\\left( z\\right) = 0 \\). The proof requires thinking about complex numbers geometrically, as discussed in Section 0.7, in the subsection \ | No |
Corollary 1.6.15 (Polynomial of degree \( k \) has \( k \) roots). Every complex polynomial \( p\left( z\right) = {z}^{k} + {a}_{k - 1}{z}^{k - 1} + \cdots + {a}_{0} \) with \( k > 0 \) can be written | \[ p\left( z\right) = {\left( z - {c}_{1}\right) }^{{k}_{1}}\cdots {\left( z - {c}_{m}\right) }^{{k}_{m}}\;\text{ with }\;{k}_{1} + \cdots + {k}_{m} = k, \]\n\nwith the \( {c}_{j} \) all distinct. This expression is unique up to permutation of the factors.\n\nProof. Let \( \widetilde{k} \) be the largest degree of a mo... | Yes |
Corollary 1.6.16 (Factoring real polynomials). Every real polynomial of degree \( k > 0 \) can be factored as a product of real polynomials of degree 1 or 2. | Proof. We will start as above: let \( \widetilde{k} \) be the largest degree of a real monic polynomial \( \widetilde{p} \) that divides \( p \) and which is a product of real polynomials of degree 1 or 2, so that we can write \( p\left( z\right) = \widetilde{p}\left( z\right) q\left( z\right) \), where \( q\left( z\ri... | No |
Lemma 1.6.17. If \( q \) is a real polynomial, and \( c \in \mathbb{C} \) is a root of \( q \), then \( \bar{c} \) is also a root of \( q \) . | Proof. In essence, the proof is the following sequence of equalities:\n\n\[ \nq\left( \bar{c}\right) = \overline{q\left( c\right) } = \overline{0} = 0. \n\] | Yes |
If \( f\left( x\right) = {x}^{2}, \) then \( {f}^{\prime }\left( x\right) = {2x} \). | This is proved by writing\n\n\[ \n{f}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {{\left( x + h\right) }^{2} - {x}^{2}}\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {{2xh} + {h}^{2}}\right) = {2x} + \mathop{\lim }\limits_{{h \rightarrow 0}}h = {2x}\text{... | Yes |
What are the partial derivatives at \( \left( \begin{array}{l} a \\ b \end{array}\right) \) of \( \mathbf{f}\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} {x}^{2}y \\ \cos y \end{matrix}\right) \) ? | \[ \overline{{}^{20}\overrightarrow{{D}_{1}\mathbf{f}}\left( \begin{array}{l} a \\ b \end{array}\right) } = \frac{\partial \mathbf{f}}{\partial {x}_{1}}\left( \begin{array}{l} a \\ b \end{array}\right) = \left\lbrack \begin{matrix} {2ab} \\ 0 \end{matrix}\right\rbrack ;\;\overrightarrow{{D}_{2}\mathbf{f}}\left( \begin{... | Yes |
The Jacobian matrix of \( f\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} {xy} \\ \sin \left( {x + y}\right) \\ {x}^{2} - {y}^{2} \end{matrix}\right) \) is | \[ \left\lbrack {\mathbf{{Jf}}\left( \begin{array}{l} x \\ y \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} y & x \\ \cos \left( {x + y}\right) & \cos \left( {x + y}\right) \\ {2x} & - {2y} \end{matrix}\right\rbrack .\;\bigtriangleup \] | Yes |
Theorem 1.7.10 (The Jacobian matrix and the derivative). If \( f \) is differentiable at \( \mathbf{a} \), then all partial derivatives of \( \mathbf{f} \) at \( \mathbf{a} \) exist, and the matrix representing \( \left\lbrack {\mathbf{D}\mathbf{f}\left( \mathbf{a}\right) }\right\rbrack \) is \( \left\lbrack {\mathbf{J... | Proof of Proposition and Definition 1.7.9 and Theorem 1.7.10. The only part of Proposition 1.7.9 that needs to be proved is uniqueness.\n\nWe know (Theorem 1.3.4) that the linear transformation \( L \) is represented by the matrix whose \( i \) th column is \( L\left( {\overrightarrow{\mathbf{e}}}_{i}\right) \), so we ... | Yes |
Proposition 1.7.11 (Differentiable implies continuous). Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \) and let \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{m} \) be a mapping; let a be a point in \( U \) . If \( \mathbf{f} \) is differentiable at \( \mathbf{a} \), then \( \mathbf{f} \) is continuous at \( \math... | Proof. Since \( \mathbf{f} \) is differentiable at \( \mathbf{a} \), the limit in equation 1.7.20 exists, which implies the much weaker statement\n\n\[ \mathop{\lim }\limits_{{\overrightarrow{\mathbf{h}} \rightarrow \overrightarrow{\mathbf{0}}}}\left( {\left( {\mathbf{f}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}... | Yes |
Proposition 1.7.14 (Computing directional derivatives from the derivative). If \( U \subset {\mathbb{R}}^{n} \) is open, and \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{m} \) is differentiable at moves in any direction \( \overrightarrow{\mathbf{v}} \), i.e., when the variable \( \mathbf{x} \) moves at constant speed f... | Proof of Proposition 1.7.14. The expression\n\n\[ \mathbf{r}\left( \overrightarrow{\mathbf{h}}\right) = \left( {\mathbf{f}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) - \mathbf{f}\left( \mathbf{a}\right) }\right) - \left\lbrack {\mathbf{D}\mathbf{f}\left( \mathbf{a}\right) }\right\rbrack \overrightarrow{\ma... | Yes |
