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The normal equations \( {A}^{\mathrm{T}}A\widehat{\mathbf{v}} = {A}^{\mathrm{T}}\mathbf{b} \) are\n\n\[ \left\lbrack \begin{array}{llll} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 4 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 3 \\ 1 & 4 \end{array}\right\rbrack \left\lbrack \begin{array}{l} \widehat... | After multiplication this matrix \( {A}^{\mathrm{T}}A \) is square and symmetric and positive definite :\n\n\[ {\mathbf{A}}^{\mathbf{T}}\mathbf{A}\widehat{\mathbf{v}} = {\mathbf{A}}^{\mathbf{T}}\mathbf{b}\;\left\lbrack \begin{matrix} 4 & 8 \\ 8 & {26} \end{matrix}\right\rbrack \left\lbrack \begin{matrix} \widehat{C} \\... | Yes |
Example 2 Suppose we have two equations for one unknown \( \mathbf{v} \) . Thus \( n = 1 \) but \( m = 2 \) (probably there is no solution). One unknown means only one column in \( A \) :\n\n\[ A\mathbf{v} = \mathbf{b}\;\text{ is }\;\left\lbrack \begin{array}{l} {a}_{1} \\ {a}_{2} \end{array}\right\rbrack \mathbf{v} = ... | The matrix \( A \) is 2 by 1 . The squared error is \( E = {e}_{1}^{2} + {e}_{2}^{2} = {\left( 1 - 2\mathbf{v}\right) }^{2} + {\left( 8 - 3\mathbf{v}\right) }^{2} \) .\n\nSum of squares\n\[ E\left( \mathbf{v}\right) = {\left( {b}_{1} - {a}_{1}\mathbf{v}\right) }^{2} + {\left( {b}_{2} - {a}_{2}\mathbf{v}\right) }^{2}. \... | Yes |
Example 3 The special case \( {a}_{1} = {a}_{2} = 1 \) has two measurements \( v = {b}_{1} \) and \( v = {b}_{2} \) of the same quantity (like pulse rate or blood pressure). The matrix has \( {A}^{\mathrm{T}} = \left\lbrack \begin{array}{ll} 1 & 1 \end{array}\right\rbrack \) . To minimize \( {\left( v - {b}_{1}\right) ... | \[ \text{If}{a}_{1} = {a}_{2} = 1\text{then}{A}^{\mathrm{T}}A = 2\text{and}{A}^{\mathrm{T}}\mathbf{b} = {b}_{1} + {b}_{2}\text{and}\widehat{v} = \left( {{b}_{1} + {b}_{2}}\right) /2\text{.} \] | Yes |
If \( A = \left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{array}\right\rbrack \) and \( \mathbf{b} = \left\lbrack \begin{array}{l} 6 \\ 0 \\ 0 \end{array}\right\rbrack \) find \( \widehat{\mathbf{v}} \) and \( \mathbf{p} \) and the matrix \( P \) . | Solution Compute the square matrix \( {A}^{\mathrm{T}}A \) and also the vector \( {A}^{\mathrm{T}}\mathbf{b} \) :\n\n\[ \n{A}^{\mathrm{T}}A = \left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{array}\right\rbrack = \left\lbrack \be... | Yes |
Example 5 Your pulse rate \( v \) is measured twice. Using unweighted least squares \( \left( {{w}_{1} = {w}_{2} = 1}\right) \), the best estimate is \( \widehat{v} = \frac{1}{2}\left( {{b}_{1} + {b}_{2}}\right) \) . Example 3 finds that least square solution \( \widehat{v} \) to two equations \( v = {b}_{1} \) and \( ... | We should weight the two measurements by \( {w}_{1} = 1/{\sigma }_{1} \) and \( {w}_{2} = 1/{\sigma }_{2} \) :\n\n\[ \text{With weights}\;\begin{array}{l} {w}_{1}v = {w}_{1}{b}_{1} \\ {w}_{2}v = {w}_{2}{b}_{2} \end{array}\;\widehat{v} = \frac{{w}_{1}{b}_{1} + {w}_{2}{b}_{2}}{{w}_{1}^{2} + {w}_{2}^{2}} \]\n\n(16)\n\nWhe... | Yes |
Are these matrices positive definite ? When their eigenvalues are positive, construct matrices \( A \) with \( S = {A}^{\mathrm{T}}A \) and find the positive energy \( {\mathbf{x}}^{\mathrm{T}}S\mathbf{x} \) . | Solution The answers are yes, yes, and no. The eigenvalues of those matrices \( S \) are\n\n(a) 4 and 1 : positive (b) 9 and 1 : positive (c) 9 and -1 : not positive.\n\nA quicker test than eigenvalues uses two determinants : the 1 by 1 determinant \( {S}_{11} \) and the 2 by 2 determinant of \( S \) . Example (b) has ... | Yes |
Choose \( Q = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack \) to multiply \( \sqrt{S} \) . Then \( A = Q\sqrt{S} \) . | \[ A = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 2 & 0 \end{array}\right\rbrack \;\text{ has }\;S = {A}^{\mathrm{T}}A = \left\lbrack \begin{array}{ll} 4 & 0 \\ 0 & 1 \end{arr... | Yes |
Example 2 The first two matrices are singular and positive semidefinite-but not the third : (d) \[ S = \left\lbrack \begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right\rbrack \] \[ S = \left\lbrack \begin{array}{ll} 4 & 4 \\ 4 & 4 \end{array}\right\rbrack \] \[ \text{(f)}S = \left\lbrack \begin{array}{rr} - 4 & 4 \\ 4 &... | The eigenvalues are 1,0 and 8,0 and \( - 8,0 \) . The energies \( {\mathbf{x}}^{\mathrm{T}}S\mathbf{x} \) are \( {x}_{2}^{2} \) and \( 4{\left( {x}_{1} + {x}_{2}\right) }^{2} \) and \( - 4{\left( {x}_{1} - {x}_{2}\right) }^{2} \) . So the third matrix is actually negative semidefinite. | Yes |
Example 3 Find the input and output eigenvectors \( \\mathbf{v} \) and \( \\mathbf{u} \) for the rectangular matrix \( A \) : | Solution Compute \( S = {A}^{\mathrm{T}}A \) and its unit eigenvectors \( {\\mathbf{v}}_{1},{\\mathbf{v}}_{2},{\\mathbf{v}}_{3} \) . The eigenvalues \( {\\sigma }^{2} \) are \( 8,2,0 \) so the positive singular values are \( {\\sigma }_{1} = \\sqrt{8} \) and \( {\\sigma }_{2} = \\sqrt{2} \) :\n\n\[ {A}^{\mathrm{T}}A = ... | Yes |
Example 1 \( u = {xy} \) solves Laplace’s equation \( {u}_{xx} + {u}_{yy} = \mathbf{0} \) . | Four solutions are easy to find: \( u = \mathbf{1},\mathbf{x},\mathbf{y},\mathbf{{xy}} \) . For those four, \( {u}_{xx} \) and \( {u}_{yy} \) are both zero. | Yes |
Find the Fourier sine coefficients \( {b}_{k} \) of the odd square wave \( {SW}\left( x\right) \) . | For \( k = 1,2,\ldots \) use formula (6) with \( S\left( x\right) = 1 \) between 0 and \( \pi \) :\n\n\[ \n{b}_{k} = \frac{2}{\pi }{\int }_{0}^{\pi }\sin {kxdx} = \frac{2}{\pi }{\left\lbrack \frac{-\cos {kx}}{k}\right\rbrack }_{0}^{\pi } = \frac{2}{\pi }\left\{ {\frac{2}{1},\frac{0}{2},\frac{2}{3},\frac{0}{4},\frac{2}{... | Yes |
Find the cosine coefficients of the ramp \( {RR}\left( x\right) \) and the up-down \( {UD}\left( x\right) \) . | The simplest way is to start with the sine series for the square wave :\n\n\[ \n{SW}\left( x\right) = \frac{4}{\pi }\left\lbrack {\frac{\sin x}{1} + \frac{\sin {3x}}{3} + \frac{\sin {5x}}{5} + \frac{\sin {7x}}{7} + \cdots }\right\rbrack = \text{ slope of }{RR} \n\] \n\nTake the derivative of every term to produce cosin... | Yes |
Example 3 Find the (cosine) coefficients of the delta function \( \\delta \\left( x\\right) \\), made \( {2\\pi } \\) -periodic. | Solution The spike in \( \\delta \\left( x\\right) \\) occurs at \( x = 0 \\) . All the integrals are 1, because the cosine of 0 is 1 . We divide by \( {2\\pi } \\) for \( {a}_{0} \\) and by \( \\pi \\) for the other cosine coefficients \( {a}_{k} \\) .\n\n\[ \n\\text{Average}{a}_{0} = \\frac{1}{2\\pi }{\\int }_{-\\pi ... | Yes |
Example 4 Find the \( a \) ’s and \( b \) ’s if \( F\left( x\right) = \) tall box \( = \left\{ \begin{array}{ll} 1/h & \text{ for }0 < x < h \\ 0 & \text{ for }h < x < {2\pi } \end{array}\right. \) | Solution The integrals for \( {a}_{0} \) and \( {a}_{k} \) and \( {b}_{k} \) stop at \( x = h \) where \( F\left( x\right) \) drops to zero. The coefficients decay like \( 1/k \) because of the jump at \( x = 0 \) and the drop at \( x = h \) :\n\nCoefficients of square pulse \( \;{a}_{0} = \frac{1}{2\pi }{\int }_{0}^{h... | Yes |
Example 6 Find \( {c}_{k} \) for the \( {2\pi } \) -periodic shifted box \( F\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ for }s \leq x \leq s + h \\ 0 & \text{ elsewhere in }\left\lbrack {-\pi ,\pi }\right\rbrack \end{array}\right. \) | Solution The integrals (23) have \( F = 1 \) from \( s \) to \( s + h \) :\n\n\[ \n{c}_{k} = \frac{1}{2\pi }{\int }_{s}^{s + h}1 \cdot {e}^{-{ikx}}{dx} = \frac{1}{2\pi }{\left\lbrack \frac{{e}^{-{ikx}}}{-{ik}}\right\rbrack }_{s}^{s + h} = {e}^{-{iks}}\left( \frac{1 - {e}^{-{ikh}}}{2\pi ik}\right) .\n\]\n\n(24)\n\nNotic... | Yes |
Example 7 A centered box has shift \( s = - h/2 \) . It becomes balanced around \( x = 0 \) . This even function equals 1 on the interval from \( - h/2 \) to \( h/2 \) : | \[ \text{Centered by}s = - \frac{h}{2}\;{c}_{k} = {e}^{{ikh}/2}\frac{1 - {e}^{-{ikh}}}{2\pi ik} = \frac{1}{2\pi }\frac{\sin \left( {{kh}/2}\right) }{k/2}\text{.} \] Divide by \( h \) for a tall box. The ratio of \( \sin \left( {{kh}/2}\right) \) to \( {kh}/2 \) is called the \ | No |
Example 8 Point source \( {u}_{0} = \delta \left( \theta \right) \) . The boundary is held at \( {u}_{0} = 0 \), except for the source at \( x = 1, y = 0 \) (where \( \theta = 0 \) ). Find the temperature \( u\left( {r,\theta }\right) \) inside the circle. | Delta function \( \;{u}_{0}\left( \theta \right) = \frac{1}{2\pi } + \frac{1}{\pi }\left( {\cos \theta + \cos {2\theta } + \cos {3\theta } + \cdots }\right) = \frac{1}{2\pi }\mathop{\sum }\limits_{{-\infty }}^{\infty }{e}^{in\theta } \)\n\nInside the circle, each \( \cos {n\theta } \) is multiplied by \( {r}^{n} \) to ... | Yes |
The delta vector \( \mathbf{f} = \left( {1,0,0,\ldots }\right) \) is like a delta function \( \delta \left( x\right) \) . The Fourier coefficients of a delta function are all equal to \( {c}_{k} = 1/{2\pi } \) . The discrete coefficients of a delta vector are all equal to \( {c}_{k} = 1/N \) . The transform of \( f \) ... | \[ \text{Fourier transform}{F}^{-1}f = c\sqrt{\frac{1}{N}\left\lbrack \begin{matrix} 1 & 1 & \cdot & \cdot & 1 \\ 1 & \bar{w} & \cdot & \cdot & {\bar{w}}^{N - 1} \\ 1 & {\bar{w}}^{2} & \cdot & \cdot & {\bar{w}}^{2\left( {N - 1}\right) } \\ \cdot & \cdot & \cdot & \cdot & \cdot \end{matrix}\right\rbrack }\left\lbrack \b... | Yes |
The shifted vector \( \mathbf{f} = \left( {0,1,0,\ldots }\right) \) is like a shifted delta function \( \delta \left( {x - \frac{2\pi }{N}}\right) \) . The shifted vector \( \mathbf{f} \) picks out the next column \( \left( {1,\bar{w},{\bar{w}}^{2},\ldots }\right) \) of \( {F}^{-1} \) in equation (3). | The shifted delta function chooses the (same) values of \( {c}_{k} = {e}^{-{ikx}} \) at \( x = {2\pi }/N \) . | No |
Example 3 The constant vector \( c = \left( {1,1,\ldots }\right) /N \) transforms back to the delta vector! | \[ \text{Fourier matrix}{Fc} = f\;\left\lbrack \begin{matrix} 1 & 1 & . & . & 1 \\ 1 & w & . & . & {w}^{N - 1} \\ 1 & {w}^{2} & . & . & {w}^{2\left( {N - 1}\right) } \\ . & . & . & . & . \end{matrix}\right\rbrack \frac{1}{N}\left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \\ . \end{array}\right\rbrack = \left\lbrack \begin{ar... | Yes |
Example 4 Sample \( \cos x \) and \( \sin x \) at \( 0,\pi /2,\pi ,{3\pi }/2 \) to get discrete vectors \( \cos \) and \( \sin \) . Transform those vectors by \( {F}^{-1} \) . Invert their transforms by \( F \) . | To transform \( x \) -space to \( k \) -space, we multiply \( \mathbf{f} \) by \( {F}^{-1} \) . For \( N = 4 \), this matrix contains powers of \( \bar{w} = - i \) . We remember to divide by \( N = 4 \) :\n\n\[ \n{F}^{-1}\cos = \frac{1}{4}\left\lbrack \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & - i & - 1 & i \\ 1 & - 1 & ... | Yes |
Example 2 (Finite bar) Suppose the bar goes from \( x = 0 \) to \( x = 1 \) . The initial condition \( u\left( {0, x}\right) = 1 \) tells us the (constant) temperature along the bar at time \( t = 0 \) . We also need boundary conditions like \( u\left( {t,0}\right) = 0 \) and \( u\left( {t,1}\right) = 0 \) at the ends ... | ## Solution by Fourier Series\n\nEverything comes from choosing the right form for the solution \( u\left( {t, x}\right) \) . Here it is :\n\n\[ \text{Sine series}u\left( {t, x}\right) = {b}_{1}\left( t\right) \sin {\pi x} + {b}_{2}\left( t\right) \sin {2\pi x} + \cdots = \mathop{\sum }\limits_{{k = 1}}^{\infty }{b}_{k... | Yes |