Let us compute the derivative in the direction \( \overrightarrow{\mathbf{v}} = \left\lbrack \begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right\rbrack \) of the function \( f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {xy}\sin z \), evaluated at \( \mathbf{a} = \left( \begin{matrix} 1 \\ 1 \\ \pi /2 \end{matri... | By Theorem 1.8.1, this function is differentiable. Its derivative is the \( 1 \times 3 \) matrix \( \left\lbrack {y\sin z\;x\sin z\;{xy}\cos z}\right\rbrack \) ; evaluated at a, it is \( \left\lbrack \begin{array}{lll} 1 & 1 & 0 \end{array}\right\rbrack \) . So the directional derivative is \( \left\lbrack \begin{array... | Yes |
Example 1.7.16 (Directional derivative of an \( {\mathbb{R}}^{m} \) -valued function). Let \( \mathbf{f} : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) be the function \( \mathbf{f}\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} {xy} \\ {x}^{2} - {y}^{2} \end{matrix}\right) \) . At the poi... | The derivative of \( \mathbf{f} \) is\n\n\[ \left\lbrack {\operatorname{Df}\left( \begin{array}{l} x \\ y \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} y & x \\ {2x} & - {2y} \end{matrix}\right\rbrack ,\;\text{ so }\;\left\lbrack {\operatorname{Df}\left( \begin{array}{l} 1 \\ 1 \end{array}\right) }\ri... | Yes |
In this case we can compute the derivative without computing the Jacobian matrix. We shall see that \( S \) is differentiable and that its derivative \( \left\lbrack {\mathbf{D}S\left( A\right) }\right\rbrack \) is the linear transformation that maps \( H \) to \( {AH} + {HA} \) | The assertion is that\n\n\[ \mathop{\lim }\limits_{{H \rightarrow \left\lbrack 0\right\rbrack }}\frac{1}{\left| H\right| }\left( {\left( {S\left( {A + H}\right) - S\left( A\right) }\right) - \underset{\begin{matrix} \text{ increment } \\ \text{ to mapping } \end{matrix}}{\left( AH + HA\right) }}\right) = \left\lbrack 0... | Yes |
Proposition 1.7.18 (Derivative of inverse function for matrices). If \( f \) is the function\n\n\[ f\left( A\right) = {A}^{-1} \]\n\ndefined on the set of invertible matrices in \( \operatorname{Mat}\left( {n, n}\right) \), then \( f \) is differentiable, and\n\n\[ \left\lbrack {\mathbf{D}f\left( A\right) }\right\rbrac... | Proof. For \( f \) to be differentiable, it must be defined on an open subset of \( \operatorname{Mat}\left( {n, n}\right) \) (see Definition 1.7.1). We proved that the set of invertible matrices is open in Corollary 1.5.40. Now we need to show that\n\n\[ \mathop{\lim }\limits_{{H \rightarrow \left\lbrack 0\right\rbrac... | Yes |
1. Any polynomial function \( {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is differentiable on all of \( {\mathbb{R}}^{n} \) .\n\n2. Any rational function is differentiable on the subset of \( {\mathbb{R}}^{n} \) where the denominator does not vanish. | Proof of Theorem 1.8.1. Proving most parts of Theorem 1.8.1 is straightforward; parts 5 and 6 are a bit tricky.\n\n1. If \( \mathbf{f} \) is a constant function, then \( \mathbf{f}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) = \mathbf{f}\left( \mathbf{a}\right) \), so the derivative \( \left\lbrack {\mathbf... | No |
Theorem 1.8.3 (Chain rule). Let \( U \subset {\mathbb{R}}^{n}, V \subset {\mathbb{R}}^{m} \) be open sets, let \( \mathbf{g} : U \rightarrow V \) and \( \mathbf{f} : V \rightarrow {\mathbb{R}}^{p} \) be mappings, and let \( \mathbf{a} \) be a point of \( U \) . If \( \mathbf{g} \) is differentiable at \( \mathbf{a} \) ... | \[ \left\lbrack {\mathbf{D}\left( {\mathbf{f} \circ \mathbf{g}}\right) \left( \mathbf{a}\right) }\right\rbrack = \left\lbrack {\mathbf{D}\mathbf{f}\left( {\mathbf{g}\left( \mathbf{a}\right) }\right) }\right\rbrack \circ \left\lbrack {\mathbf{D}\mathbf{g}\left( \mathbf{a}\right) }\right\rbrack . \] | Yes |
Define \( \mathbf{g} : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) and \( f : {\mathbb{R}}^{3} \rightarrow \mathbb{R} \) by\n\n\[ f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {x}^{2} + {y}^{2} + {z}^{2};\;\mathbf{g}\left( t\right) = \left( \begin{matrix} t \\ {t}^{2} \\ {t}^{3} \end{matrix}\right) . \]\n\nT... | \[ \left\lbrack {\mathbf{D}\left( {f \circ \mathbf{g}}\right) \left( t\right) }\right\rbrack = \left\lbrack {\mathbf{D}f\left( {\mathbf{g}\left( t\right) }\right) }\right\rbrack \left\lbrack {\mathbf{D}\mathbf{g}\left( t\right) }\right\rbrack = \left\lbrack {{2t},2{t}^{2},2{t}^{3}}\right\rbrack \left\lbrack \begin{matr... | Yes |