Example 3 (Insulated bar) No heat escapes from the ends of the bar. The boundary conditions change to \( \partial u/\partial x = 0 \) at those ends. The basis functions change to cosines. The series (8) becomes a Fourier cosine series. | Initial condition \( \;u\left( {0, x}\right) = \sum {a}_{k}\left( 0\right) \cos {k\pi x} \)\n\nEquation for the \( {a}_{k}\;d{a}_{k}/{dt} = - {k}^{2}{\pi }^{2}{a}_{k} \) for \( k = 0,1,2,\ldots \)\n\nNotice that \( k = 0 \) is included. The first basis function is \( \cos {0\pi x} = 1 \) . Its coefficient is controlled... | Yes |
Example 4 (Circular bar) Now sines and cosines are both included. The basis functions can also be complex exponentials \( {e}^{ikx} \) . Again \( u \) goes to a constant steady state \( {c}_{0} \) : | \[ u\left( {t, x}\right) = \mathop{\sum }\limits_{{-\infty }}^{\infty }{c}_{k}\left( t\right) {e}^{ik\pi x}\;\text{ and }\;\frac{d{c}_{k}}{dt} = - {k}^{2}{\pi }^{2}{c}_{k}. \] | Yes |
Now solve Example 1, which started from \( u = - 1 \) for negative \( x \) and \( u = 1 \) for positive \( x \) . | Here is the key idea for Example 1. The derivative of the jump from -1 to 1 at \( x = 0 \) is \( {du}/{dx} = \mathbf{2}\delta \left( x\right) \) . The solution starting from \( {2\delta }\left( x\right) \) has \( {du}/{dx} = {2U} \), which cancels \( \sqrt{4} \) in (14). Then integrate \( {2U} \) to undo the derivative... | Yes |
Example 7 (Smokestack problem) We backed away from the heat equation in \( 2 + 1 \) dimensions. The solution to \( {\mathbf{u}}_{t} = {\mathbf{u}}_{xx} + {\mathbf{u}}_{yy} \) involves three variables \( t, x, y \) . Put a smokestack at the center point \( x = y = 0 \), and suppose there is no wind. Then nothing depends... | The heat equation is not quite \( {u}_{t} = {u}_{rr} \) because \( r = \) constant is curved (a circle). The correct radial equation is perfect for separation of variables \( u = B\left( t\right) A\left( r\right) \) . \n\n\[ \frac{\partial u}{\partial t} = \frac{{\partial }^{2}u}{\partial {r}^{2}} + \frac{1}{r}\frac{\p... | Yes |
Start from rest (velocity \( {v}_{0} = 0 \) ) with a sine wave \( {u}_{0}\left( x\right) = \sin {\omega x} \) . That wave splits into two waves : | \[ u\left( {t, x}\right) = \frac{{u}_{0}\left( {x + {ct}}\right) + {u}_{0}\left( {x - {ct}}\right) }{2} = \frac{1}{2}\sin \left( {{\omega x} + {c\omega t}}\right) + \frac{1}{2}\sin \left( {{\omega x} - {c\omega t}}\right) . \]\n\nThe trigonometry formula \( \sin A + \sin B = 2\sin \frac{A + B}{2}\cos \frac{A - B}{2} \)... | Yes |
Example 2 Suppose the length is \( L = 3 \) and the initial displacement is a hat function : | \[ {u}_{0}\left( x\right) = x\text{ for }0 \leq x \leq 1\text{ and }{u}_{0}\left( x\right) = \frac{1}{2}\left( {3 - x}\right) \text{ for }1 \leq x \leq 3. \] The integrals in (18) lead in Mathematica to \( {b}_{k} = 3/2{k}^{2}{\pi }^{2} \) . The decay rate is \( 1/{k}^{2} \) for this function \( {u}_{0}\left( x\right) ... | Yes |
Example 1 Solve \( {y}^{\prime } - y = 2{e}^{-t} \) starting from \( y\left( 0\right) = 1 \) . | Solution Take the Laplace transform of both sides. We know \( \mathcal{L}\left\lbrack {2{e}^{-t}}\right\rbrack = 2/\left( {s + 1}\right) \) :\n\n\[ s\mathcal{L}\left\lbrack y\right\rbrack - y\left( 0\right) - \mathcal{L}\left\lbrack y\right\rbrack = \mathcal{L}\left\lbrack {2{e}^{-t}}\right\rbrack \text{ is the same as... | No |
Solve the equation \( {y}^{\prime \prime } + y = \frac{1}{2}\sin {2t} \) starting from rest: \( y\left( 0\right) = {y}^{\prime }\left( 0\right) = 0 \) . | The transform of \( {y}^{\prime \prime } \) is \( {s}^{2}Y\left( s\right) \) from (7) :\n\n\[ \n{s}^{2}Y\left( s\right) + Y\left( s\right) = \frac{1}{{s}^{2} + {2}^{2}}\text{ and then }Y\left( s\right) = \frac{1}{\left( {{s}^{2} + 1}\right) \left( {{s}^{2} + 4}\right) } \n\]\n\nPartial fractions will rewrite that trans... | Yes |
Example 4 (Unit step function) Solve \( {y}^{\prime } - {ay} = H\left( {t - T}\right) = \left\{ \begin{array}{ll} \mathbf{0} & t < T \\ \mathbf{1} & t \geq T \end{array}\right\} \) . | The transform of every term (with \( y\left( 0\right) = 1 \) ) will give the transform \( Y\left( s\right) \) of the solution :\n\n\[ \n{sY}\left( s\right) - 1 - {aY}\left( s\right) = \frac{{e}^{-{sT}}}{s}\;Y\left( s\right) = \frac{1}{s - a} + \frac{{e}^{-{sT}}}{\left( {s - a}\right) s}. \n\]\n\nThe inverse transform o... | Yes |
Solve \(\mathbf{y}^{\prime } - \mathbf{a}\mathbf{y} = \left\{ {\begin{array}{ll} \mathbf{1} & t \leq T \\ \mathbf{0} & t > T \end{array}\;\text{with}y\left( 0\right) = 1}\right. \text{. | Solution Instead of the previous \( H\left( {t - T}\right) \), this new driving function is \( 1 - H\left( {t - T}\right) \). The step function drops from 1 to 0. We still take the Laplace transform of every term in the differential equation:\n\n\[{sY}\left( s\right) - 1 - {aY}\left( s\right) = \text{ transform of }\le... | Yes |
The autocorrelation of a vector \( c \) is the convolution \( c * {c}^{\prime } \) . That vector \( {c}^{\prime } \) is the reverse of \( c \) . The components of \( {c}^{\prime } \) are the Fourier coefficients \( {\bar{c}}_{-k} \) of \( \overline{f\left( x\right) } \) . So autocorrelation \( c * {c}^{\prime } \) give... | \[ f\bar{f} = \left( {1 + {e}^{ix}}\right) \left( {1 + {e}^{-{ix}}}\right) = \mathbf{1}{e}^{-{ix}} + \mathbf{2} + \mathbf{1}{e}^{ix}\;\mathbf{c} * {\mathbf{c}}^{\prime } = \left( {0,1,1}\right) * \left( {1,1,0}\right) = \left( {\mathbf{1},\mathbf{2},\mathbf{1}}\right) . \] | Yes |
Solve the equation \( {y}^{\prime } - {ay} = f\left( t\right) \) by convolution, starting from \( y\left( 0\right) = 0 \) . | Take the Laplace transform of both sides, and divide to find \( Y\left( s\right) \) :\n\n\[ \n{sY}\left( s\right) - {aY}\left( s\right) = F\left( s\right) \;\text{ gives }\;Y\left( s\right) = \frac{F\left( s\right) }{s - a} = \mathbf{G}\left( s\right) \mathbf{F}\left( s\right) .\n\]\n\nThe transform \( F\left( s\right)... | Yes |
Example 4 (Fourier) Solve the equation \( - {y}^{\prime \prime } + y = f\left( x\right) \) for \( - \infty < x < \infty \) . | Solution This is a boundary value problem, with \( y = 0 \) at the endpoints \( x = - \infty \) and \( x = \infty \) . Take the Fourier transform of every term, so the two derivatives in \( {y}^{\prime \prime } \) become multiplications by \( {ik} \) :\n\n\[ - {y}^{\prime \prime } + y = f\left( x\right) \; - {\left( ik... | Yes |
Example 5 Solve the equation \( {y}^{\prime \prime } + y = f\left( t\right) \) starting from \( y\left( 0\right) = {y}^{\prime }\left( 0\right) = 0 \) . | Solution Take the Laplace transform of both sides, and divide by \( {s}^{2} + 1 \) to find \( Y\left( s\right) \) :\n\n\[ \n{s}^{2}Y\left( s\right) + Y\left( s\right) = F\left( s\right) \text{ gives }Y\left( s\right) = \frac{F\left( s\right) }{{s}^{2} + 1} = F\left( s\right) G\left( s\right) .\n\]\n\n(10)\n\nThe transf... | Yes |
The vectors \( \mathbf{v} = \left( {4,2}\right) \) and \( \mathbf{w} = \left( {-1,2}\right) \) have a zero dot product: | Dot product is zero\n\[ \left\lbrack \begin{array}{l} 4 \\ 2 \end{array}\right\rbrack \cdot \left\lbrack \begin{array}{r} - 1 \\ 2 \end{array}\right\rbrack = - 4 + 4 = 0 \] | Yes |
Put a weight of 4 at the point \( x = - 1 \) (left of zero) and a weight of 2 at the point \( x = 2 \) (right of zero). The \( x \) axis will balance on the center point (like a see-saw). | The weights balance because the dot product is \( \left( 4\right) \left( {-1}\right) + \left( 2\right) \left( 2\right) = 0 \) . | Yes |
Find \( \cos \theta \) for \( v = \left\lbrack \begin{array}{l} 2 \\ 1 \end{array}\right\rbrack \) and \( w = \left\lbrack \begin{array}{l} 1 \\ 2 \end{array}\right\rbrack \) and check both inequalities. | Solution The dot product is \( \mathbf{v} \cdot \mathbf{w} = 4 \) . Both \( \mathbf{v} \) and \( \mathbf{w} \) have length \( \sqrt{5} \) . The cosine is \( 4/5 \) .\n\n\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\parallel \mathbf{v}\parallel \parallel \mathbf{w}\parallel } = \frac{4}{\sqrt{5}\sqrt{5}} = \frac{... | Yes |
The dot product of \( \\mathbf{v} = \\left( {a, b}\\right) \) and \( \\mathbf{w} = \\left( {b, a}\\right) \) is \( {2ab} \) . Both lengths are \( \\sqrt{{a}^{2} + {b}^{2}} \) . The Schwarz inequality \( \\mathbf{v} \\cdot \\mathbf{w} \\leq \\parallel \\mathbf{v}\\parallel \\parallel \\mathbf{w}\\parallel \) says that \... | This is more famous if we write \( x = {a}^{2} \) and \( y = {b}^{2} \) . The \ | No |
Here are 3 by 3 matrices \( A \) and \( I = \) identity, with three 1 ’s and six 0 ’s : | \[ {Ax} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right\rbrack \left\lbrack \begin{array}{l} 4 \\ 5 \\ 6 \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 4 \\ 4 \\ 4 \end{array}\right\rbrack \;{Ix} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end... | Yes |
\[ x - {2y} = 1\;\text{Subtract 3 times} \] \[ {3x} - {6y} = {11}\;\text{ eqn. }1\text{ from eqn. }2\;\mathbf{0}y = 8. \] | There is no solution to \( {0y} = 8 \) . Normally we divide the right side 8 by the second pivot, but this system has no second pivot. (Zero is never allowed as a pivot!) The row and column pictures in Figure 2.6 show why failure was unavoidable. If there is no solution, elimination will discover that fact by reaching ... | Yes |
Example 2 Failure with infinitely many solutions. Change \( b = \\left( {1,{11}}\\right) \) to \( \\left( {1,3}\\right) \) . | \[ x - {2y} = 1 \] Still only \[ {3x} - {6y} = 3\\;\\text{ eqn. }1\\text{ from eqn. }2 \] \[ \\mathbf{0}y = \\mathbf{0}\\text{.} \] one pivot. Every \( y \) satisfies \( {0y} = 0 \) . There is really only one equation \( x - {2y} = 1 \) . The unknown \( y \) is \ | No |
Example 3 Temporary failure (zero in pivot). A row exchange produces two pivots: | \[ {0x} + {2y} = 4\;\text{Exchange the} \] \[ {3x} - {2y} = 5 \] Permutation \[ {3x} - {2y} = 5\;\text{two equations} \] \[ {2y} = 4\text{.} \] The new system is already triangular. This small example is ready for back substitution. The last equation gives \( y = 2 \), and then the first equation gives \( x = 3 \) . Th... | Yes |
This matrix has \( {a}_{ij} = {2i} + j \) . Then \( {a}_{11} = 3 \) . Also \( {a}_{12} = 4 \) and \( {a}_{21} = 5 \) . | \[ \left\lbrack \begin{array}{ll} 3 & 4 \\ 5 & 6 \end{array}\right\rbrack \left\lbrack \begin{array}{l} 2 \\ 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 3 \cdot 2 + 4 \cdot 1 \\ 5 \cdot 2 + 6 \cdot 1 \end{array}\right\rbrack \;\left\lbrack \begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \end... | Yes |
The matrix \( {E}_{31} \) has \( - \ell \) in the 3,1 position: | \[ \text{Identity}I = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \;\text{Elimination}\;{E}_{31} = \left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - \ell & 0 & 1 \end{array}\right\rbrack \text{.} \]\n\nWhen you multiply \( I \) times \( \mathbf{b} \), you get \... | Yes |
Square matrices can be multiplied if and only if they have the same size: | \[ \left\lbrack \begin{array}{rr} 1 & 1 \\ 2 & - 1 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 2 & 2 \\ 3 & 4 \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} 5 & 6 \\ 1 & 0 \end{array}\right\rbrack \] The first dot product is \( 1 \cdot 2 + 1 \cdot 3 = 5 \) . Three more dot products give 6,1, and ... | Yes |