Let \( \mathbf{f} : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) be given by \( \mathbf{f}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left( \begin{matrix} {z}^{2} \\ {xy} - 3 \\ {2y} \end{matrix}\right) \) . What is the derivative of \( \mathbf{f} \circ \mathbf{f} \), evaluated at \( \left( \begin{arr... | We need to compute\n\n\[ \left\lbrack {\mathbf{D}\left( {\mathbf{f} \circ \mathbf{f}}\right) \left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) }\right\rbrack = \left\lbrack {\mathbf{D}\mathbf{f}\left( {\mathbf{f}\left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) }\right) }\right\rbrack \left\lbrack {\mathbf{D}... | Yes |
Example 1.8.6 (Composition of linear transformations). Here is a case where it is easier to think of the derivative as a linear transformation | Since \( \\mathbf{f} \\circ \\mathbf{f}\\left( \\begin{array}{l} x \\\\ y \\\\ z \\end{array}\\right) = \\mathbf{f}\\left( \\begin{matrix} {z}^{2} \\\\ {xy} - 3 \\\\ {2y} \\end{matrix}\\right) = \\left( \\begin{matrix} 4{y}^{2} \\\\ {xy}{z}^{2} - 3{z}^{2} - 3 \\\\ {2xy} - 6 \\end{matrix}\\right) \\), the derivative of ... | Yes |
Corollary 1.9.2. If \( f \) is a function as defined in Theorem 1.9.1, then\n\n\[ \left| {f\left( \mathbf{b}\right) - f\left( \mathbf{a}\right) }\right| \leq \left( {\mathop{\sup }\limits_{{\mathbf{c} \in \left\lbrack {\mathbf{a},\mathbf{b}}\right\rbrack }}\left| \left\lbrack {\mathbf{D}f\left( \mathbf{c}\right) }\righ... | ## Proof of Corollary 1.9.2. This follows immediately from Theorem 1.9.1 and Proposition 1.4.11. | Yes |
Example 1.9.3 (Nondifferentiable function with Jacobian matrix). This happens even for the innocent-looking function\n\n\\[ \nf\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\frac{{x}^{2}y}{{x}^{2} + {y}^{2}} \n\\]\n\nshown in Figure 1.9.1. Actually, we should write this function as\n\n\\[ \nf\\left( \\begi... | You have probably learned to be suspicious of functions that are defined by different formulas for different values of the variable. In this case, the value at \\( \\left( \\begin{array}{l} 0 \\\\ 0 \\end{array}\\right) \\) is really natural, in the sense that as \\( \\left( \\begin{array}{l} x \\\\ y \\end{array}\\rig... | No |
Consider the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) defined by\n\n\[ f\left( x\right) = \frac{x}{2} + {x}^{2}\sin \frac{1}{x} \]\n\nTo be precise, one should add \( f\left( 0\right) = 0 \), since \( \sin 1/x \) is not defined there, but this was the only reasonable value, since\n\n\[ \mathop{\lim }\limits... | We can compute the derivative at 0 using the definition of the derivative:\n\n\[ {f}^{\prime }\left( 0\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {\frac{0 + h}{2} + {\left( 0 + h\right) }^{2}\sin \frac{1}{0 + h}}\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {\frac{h}{2} +... | Yes |
Let us go back to the function of Example 1.9.3:\n\n\[ f\left( \begin{array}{l} x \\ y \end{array}\right) = \left\{ \begin{array}{ll} \frac{{x}^{2}y}{{x}^{2} + {y}^{2}} & \text{ if }\left( \begin{array}{l} x \\ y \end{array}\right) \neq \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \\ 0 & \text{ if }\left( \begin{a... | Both partial derivatives are 0 at the origin. Away from the origin - that is, if \( \left( \begin{array}{l} x \\ y \end{array}\right) \neq \left( \begin{array}{l} 0 \\ 0 \end{array}\right) - \) then\n\n\[ {D}_{1}f\left( \begin{array}{l} x \\ y \end{array}\right) = \frac{\left( {{x}^{2} + {y}^{2}}\right) \left( {2xy}\ri... | Yes |
Theorem 1.9.8 (Criterion for differentiability). If \( U \) is an open subset of \( {\mathbb{R}}^{n} \), and \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{m} \) is a \( {C}^{1} \) mapping, then \( \mathbf{f} \) is differentiable on \( U \), and its derivative is given by its Jacobian matrix. | Proof of Theorem 1.9.8. This is an application of Theorem 1.9.1, the mean value theorem. What we need to show is that \[ \mathop{\lim }\limits_{{\overrightarrow{\mathbf{h}} \rightarrow \overrightarrow{\mathbf{0}}}}\frac{1}{\left| \overrightarrow{\mathbf{h}}\right| }\left( {\mathbf{f}\left( {\mathbf{a} + \overrightarrow... | Yes |
Example 1.9.9. Here we work out the preceding computation when \( f \) is a scalar-valued function on \( {\mathbb{R}}^{2} \) :\n\n\[ f\left( \begin{array}{l} {a}_{1} + {h}_{1} \\ {a}_{2} + {h}_{2} \end{array}\right) - f\left( \begin{array}{l} {a}_{1} \\ {a}_{2} \end{array}\right) \] | \[ = f\left( \begin{array}{l} {a}_{1} + {h}_{1} \\ {a}_{2} + {h}_{2} \end{array}\right) \overset{0}{\overbrace{-f\left( \begin{matrix} {a}_{1} \\ {a}_{2} + {h}_{2} \end{matrix}\right) + f\left( \begin{matrix} {a}_{1} \\ {a}_{2} + {h}_{2} \end{matrix}\right) }} - f\left( \begin{array}{l} {a}_{1} \\ {a}_{2} \end{array}\r... | Yes |
Theorem 2.1.2 (Solutions of \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{b}} \) unchanged by row operations). If the matrix \( \left\lbrack {A \mid \overrightarrow{\mathbf{b}}}\right\rbrack \) representing a system of linear equations \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{b}} \) can be ... | Proof. Row operations consist of multiplying one equation by a nonzero number, adding a multiple of one equation to another, and exchanging two equations. Any solution of \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{b}} \) is thus a solution of \( {A}^{\prime }\overrightarrow{\mathbf{x}} = {\overrightarrow... | No |