Suppose \( A \) is a row vector (1 by 3) and \( B \) is a column vector (3 by 1 ). Then \( {AB} \) is 1 by 1 (only one entry, the dot product). On the other hand \( B \) times \( A \) (a column times a row) is a full 3 by 3 matrix. This multiplication is allowed! | Column times row\n\[ \left\lbrack \begin{array}{l} 0 \\ 1 \\ 2 \end{array}\right\rbrack \left\lbrack \begin{array}{lll} 1 & 2 & 3 \end{array}\right\rbrack = \left\lbrack \begin{array}{lll} 0 & 0 & 0 \\ 1 & 2 & 3 \\ 2 & 4 & 6 \end{array}\right\rbrack \] | Yes |
Example 3 (Important special case) Let the blocks of \( A \) be its \( n \) columns. Let the blocks of \( B \) be its \( n \) rows. Then block multiplication \( {AB} \) adds up columns times rows: | \[ \begin{array}{l} \text{ Columns } \\ \text{ times } \\ \text{ rows } \end{array}\;\left\lbrack \begin{matrix} \mid & & \mid \\ {a}_{1} & \cdots & {a}_{n} \\ \mid & & \mid \end{matrix}\right\rbrack \left\lbrack \begin{matrix} - & {b}_{1} & - \\ & \vdots & \\ - & {b}_{n} & - \end{matrix}\right\rbrack = \left\lbrack {{... | Yes |
Example 4 (Elimination by blocks) Suppose the first column of \( A \) contains \( 1,3,4 \) . To change 3 and 4 to 0 and 0 , multiply the pivot row by 3 and 4 and subtract. Those row operations are really multiplications by elimination matrices \( {E}_{21} \) and \( {E}_{31} \) : | \[ \text{One at a time}\;{E}_{21} = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \;\text{and}\;{E}_{31} = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 4 & 0 & 1 \end{matrix}\right\rbrack \text{.} \] The \ | Yes |
The 2 by 2 matrix \( A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 1 & 2 \end{array}\right\rbrack \) is not invertible. | It fails the test in Note 5, because \( {ad} - {bc} \) equals \( 2 - 2 = 0 \) . It fails the test in Note 3, because \( {Ax} = \mathbf{0} \) when \( \mathbf{x} = \left( {2, - 1}\right) \) . It fails to have two pivots as required by Note 1 . Elimination turns the second row of this matrix \( A \) into a zero row. | Yes |
Example 2 Inverse of an elimination matrix. If \( E \) subtracts 5 times row 1 from row 2, then \( {E}^{-1} \) adds 5 times row 1 to row 2 : | Multiply \( E{E}^{-1} \) to get the identity matrix \( I \) . Also multiply \( {E}^{-1}E \) to get \( I \) . We are adding and subtracting the same 5 times row 1 . If \( {AC} = I \) then automatically \( {CA} = I \) . | Yes |
Example 3 Suppose \( F \) subtracts 4 times row 2 from row 3, and \( {F}^{-1} \) adds it back:\n\n\[ F = \left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \end{array}\right\rbrack \;\text{ and }\;{F}^{-1} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{array}\right\rbrack ... | \[ {FE} = \left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ - 5 & 1 & 0 \\ \mathbf{{20}} & - 4 & 1 \end{array}\right\rbrack \;\text{ is inverted by }\;{E}^{-1}{F}^{-1} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ \mathbf{5} & 1 & 0 \\ \mathbf{0} & \mathbf{4} & 1 \end{array}\right\rbrack . \] | Yes |
Example 4 Find \( {A}^{-1} \) by Gauss-Jordan elimination starting from \( A = \left\lbrack \begin{array}{ll} 2 & 3 \\ 4 & 7 \end{array}\right\rbrack \) . | \[ \left\lbrack \begin{array}{ll} \mathbf{A} & I \end{array}\right\rbrack = \left\lbrack \begin{array}{llll} \mathbf{2} & \mathbf{3} & 1 & 0 \\ \mathbf{4} & \mathbf{7} & 0 & 1 \end{array}\right\rbrack \rightarrow \left\lbrack \begin{array}{rrrr} 2 & 3 & 1 & 0 \\ \mathbf{0} & \mathbf{1} & - \mathbf{2} & \mathbf{1} \end{... | Yes |
If \( A \) is invertible and upper triangular, so is \( {A}^{-1} \) . | Start with \( A{A}^{-1} = I \). \n\n1 A times column \( j \) of \( {A}^{-1} \) equals column \( j \) of \( I \), ending with \( n - j \) zeros. \n\n2 Back substitution keeps those \( n - j \) zeros at the end of column \( j \) of \( {A}^{-1} \). \n\n3 Put those columns \( {\left\lbrack *\ldots * 0\ldots 0\right\rbrack ... | Yes |
Example 6 If \( L \) is lower triangular with 1 ’s on the diagonal, so is \( {L}^{-1} \) . | A triangular matrix is invertible if and only if no diagonal entries are zero.\n\nHere \( L \) has 1’s so \( {L}^{-1} \) also has 1’s. Use the Gauss-Jordan method to construct \( {L}^{-1} \) from \( {E}_{32},{E}_{31},{E}_{21} \) . Notice how \( {L}^{-1} \) contains the strange entry 11, from 3 times 5 minus 4 .\n\nGaus... | Yes |
Example 1 Elimination subtracts \( \frac{1}{2} \) times row 1 from row 2 . The last step subtracts \( \frac{2}{3} \) times row 2 from row 3 . The lower triangular \( L \) has \( {\ell }_{21} = \frac{1}{2} \) and \( {\ell }_{32} = \frac{2}{3} \) . Multiplying \( {LU} \) produces \( A \) : | \[ A = \left\lbrack \begin{array}{lll} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right\rbrack = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ 0 & \frac{2}{3} & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{lll} 2 & 1 & 0 \\ 0 & \frac{3}{2} & 1 \\ 0 & 0 & \frac{4}{3} \end{array}\right... | Yes |
The key reason why A equals LU: Ask yourself about the pivot rows that are subtracted from lower rows. Are they the original rows of \( A \) ? No, elimination probably changed them. Are they rows of \( U \) ? Yes, the pivot rows never change again. When computing the third row of \( U \), we subtract multiples of earli... | \[ \text{Row 3 of}U = \left( {\operatorname{Row}3\text{of}A}\right) - {\ell }_{31}\left( {\operatorname{Row}1\text{of}U}\right) - {\ell }_{32}\left( {\operatorname{Row}2\text{of}U}\right) \text{.} \] (1) Rewrite this equation to see that the row \( \left\lbrack \begin{array}{lll} {\ell }_{31} & {\ell }_{32} & 1 \end{ar... | Yes |
Example 3 Forward elimination (downward) on \( {Ax} = b \) ends at \( {Ux} = c \) : | \[ \begin{array}{l} {Ax} = b\;u + {2v} = 5\;\text{ becomes }\;u + {2v} = 5\;{Ux} = c \\ u = 1\;v = 1\;{Ux} = c \\ \end{array} \] The multiplier was 4, which is saved in \( L \) . The right side used that 4 to change 21 to 1 : \( {Lc} = b \) The lower triangular system \( \left\lbrack \begin{array}{ll} 1 & 0 \\ 4 & 1 \e... | Yes |
The inverse of \( A = \left\lbrack \begin{array}{ll} 1 & 0 \\ 6 & 1 \end{array}\right\rbrack \) is \( {A}^{-1} = \left\lbrack \begin{array}{rr} 1 & 0 \\ - 6 & 1 \end{array}\right\rbrack \) . The transpose is \( {A}^{\mathrm{T}} = \left\lbrack \begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right\rbrack \) . | \[ {\left( {A}^{-1}\right) }^{\mathrm{T}}\text{and}{\left( {A}^{\mathrm{T}}\right) }^{-1}\text{are both equal to}\left\lbrack \begin{matrix} 1 & - 6 \\ 0 & 1 \end{matrix}\right\rbrack \text{.} \] | Yes |
Multiply \( A = \left\lbrack \begin{array}{rrr} - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{array}\right\rbrack \) and \( {A}^{\mathrm{T}} = \left\lbrack \begin{array}{rr} - 1 & 0 \\ 1 & - 1 \\ 0 & 1 \end{array}\right\rbrack \) in both orders. | \[ A{A}^{\mathrm{T}} = \left\lbrack \begin{array}{rr} 2 & - 1 \\ - 1 & 2 \end{array}\right\rbrack \text{and}{A}^{\mathrm{T}}A = \left\lbrack \begin{array}{rrr} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}\right\rbrack \text{are both symmetric matrices.} \] | Yes |
There are six 3 by 3 permutation matrices. Here they are without the zeros: | \[ I = \left\lbrack \begin{array}{lll} 1 & & \\ & 1 & \\ & & 1 \end{array}\right\rbrack \;{P}_{21} = \left\lbrack \begin{array}{lll} & 1 & \\ 1 & & \\ & & 1 \end{array}\right\rbrack \;{P}_{32}{P}_{21} = \left\lbrack \begin{array}{lll} & 1 & \\ & & 1 \\ 1 & & \end{array}\right\rbrack \] \[ {P}_{31} = \left\lbrack \begin... | Yes |
Keep only the vectors \( \left( {x, y}\right) \) whose components are positive or zero (this is a quarter-plane). | The vector \( \left( {2,3}\right) \) is included but \( \left( {-2, - 3}\right) \) is not. So rule (ii) is violated when we try to multiply by \( c = - 1 \) . The quarter-plane is not a subspace. | Yes |
Inside the vector space \( \mathbf{M} \) of all 2 by 2 matrices, here are two subspaces:\n\n(U) All upper triangular matrices \( \left\lbrack \begin{array}{ll} a & b \\ 0 & d \end{array}\right\rbrack \; \) (D) All diagonal matrices \( \left\lbrack \begin{array}{ll} a & 0 \\ 0 & d \end{array}\right\rbrack \) .\n\nAdd an... | Multiples of the identity matrix also form a subspace. \( {2I} + {3I} \) is in this subspace, and so is 3 times \( {4I} \) . The matrices \( {cI} \) form a \ | No |
Example 5 Describe the column spaces (they are subspaces of \( {\mathbf{R}}^{2} \) ) for\n\n\[ I = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right\rbrack \;\text{ and }\;A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{array}{lll} 1 & 2 &... | Solution The column space of \( I \) is the whole space \( {\mathbf{R}}^{2} \) . Every vector is a combination of the columns of \( I \) . In vector space language, \( \mathbf{C}\left( I\right) \) is \( {\mathbf{R}}^{2} \) .\n\nThe column space of \( A \) is only a line. The second column \( \left( {2,4}\right) \) is a... | Yes |
Example 4 When all rows are multiples of one pivot row, the rank is \( r = 1 \) : | \[ \left\lbrack \begin{array}{lll} 1 & 3 & 4 \\ 2 & 6 & 8 \end{array}\right\rbrack \text{ and }\left\lbrack \begin{array}{ll} 0 & 3 \\ 0 & \mathbf{5} \end{array}\right\rbrack \text{ and }\left\lbrack \begin{array}{l} 5 \\ 2 \end{array}\right\rbrack \text{ and }\left\lbrack 6\right\rbrack \text{ all have rank }1. \] For... | Yes |
The columns of this \( A \) are dependent. \( {Ax} = \mathbf{0} \) has a nonzero solution: | \[ {Ax} = \left\lbrack \begin{array}{lll} 1 & 0 & 3 \\ 2 & 1 & 5 \\ 1 & 0 & 3 \end{array}\right\rbrack \left\lbrack \begin{array}{r} - 3 \\ 1 \\ 1 \end{array}\right\rbrack \text{ is } - 3\left\lbrack \begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right\rbrack + 1\left\lbrack \begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right\rbr... | Yes |
Example 5 Describe the column space and the row space of \( A \) . | \[ A = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 7 \\ 3 & 5 \end{array}\right\rbrack \text{ and }{A}^{\mathrm{T}} = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 7 & 5 \end{array}\right\rbrack \text{. Here }m = 3\text{ and }n = 2. \]\n\nThe column space of \( A \) is the plane in \( {\mathbf{R}}^{3} \) spanned by ... | Yes |
The columns of every invertible \( n \) by \( n \) matrix give a basis for \( {\mathbf{R}}^{n} \) | The only solution to \( {Ax} = \mathbf{0} \) is \( \mathbf{x} = {A}^{-1}\mathbf{0} = \mathbf{0} \) . The columns are independent. They span the whole space \( {\mathbf{R}}^{n} \) -because every vector \( \mathbf{b} \) is a combination of the columns. \( A\mathbf{x} = \mathbf{b} \) can always be solved by \( \mathbf{x} ... | Yes |
Example 8 This matrix is not invertible. Its columns are not a basis for anything! | \[ \text{One pivot column}\;A = \left\lbrack \begin{array}{ll} 2 & 4 \\ 3 & 6 \end{array}\right\rbrack \text{reduces to}R = \left\lbrack \begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array}\right\rbrack \text{.} \] Column 1 of \( A \) is the pivot column. That column alone is a basis for its column space. The second column of ... | Yes |