To solve the system of equations 2.1.1 we can use row operations to bring the matrix\n\n\\[ \n\\left\\lbrack \\begin{array}{rrrr} 2 & 1 & 3 & 1 \\\\ 1 & - 1 & 0 & 1 \\\\ 2 & 0 & 1 & 1 \\end{array}\\right\\rbrack \\text{ to the form }\\left\\lbrack \\begin{array}{rrrr} 1 & 0 & 0 & 1/3 \\\\ 0 & 1 & 0 & - 2/3 \\\\ 0 & 0 &... | In this case, the solution can just be read off the matrix. If we put the unknowns back in the matrix, we get\n\n\\[ \n\\left\\lbrack \\begin{array}{rrrr} x & 0 & 0 & 1/3 \\\\ 0 & y & 0 & - 2/3 \\\\ 0 & 0 & z & 1/3 \\end{array}\\right\\rbrack\n\\]\n\n\\[ \nx = 1/3\n\\]\n\n\\[ \ny = - 2/3\n\\]\n\n\\[ \nz = 1/3\n\\]\n | Yes |
Example 2.1.5 (Matrices in echelon form). Clearly, the identity matrix is in echelon form. So are the following matrices, in which the pivotal 1's are underlined: | \[ \left\lbrack \begin{array}{rrrr} \underline{1} & 0 & 0 & 3 \\ 0 & \underline{1} & 0 & - 2 \\ 0 & 0 & \underline{1} & 1 \end{array}\right\rbrack ,\;\left\lbrack \begin{array}{rrrr} \underline{1} & 1 & 0 & 0 \\ 0 & 0 & \underline{1} & 0 \\ 0 & 0 & 0 & \underline{1} \end{array}\right\rbrack ,\;\left\lbrack \begin{array... | Yes |
Given any matrix \( A \), there exists a matrix \( \widetilde{A} \) in echelon form that can be obtained from \( A \) by row operations. | The proof of part 1 is an explicit algorithm for computing \( \widetilde{A} \). Called row reduction or Gaussian elimination (or several other names), it is the main tool for solving linear equations. | No |
Example 2.1.8 (Row reduction). Here we row reduce a matrix. The \( R \) ’s refer in each case to the rows of the immediately preceding matrix. For example, the second row of the second matrix is labeled \( {R}_{1} + {R}_{2} \), because that row is obtained by adding the first and second rows of the preceding matrix. Ag... | \[ \left\lbrack \begin{array}{rrrr} 1 & 2 & 3 & 1 \\ - 1 & 1 & 0 & 2 \\ 1 & 0 & 1 & 2 \end{array}\right\rbrack \rightarrow \underset{{R}_{1} + {R}_{2}}{{R}_{1} + {R}_{2}}\left\lbrack \begin{array}{rrrr} 1 & 2 & 3 & 1 \\ 0 & 3 & 3 & 3 \\ 0 & - 2 & - 2 & 1 \end{array}\right\rbrack \rightarrow {R}_{2}/3\left\lbrack \begin... | Yes |
Example 2.1.9 (Thresholding to minimize round-off errors). If you are computing to 10 significant digits, then \( 1 + {10}^{-{10}} = {1.0000000001} = 1 \) . So consider the system of equations\n\n\[ \n{10}^{-{10}}x + {2y} = 1 \n\]\n\n\[ \nx + y = 1 \n\] | the solution of which is\n\n\[ \nx = \frac{1}{2 - {10}^{-{10}}},\;y = \frac{1 - {10}^{-{10}}}{2 - {10}^{-{10}}}. \n\]\n\nIf you are computing to 10 significant digits, this is \( x = y = {.5} \) . If you use \( {10}^{-{10}} \) as a pivot, the row reduction, to 10 significant digits, goes as follows:\n\n\[ \n\left\lbrac... | Yes |
Theorem 2.2.1 (Solutions to linear equations). Represent the system \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{b}} \), involving \( m \) linear equations in \( n \) unknowns, by the \( m \times \left( {n + 1}\right) \) matrix \( \left\lbrack {A \mid \overrightarrow{\mathbf{b}}}\right\rbrack \), which row... | 1. If the row-reduced vector \( \widetilde{\mathbf{b}} \) contains a pivotal 1, the system has no solutions.\n\n2. If \( \widetilde{\mathbf{b}} \) does not contain a pivotal 1, then solutions are uniquely determined by the values of the nonpivotal variables:\n\na. If each column of \( \widetilde{A} \) contains a pivota... | Yes |
Example 2.2.2 (A system with no solutions). Let us solve\n\n\[ \n{2x} + y + {3z} = 1 \n\]\n\n\[ \nx - y = 1 \n\]\n\n\[ \nx + y + {2z} = 1\text{.} \n\] | The matrix\n\n\[ \n\left\lbrack \begin{array}{rrrr} 2 & 1 & 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 1 & 1 & 2 & 1 \end{array}\right\rbrack \;\text{ row reduces to }\;\left\lbrack \begin{matrix} \underline{1} & 0 & 1 & 0 \\ 0 & \underline{1} & 1 & 0 \\ 0 & 0 & 0 & \underline{1} \end{matrix}\right\rbrack ,\n\]\n\nso the equations ar... | Yes |
Example 2.2.3 (A system with infinitely many solutions). Let us solve\n\n\[ \n{2x} + y + {3z} = 1 \n\]\n\n\[ \nx - y\; = 1 \n\]\n\n\[ \nx + y + {2z} = 1/3\text{.} \n\] | Example 2.2.3: In this case, the solutions form a family that depends on the single nonpivotal variable, \( z \) ; the matrix \( \widetilde{A} \) has one column that does not contain a pivotal 1.\n\n\[ \n\left\lbrack \begin{matrix} 2 & 1 & 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 1 & 1 & 2 & 1/3 \end{matrix}\right\rbrack \;\text{ r... | Yes |
Suppose we want to know what solutions, if any, exist for the system of equations\n\n\[ \n{x}_{1} + {x}_{2} = {a}_{1} \n\]\n\n\[ \n{x}_{2} + {x}_{3} = {a}_{2} \n\]\n\n\[ \n{x}_{3} + {x}_{4} = {a}_{3} \n\]\n\n\[ \n{x}_{4} + {x}_{1} = {a}_{4} \n\] | Row operations bring the matrix\n\n\[ \n\left\lbrack \begin{matrix} 1 & 1 & 0 & 0 & {a}_{1} \\ 0 & 1 & 1 & 0 & {a}_{2} \\ 0 & 0 & 1 & 1 & {a}_{3} \\ 1 & 0 & 0 & 1 & {a}_{4} \end{matrix}\right\rbrack \;\text{ to }\;\left\lbrack \begin{matrix} 1 & 0 & 0 & 1 & {a}_{1} + {a}_{3} - {a}_{2} \\ 0 & 1 & 0 & - 1 & {a}_{2} - {a}... | Yes |
Corollary 2.2.9 (Solving several systems of equations simultaneously). Several systems of \( n \) linear equations in \( n \) unknowns, with the same coefficients (e.g. \( A\overrightarrow{\mathbf{x}} = {\overrightarrow{\mathbf{b}}}_{1},\ldots, A\overrightarrow{\mathbf{x}} = {\overrightarrow{\mathbf{b}}}_{k} \) ) can b... | Proof. If \( A \) row reduces to the identity, the row reduction is completed by the time one has dealt with the last row of \( A \) . The row operations needed to turn \( A \) into \( \widetilde{A} \) affect each \( {\overrightarrow{\mathbf{b}}}_{i} \), but the \( {\overrightarrow{\mathbf{b}}}_{i} \) do not affect eac... | Yes |
Proposition 2.3.1 (Solving equations with matrix inverse). If \( A \) has an inverse \( {A}^{-1} \), then for any \( \overrightarrow{\mathbf{b}} \) the equation \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{b}} \) has a unique solution, namely \( \overrightarrow{\mathbf{x}} = {A}^{-1}\overrightarrow{\mathbf... | The following verifies that \( {A}^{-1}\overrightarrow{\mathbf{b}} \) is a solution:\n\n\[ A\left( {{A}^{-1}\overrightarrow{\mathbf{b}}}\right) = \left( {A{A}^{-1}}\right) \overrightarrow{\mathbf{b}} = I\overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}}. \]\n\nThis makes use of the associativity of matrix multip... | Yes |
Consider the matrix of Examples 2.2.2 and 2.2.3, for two systems of linear equations, neither of which has a unique solution:\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & 1 & 3 \\ 1 & - 1 & 0 \\ 1 & 1 & 2 \end{array}\right\rbrack \]\n\nThis matrix has no inverse because | \[ \left\lbrack \begin{array}{rrrrrrr} 2 & 1 & 3 & & 1 & 0 & 0 \\ 1 & - 1 & 0 & & 0 & 1 & 0 \\ 1 & 1 & 2 & & 0 & 0 & 1 \end{array}\right\rbrack \text{ row reduces to }\left\lbrack \begin{array}{rrrrrrr} \underline{1} & 0 & 1 & & 1 & 0 & - 1 \\ 0 & \underline{1} & 1 & & - 1 & 0 & 2 \\ 0 & 0 & 0 & & - 2 & 1 & 3 \end{arra... | Yes |
Proposition 2.3.7 (Square matrix approximated by invertible matrices). Any square matrix \( A \) can be approximated by a sequence of invertible matrices. | Proof. Set \( \widetilde{A} = {E}_{k}\cdots {E}_{1}A \), with \( \widetilde{A} \) upper triangular (i.e., row reduce \( A \) until it is either the identity or has rows of 0 's at the bottom). If any diagonal entries of \( \widetilde{A} \) are 0, change them to \( 1/n \) ; denote the resulting invertible matrix \( {\wi... | Yes |
Theorem 2.4.5 (Linear independence and span). Let \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k} \) be vectors in \( {\mathbb{R}}^{n} \) ; let \( A \) be the \( n \times k \) matrix \( \left\lbrack {{\overrightarrow{\mathrm{v}}}_{1},\ldots {\overrightarrow{\mathrm{v}}}_{k}}\right\rbrack ... | Proof. 1. The vectors \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k} \) are linearly independent if and only if the only solution to \( A\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}} \) is \( \overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}} \) . Thus part 1 follows i... | Yes |
Given the vectors\n\n\[ \n{\overrightarrow{\mathbf{w}}}_{1} = \left\lbrack \begin{array}{l} 2 \\ 1 \\ 1 \end{array}\right\rbrack ,\;{\overrightarrow{\mathbf{w}}}_{2} = \left\lbrack \begin{array}{r} 1 \\ - 1 \\ 1 \end{array}\right\rbrack ,\;{\overrightarrow{\mathbf{w}}}_{3} = \left\lbrack \begin{array}{l} 3 \\ 0 \\ 2 \e... | \[ \n\left\lbrack {A \mid \overrightarrow{\mathbf{v}}}\right\rbrack = \left\lbrack {{\overrightarrow{\mathbf{w}}}_{1},{\overrightarrow{\mathbf{w}}}_{2},{\overrightarrow{\mathbf{w}}}_{3} \mid \overrightarrow{\mathbf{v}}}\right\rbrack = \left\lbrack \begin{array}{rrrr} 2 & 1 & 3 & 3 \\ 1 & - 1 & 0 & 3 \\ 1 & 1 & 2 & 1 \e... | Yes |
Are the vectors \(\overrightarrow{\mathbf{w}}_{1} = \left\lbrack \begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right\rbrack ,\;\overrightarrow{\mathbf{w}}_{2} = \left\lbrack \begin{array}{r} - 2 \\ 1 \\ 2 \end{array}\right\rbrack ,\;\overrightarrow{\mathbf{w}}_{3} = \left\lbrack \begin{array}{r} - 1 \\ 1 \\ - 1 \end{array}\... | The matrix \(\left\lbrack \begin{matrix} 1 & - 2 & - 1 \\ 2 & 1 & 1 \\ 3 & 2 & - 1 \\ \overrightarrow{\mathbf{w}}_{1} & \overrightarrow{\mathbf{w}}_{2} & \overrightarrow{\mathbf{w}}_{3} \end{matrix}\right\rbrack \text{ row reduces to }\left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right... | Yes |