If \( {\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{m} \) and \( {\mathbf{w}}_{1},\ldots ,{\mathbf{w}}_{n} \) are both bases for the same vector space, then \( m = n \). | Suppose that there are more \( w \)’s than \( v \)’s. From \( n > m \) we want to reach a contradiction. The \( \mathbf{v} \)’s are a basis, so \( {\mathbf{w}}_{1} \) must be a combination of the \( \mathbf{v} \)’s. If \( {\mathbf{w}}_{1} \) equals \( {a}_{11}{\mathbf{v}}_{1} + \cdots + {a}_{m1}{\mathbf{v}}_{m} \), thi... | Yes |
Two walls look perpendicular but those two subspaces are not orthogonal! The meeting line is in both \( V \) and \( W \) -and this line is not perpendicular to itself. Two planes (dimensions 2 and 2 in \( {\mathbf{R}}^{3} \) ) cannot be orthogonal subspaces. | When a vector is in two orthogonal subspaces, it must be zero. It is perpendicular to itself. It is \( \mathbf{v} \) and it is \( \mathbf{w} \), so \( {\mathbf{v}}^{\mathrm{T}}\mathbf{v} = 0 \) . This has to be the zero vector. | No |
The rows of \( A \) are perpendicular to \( \mathbf{x} = \left( {1,1, - 1}\right) \) in the nullspace: | \[ A\mathbf{x} = \left\lbrack \begin{array}{lll} 1 & 3 & 4 \\ 5 & 2 & 7 \end{array}\right\rbrack \left\lbrack \begin{array}{r} 1 \\ 1 \\ - 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rbrack \;\text{ gives the dot products }\begin{array}{l} 1 + 3 - 4 = 0 \\ 5 + 2 - 7 = 0 \end{arra... | Yes |
Example 1 Project \( b = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \) onto \( a = \left\lbrack \begin{array}{l} 1 \\ 2 \\ 2 \end{array}\right\rbrack \) to find \( p = \widehat{x}a \) in Figure 4.6. | Solution The number \( \widehat{x} \) is the ratio of \( {\mathbf{a}}^{\mathrm{T}}\mathbf{b} = 5 \) to \( {\mathbf{a}}^{\mathrm{T}}\mathbf{a} = 9 \) . So the projection is \( \mathbf{p} = \frac{5}{9}\mathbf{a} \) .\n\nThe error vector between \( b \) and \( p \) is \( e = b - p \) . Those vectors \( p \) and \( e \) wi... | Yes |
Example 2 Find the projection matrix \( P = \frac{a{a}^{\mathrm{T}}}{{a}^{\mathrm{T}}a} \) onto the line through \( a = \left\lbrack \begin{array}{l} 1 \\ 2 \end{array}\right\rbrack \) . | Solution Multiply column \( \mathbf{a} \) times row \( {\mathbf{a}}^{\mathrm{T}} \) and divide by \( {\mathbf{a}}^{\mathrm{T}}\mathbf{a} = 9 \) :\n\n\[ \n\text{Projection matrix}\;P = \frac{a{a}^{\mathrm{T}}}{{a}^{\mathrm{T}}a} = \frac{1}{9}\left\lbrack \begin{array}{l} 1 \\ 2 \\ 2 \end{array}\right\rbrack \left\lbrack... | Yes |
If \( A = \left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{array}\right\rbrack \) and \( b = \left\lbrack \begin{array}{l} 6 \\ 0 \\ 0 \end{array}\right\rbrack \) find \( \widehat{x} \) and \( p \) and \( P \) . | Compute the square matrix \( {A}^{\mathrm{T}}A \) and also the vector \( {A}^{\mathrm{T}}\mathbf{b} \):\n\n\[ \n{A}^{\mathrm{T}}A = \left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{array}\right\rbrack = \left\lbrack \begin{array}... | Yes |
Example 1 A crucial application of least squares is fitting a straight line to \( m \) points. Start with three points: Find the closest line to the points \( \left( {0,6}\right) ,\left( {1,0}\right) \), and \( \left( {2,0}\right) \) . | No straight line \( b = C + {Dt} \) goes through those three points. We are asking for two numbers \( C \) and \( D \) that satisfy three equations: \( n = 2 \) and \( m = 3 \) . Here are the three equations at \( t = 0,1,2 \) to match the given values \( b = 6,0,0 \) :\n\n\( t = 0\; \) The first point is on the line \... | Yes |
Example 2 A has orthogonal columns when the measurement times \( {t}_{i} \) add to zero. | Suppose \( b = 1,2,4 \) at times \( t = - 2,0,2 \) . Those times add to zero. The columns of \( A \) have zero dot product: \( \left( {1,1,1}\right) \) is orthogonal to \( \left( {-2,0,2}\right) \) :\n\n\[ \n\begin{array}{l} C + D\left( {-2}\right) = 1 \\ C + D\left( 0\right) = 2 \\ C + D\left( 2\right) = 4 \end{array}... | Yes |
For a parabola \( b = C + {Dt} + E{t}^{2} \) to go through the three heights \( b = 6,0,0 \) when \( t = 0,1,2 \), the equations for \( C, D, E \) are | \[ C + D \cdot 0 + E \cdot {0}^{2} = 6 \] \[ C + D \cdot 1 + E \cdot {1}^{2} = 0 \] \[ C + D \cdot 2 + E \cdot {2}^{2} = 0. \] This is \( {Ax} = b \) . We can solve it exactly. Three data points give three equations and a square matrix. The solution is \( \mathbf{x} = \left( {C, D, E}\right) = \left( {\mathbf{6}, - \ma... | Yes |
Example 1 (Rotation) \( Q \) rotates every vector in the plane by the angle \( \theta \) : | \[ Q = \left\lbrack \begin{array}{rr} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array}\right\rbrack \;\text{ and }\;{Q}^{\mathrm{T}} = {Q}^{-1} = \left\lbrack \begin{array}{rr} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array}\right\rbrack . \]\n\nThe columns of \( Q \) are orthog... | Yes |
Example 2 (Permutation) These matrices change the order to \( \\left( {y, z, x}\\right) \) and \( \\left( {y, x}\\right) \) : | \[ \left\\lbrack \\begin{array}{lll} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 1 & 0 & 0 \\end{array}\\right\\rbrack \\left\\lbrack \\begin{array}{l} x \\\\ y \\\\ z \\end{array}\\right\\rbrack = \\left\\lbrack \\begin{array}{l} y \\\\ z \\\\ x \\end{array}\\right\\rbrack \\;\\text{ and }\\;\\left\\lbrack \\begin{array}{ll} 0 & 1 ... | Yes |
If \( \mathbf{u} \) is any unit vector, set \( Q = I - 2\mathbf{u}{\mathbf{u}}^{\mathrm{T}} \). Notice that \( \mathbf{u}{\mathbf{u}}^{\mathrm{T}} \) is a matrix while \( {\mathbf{u}}^{\mathrm{T}}\mathbf{u} \) is the number \( \parallel \mathbf{u}{\parallel }^{2} = 1 \). Then \( {Q}^{\mathrm{T}} \) and \( {Q}^{-1} \) b... | \n\[
{Q}^{\mathrm{T}} = I - {2u}{\mathbf{u}}^{\mathrm{T}} = Q\;\text{ and }\;{Q}^{\mathrm{T}}Q = I - {4u}{\mathbf{u}}^{\mathrm{T}} + {4u}{\mathbf{u}}^{\mathrm{T}}\mathbf{u}{\mathbf{u}}^{\mathrm{T}} = I.