Example 2.4.8. We can make the collection of vectors in Example 2.4.7 linearly dependent, by adding a vector that is a linear combination of some of them, say \( {\overrightarrow{\mathbf{w}}}_{4} = 2{\overrightarrow{\mathbf{w}}}_{2} + {\overrightarrow{\mathbf{w}}}_{3} \) : | \[ \left\lbrack \begin{array}{rrrrr} 1 & - 2 & - 1 & - 5 & - 7 \\ 2 & 1 & 1 & 3 & - 2 \\ 3 & 2 & - 1 & 3 & 1 \end{array}\right\rbrack \text{ row reduces to }\left\lbrack \begin{array}{rrrrr} 1 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 2 & 3 \\ 0 & 0 & 1 & 1 & - 1 \end{array}\right\rbrack . \] | Yes |
Theorem 2.4.10. In \( {\mathbb{R}}^{n} \) , \( n + 1 \) vectors are never linearly independent, and \( n - 1 \) vectors in \( {\mathbb{R}}^{n} \) never span \( {\mathbb{R}}^{n} \) . | Proof. This follows immediately from Theorem 2.4.5. When we use \( n + 1 \) vectors in \( {\mathbb{R}}^{n} \) to form a matrix \( A \), the matrix is \( n \times \left( {n + 1}\right) \) ; when we row reduce \( A \) to \( \widetilde{A} \), at least one column of \( \widetilde{A} \) contains no pivotal 1, since there ca... | Yes |
Proposition 2.5.3. Let \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) be a linear transformation. The system of linear equations \( T\left( \overrightarrow{\mathbf{x}}\right) = \overrightarrow{\mathbf{b}} \) has\n\n1. at most one solution for every \( \overrightarrow{\mathbf{b}} \in {\mathbb{R}}^{m} \) if and ... | Proof. 1. If the kernel of \( T \) is not \( \{ \overrightarrow{\mathbf{0}}\} \), then there is more than one solution to \( T\left( \overrightarrow{\mathbf{x}}\right) = \overrightarrow{\mathbf{0}} \) . (Of course, one solution is \( \overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}} \) ).\n\nIn the other direct... | Yes |
Example 2.5.5 (Finding a basis for the image). Consider the matrix \( A \) below, which describes a linear transformation from \( {\mathbb{R}}^{5} \) to \( {\mathbb{R}}^{4} \) : | \[ A = \left\lbrack \begin{array}{rrrrr} 1 & 2 & 4 & - 1 & 2 \\ - 1 & 0 & - 2 & - 1 & 1 \\ 2 & 0 & 4 & 2 & 1 \\ 1 & 1 & 3 & 0 & 2 \end{array}\right\rbrack \text{, which row reduces to}\widetilde{A} = \left\lbrack \begin{array}{rrrrr} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \en... | No |
Example 2.5.7 (Finding a basis for the kernel). The third and fourth columns of \( A \) in Example 2.5.5 are nonpivotal, so \( {k}_{1} = 3 \) and \( {k}_{2} = 4 \) . The system has a unique solution for any values we choose of the third and fourth unknowns. In particular, there is a unique vector \( {\overrightarrow{\m... | \[ {\overrightarrow{\mathbf{v}}}_{1} = \left\lbrack \begin{matrix} - \\ - \\ 1 \\ 0 \\ - \end{matrix}\right\rbrack ,\;{\overrightarrow{\mathbf{v}}}_{2} = \left\lbrack \begin{matrix} - \\ - \\ 0 \\ 1 \\ - \end{matrix}\right\rbrack . \] The first, second, and fifth entries of \( {\overrightarrow{\mathbf{v}}}_{1} \) and \... | Yes |
Corollary 2.5.10 (Deducing existence from uniqueness). Let \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be a linear transformation. Then the equation \( T\left( \overrightarrow{\mathbf{x}}\right) = \overrightarrow{\mathbf{b}} \) has a solution for every \( \overrightarrow{\mathbf{b}} \in {\mathbb{R}}^{n} \) ... | Proof. Saying that \( T\left( \overrightarrow{\mathbf{x}}\right) = \overrightarrow{\mathbf{b}} \) has a solution for every \( \overrightarrow{\mathbf{b}} \in {\mathbb{R}}^{n} \) means that \( {\mathbb{R}}^{n} \) is the image of \( T \), so \( \dim \operatorname{img}T = n \), which is equivalent to \( \dim \) \( \ker \l... | Yes |
Proposition 2.5.11. Let \( A \) be an \( m \times n \) matrix. Then the number of linearly independent columns of \( A \) equals the number of linearly independent rows. | Proof. Call the span of the columns of a matrix \( A \) the column space of \( A \) and the span of the rows its row space. The rows of \( \widetilde{A} \) are linear combinations of the rows of \( A \), and vice versa since row operations are reversible, so if \( A \) row reduces to \( \widetilde{A} \), the row space ... | Yes |
Example 2.5.12 (Interpolation: polynomials of degree at most 2).\n\nThe matrix of \( {T}_{2} : p \mapsto \left\lbrack \begin{array}{l} p\left( 0\right) \\ p\left( 1\right) \\ p\left( 2\right) \end{array}\right\rbrack \), where \( {P}_{2} \) is identified to \( {\mathbb{R}}^{3} \) by identifying\n\n\( a + {bx} + c{x}^{2... | equivalent to the polynomial \( p\left( x\right) = a + {bx} + c{x}^{2} \) evaluated at \( x = 0, x = 1 \) , and \( x = 2 \) . The inverse of \( {T}_{2} \) is\n\n\[ {T}_{2}^{-1} = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ - 3/2 & 2 & - 1/2 \\ 1/2 & - 1 & 1/2 \end{matrix}\right\rbrack \]\n\nso if we choose (for instance) ... | Yes |
When\n\n\[ q\left( x\right) = {2x} + 3\text{ and }p\left( x\right) = {x}^{2} - 1 \]\n\nProposition 2.5.13 says that there exist polynomials \( {q}_{1} \) and \( {q}_{2} \) of degree less than 1 (i.e., numbers, which we will call \( {A}_{0} \) and \( {B}_{0} \), the subscript indicating that they are coefficients of the... | In simple cases, it’s clear how to proceed. In equation 2.5.22, to find \( {A}_{0} \) and \( {B}_{0} \), we multiply out to get a common denominator:\n\n\[ \frac{{2x} + 3}{{x}^{2} - 1} = \frac{{A}_{0}}{x + 1} + \frac{{B}_{0}}{x - 1} = \frac{{A}_{0}\left( {x - 1}\right) + {B}_{0}\left( {x + 1}\right) }{{x}^{2} - 1} = \f... | Yes |
Lemma 2.5.15. If \( {q}_{i} \neq 0 \) is a polynomial of degree \( < {n}_{i} \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}_{i}}}\left| \frac{{q}_{i}\left( x\right) }{{\left( x - {a}_{i}\right) }^{{n}_{i}}}\right| = \infty \]\n\nThat is, if \( {q}_{i} \neq 0 \), then \( {q}_{i}\left( x\right) /{\left( x - {a}... | Proof of Lemma 2.5.15. For values of \( x \) very close to \( {a}_{i} \), the denominator \( {\left( x - {a}_{i}\right) }^{{n}_{i}} \) gets very small; if all goes well the entire term then gets very big. But we have to make sure that the numerator does not get small equally fast. Let us make the change of variables \(... | Yes |
Consider the space of twice differentiable functions \( f : \mathbb{R} \rightarrow \mathbb{R} \) such that \( {D}^{2}f = 0 \) (i.e., functions of one variable whose second derivatives are 0 ; we could also write this \( {f}^{\prime \prime } = 0 \) ). This is a subspace of the vector space of Example 2.6.2 and is also a... | But since a function has a vanishing second derivative if and only if it is a polynomial of degree at most 1 , we see that this space is the set of functions \[ {f}_{a, b}\left( x\right) = a + {bx}. \] Precisely two numbers are needed to specify each element of this vector space; we could choose as our basis the consta... | Yes |
If \( A \in \operatorname{Mat}\left( {n, n}\right) \), then the transformation \( \operatorname{Mat}\left( {n, n}\right) \rightarrow \operatorname{Mat}\left( {n, n}\right) \) given by \( H \mapsto {AH} + {HA} \) is a linear transformation. | \[ \left\lbrack {\mathbf{D}S\left( A\right) }\right\rbrack H = {AH} + {HA}. \] | No |
Let \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) denote the space of continuous real-valued functions defined for \( 0 \leq x \leq 1 \) . Let \( g : \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function, and define the mapping \( {T}_{g} : \mathcal... | We first show that\n\n\[ {T}_{g}\left( {{f}_{1} + {f}_{2}}\right) = {T}_{g}\left( {f}_{1}\right) + {T}_{g}\left( {f}_{2}\right) \]\n\nwhich we do as follows: definition of addition in vector space\n\n\[ \left( {{T}_{g}\left( {{f}_{1} + {f}_{2}}\right) }\right) \left( x\right) = {\int }_{0}^{1}g\left( \begin{array}{l} x... | Yes |
Example 2.6.13 (Concrete to abstract function). Let \( {P}_{2} \) be the space of polynomials of degree at most 2, with basis \( {\mathbf{v}}_{1} = 1,{\mathbf{v}}_{2} = x,{\mathbf{v}}_{3} = {x}^{2} \) . Then \( {\Phi }_{\{ \mathbf{v}\} }\left( \left\lbrack \begin{array}{l} {a}_{1} \\ {a}_{2} \\ {a}_{3} \end{array}\righ... | \( \;\bigtriangleup \) | No |
Proposition 2.6.15 (Linear independence, span, and basis). Let \( \{ \mathbf{v}\} = {\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{n} \) be vectors in a vector space \( V, \) and let \( {\Phi }_{\{ \mathbf{v}\} } : {\mathbb{R}}^{n} \rightarrow V \) be the associated concrete-to-abstract transformation. Then\n\n1. The set \( \{... | Proof. 1. Definition 2.6.10 says that \( {\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{n} \) are linearly independent if and only if\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\mathbf{v}}_{i} = \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i}{\mathbf{v}}_{i}\;\text{ implies }{a}_{1} = {b}_{1},{a}_{2} = {b}_{2},\ldots ,{a}_{n}... | Yes |
What is the change of basis matrix \( \left\lbrack {P}_{{\mathbf{v}}^{\prime } \rightarrow \mathbf{v}}\right\rbrack \) ? | Since \( {\overrightarrow{\mathbf{v}}}_{1}^{\prime } = - {\overrightarrow{\mathbf{v}}}_{1} + {\overrightarrow{\mathbf{v}}}_{2} \), the coordinates of \( {\overrightarrow{\mathbf{v}}}_{1}^{\prime } \) expressed in terms of \( {\overrightarrow{\mathbf{v}}}_{1} \) and \( {\overrightarrow{\mathbf{v}}}_{2} \) are \( {p}_{1,... | Yes |
Example 2.6.19 (Translating into the standard basis). If the basis \( \left\{ {\overrightarrow{\mathbf{v}}}^{\prime }\right\} \) of \( {\mathbb{R}}^{2} \) consists of \( \left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack ,\left\lbrack \begin{array}{r} 1 \\ - 1 \end{array}\right\rbrack \), then \( \left\lbrac... | \[ \left\lbrack {P}_{{\overrightarrow{\mathbf{v}}}^{\prime } \rightarrow \overrightarrow{\mathbf{e}}}\right\rbrack \left\lbrack \begin{array}{l} 2 \\ 3 \end{array}\right\rbrack = \left\lbrack \begin{array}{r} 5 \\ - 1 \end{array}\right\rbrack \] | Yes |
Theorem 2.6.20 (Change of basis formula). In the first line of equation \[ {\left\lbrack T\right\rbrack }_{\left\{ {\mathbf{v}}^{\prime }\right\} ,\left\{ {\mathbf{w}}^{\prime }\right\} } = {\left\lbrack {P}_{{\mathbf{w}}^{\prime } \rightarrow \mathbf{w}}\right\rbrack }^{-1}{\left\lbrack T\right\rbrack }_{\{ \mathbf{v}... | Proof. Write everything using the \ | No |
The vector space \( \mathcal{C}\left\lbrack {0,1}\right\rbrack \) of continuous functions on \( \left\lbrack {0,1}\right\rbrack \) is infinite dimensional. | Assume functions \( {f}_{1},\ldots ,{f}_{n} \) are a basis, and pick \( n + 1 \) distinct points \( 0 = {x}_{1} < {x}_{2}\cdots < {x}_{n + 1} = 1 \) in \( \left\lbrack {0,1}\right\rbrack . \) Then given any values \( {c}_{1},\ldots ,{c}_{n + 1}, \) there certainly exists a continuous function \( f\left( x\right) \) wit... | Yes |
Fibonacci numbers are the numbers \( 1,1,2,3,5,8,{13},\ldots \) defined by \( {a}_{0} = {a}_{1} = 1 \) and \( {a}_{n + 1} = {a}_{n} + {a}_{n - 1} \) for \( n \geq 1 \). We propose to prove the formula\n\n\[ \n{a}_{n} = \frac{5 + \sqrt{5}}{10}{\left( \frac{1 + \sqrt{5}}{2}\right) }^{n} + \frac{5 - \sqrt{5}}{10}{\left( \... | Equation 2.7.1 is quite amazing: it isn't even obvious that the right side is an integer! The key to understanding it is the matrix equation\n\n\[ \n\left\lbrack \begin{matrix} {a}_{n} \\ {a}_{n + 1} \end{matrix}\right\rbrack = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right\rbrack \left\lbrack \begin{m... | No |
Example 2.7.6. If \( A = \left\lbrack \begin{array}{rrr} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}\right\rbrack \) and \( P = \left\lbrack \begin{array}{rrr} 1 & - 1 & 1 \\ 1 & 0 & - 2 \\ 1 & 1 & 1 \end{array}\right\rbrack \), then | \[ {P}^{-1}{AP} = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array}\right\rbrack \text{ is diagonal, so }\left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack ,\left\lbrack \begin{array}{r} - 1 \\ 0 \\ 1 \end{array}\right\rbrack ,\left\lbrack \begin{array}{r} 1 \\ - 2 \\ 1 \en... | Yes |
Theorem 2.7.7 (Eigenvectors with distinct eigenvalues are linearly independent). If \( A : V \rightarrow V \) is a linear transformation, and \( {\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{k}\; \) are eigenvectors of \( \;A\; \) with distinct eigenvalues \( \;{\lambda }_{1},\ldots ,{\lambda }_{k},\; \) then \( {\mathbf{v}}_... | Proof. We will prove this by contradiction. If \( {\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{k} \) are not linearly independent, then there is a first vector \( {\mathbf{v}}_{j} \) that is a linear combination of the earlier ones. Thus we can write\n\n\[ \n{\mathbf{v}}_{j} = {a}_{1}{\mathbf{v}}_{1} + \cdots + {a}_{j - 1}{\... | Yes |
Example 2.7.8 (Finding an eigenbasis). Let \( A : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) be the linear transformation \( A = \left\lbrack \begin{array}{rrr} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}\right\rbrack \), and for \( \overrightarrow{\mathbf{w}} \) use \( {\overrightarrow{\mathbf{e}}}_{... | We row reduce\n\n\[ \left\lbrack \begin{array}{rrrr} 1 & 1 & 2 & 5 \\ 0 & - 1 & - 3 & - 9 \\ 0 & 0 & 1 & 4 \end{array}\right\rbrack \text{ to get }\left\lbrack \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 4 \end{array}\right\rbrack \]\n\nwhich tells us that \( 0{\overrightarrow{\mathbf{e}}}_{1} -... | Yes |
Theorem 2.7.9. Let \( A \) be an \( n \times n \) complex matrix. There exists an eigenbasis of \( {\mathbb{C}}^{n} \) for \( A \) if and only if all the roots of all the \( {p}_{i} \) are simple. | Proof. First assume that all the roots of all the \( {p}_{i} \) are simple, and denote by \( {m}_{i} \) the degree of \( {p}_{i} \) . Let \( {E}_{i} \) be the span of \( {\overrightarrow{\mathbf{e}}}_{i}, A{\overrightarrow{\mathbf{e}}}_{i},{A}^{2}{\overrightarrow{\mathbf{e}}}_{i},\ldots \) It should be clear that this ... | Yes |
Theorem 2.7.10 (Perron-Frobenius theorem). If \( A \) is a real \( n \times n \) matrix such that \( A > \mathbf{0} \), there exists a unique real eigenvector \( \overrightarrow{\mathbf{v}} > \mathbf{0} \) with \( \left| \overrightarrow{\mathbf{v}}\right| = 1 \). This eigenvector has a simple real eigenvalue \( \lambda... | Further, \( \overrightarrow{\mathbf{v}} \) can be found by iteration: for any \( \overrightarrow{\mathbf{w}} \) with \( \overrightarrow{\mathbf{w}} \geq \mathbf{0} \), we have \( \overrightarrow{\mathbf{v}} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{A}^{k}\left( \overrightarrow{\mathbf{w}}\right) /\left| {{A}^{k... | No |
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