\] | No |
Example 4 The columns of this orthogonal \( Q \) are orthonormal vectors \( {\mathbf{q}}_{1},{\mathbf{q}}_{2},{\mathbf{q}}_{3} \) : | \[ m = n = 3\;Q = \frac{1}{3}\left\lbrack \begin{array}{rrr} - 1 & 2 & 2 \\ 2 & - 1 & 2 \\ 2 & 2 & - 1 \end{array}\right\rbrack \;\text{ has }\;{Q}^{\mathrm{T}}Q = Q{Q}^{\mathrm{T}} = I. \] The separate projections of \( \mathbf{b} = \left( {0,0,1}\right) \) onto \( {\mathbf{q}}_{1} \) and \( {\mathbf{q}}_{2} \) and \(... | Yes |
Property 3 (Linearity) When row 1 of \( A \) is multiplied by \( t \), so is row 1 of \( {AB} \) . This multiplies the determinant \( \left| {AB}\right| \) by \( t \) . So the ratio \( \left| {AB}\right| /\left| B\right| \) is multiplied by \( t \) . | Add row 1 of \( A \) to row 1 of \( {A}^{\prime } \) . Then row 1 of \( {AB} \) adds to row 1 of \( {A}^{\prime }B \) . By rule 3, determinants add. After dividing by \( \left| B\right| \), the ratios add-as desired.\n\nConclusion This ratio \( \left| {AB}\right| /\left| B\right| \) has the same three properties that d... | Yes |
Example 1 A row exchange produces pivots 4, 2, 1 and that important minus sign: | \[ A = \left\lbrack \begin{array}{lll} 0 & 0 & 1 \\ 0 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right\rbrack \;{PA} = \left\lbrack \begin{array}{lll} 4 & 5 & 6 \\ 0 & 2 & 3 \\ 0 & 0 & 1 \end{array}\right\rbrack \;\det A = - \left( 4\right) \left( 2\right) \left( 1\right) = - 8. \] The odd number of row exchanges (namely one exc... | Yes |
The first pivots of this tridiagonal matrix \( A \) are \( 2,\frac{3}{2},\frac{4}{3} \) . The next are \( \frac{5}{4} \) and \( \frac{6}{5} \) and eventually \( \frac{n + 1}{n} \) . Factoring this \( n \) by \( n \) matrix reveals its determinant: | \[ - \mathbf{1},\mathbf{2}, - \mathbf{1}\text{matrix}\;\det A = \left( \mathbf{2}\right) \left( \frac{\mathbf{3}}{\mathbf{2}}\right) \left( \frac{\mathbf{4}}{\mathbf{3}}\right) \cdots \left( \frac{\mathbf{n} + \mathbf{1}}{\mathbf{n}}\right) = \mathbf{n} + \mathbf{1}\text{.} \] | Yes |
Example 3 (Determinant of \( U \) ) When \( U \) is upper triangular, only one of the \( n \) ! products can be nonzero. This one term comes from the diagonal: \( \det U = + {u}_{11}{u}_{22}\cdots {u}_{nn} \) . | All other column orderings pick at least one entry below the diagonal, where \( U \) has zeros. As soon as we pick a number like \( {u}_{21} = 0 \), that term in equation (8) is sure to be zero. | Yes |
Example 4 Suppose \( Z \) is the identity matrix except for column 3 . Then\n\n\[ \text{The determinant of}Z = \left| \begin{array}{llll} 1 & 0 & \mathbf{a} & 0 \\ 0 & 1 & \mathbf{b} & 0 \\ 0 & 0 & \mathbf{c} & 0 \\ 0 & 0 & \mathbf{d} & 1 \end{array}\right| \text{is}\mathbf{c}\text{.} \] | (9)\n\nThe term \( \left( 1\right) \left( 1\right) \left( c\right) \left( 1\right) \) comes from the main diagonal with a plus sign. There are \( 4! = {24} \) products (choosing one factor from each row and column) but the other 23 products are zero. Reason: If we pick \( a, b \), or \( d \) from column 3, that column ... | Yes |
Example 5 Suppose \( A \) has 1’s just above and below the main diagonal. Here \( n = 4 \) :\n\n\[ A = \left\lbrack \begin{array}{llll} 0 & \mathbf{1} & 0 & 0 \\ \mathbf{1} & 0 & \mathbf{1} & 0 \\ 0 & \mathbf{1} & 0 & \mathbf{1} \\ 0 & 0 & \mathbf{1} & 0 \end{array}\right\rbrack \;\text{ and }\;P = \left\lbrack \begin{... | The only nonzero choice in the first row is column 2 . The only nonzero choice in row 4 is column 3. Then rows 2 and 3 \( {must} \) choose columns 1 and 4. In other words \( \det P = \det A. \) The determinant of \( P \) is +1 (two exchanges to reach \( 2,1,4,3 \) ). Therefore \( \det A = + 1 \) . | Yes |
The \( - 1,2, - 1 \) matrix has only two nonzeros in its first row. So only two cofactors \( {C}_{11} \) and \( {C}_{12} \) are involved in the determinant. | \[ \left| \begin{array}{rrrr} 2 & - 1 & & \\ - 1 & 2 & - 1 & \\ & - 1 & 2 & - 1 \\ & & - 1 & 2 \end{array}\right| = 2\left| \begin{array}{rrr} 2 & - 1 & \\ - 1 & 2 & - 1 \\ & - 1 & 2 \end{array}\right| - \left( {-1}\right) \left| \begin{array}{rrr} - 1 & - 1 & \\ & 2 & - 1 \\ & - 1 & 2 \end{array}\right| . \] You see 2... | Yes |
Example 7 This is the same matrix, except the first entry (upper left) is now 1 :\n\n\[ \n{B}_{4} = \left\lbrack \begin{array}{rrrr} 1 & - 1 & & \\ - 1 & 2 & - 1 & \\ & - 1 & 2 & - 1 \\ & & - 1 & 2 \end{array}\right\rbrack \n\]\n\nAll pivots of this matrix turn out to be 1 . So its determinant is 1 . How does that come... | \[ \n\det {B}_{4} = {D}_{3} - {D}_{2}\;\text{ instead of }\;\det {A}_{4} = 2{D}_{3} - {D}_{2}. \n\]\n\nThe determinant of \( {B}_{4} \) is \( 4 - 3 = 1 \) . The determinant of every \( {B}_{n} \) is \( n - \left( {n - 1}\right) = 1 \) . | Yes |
Solving \( 3{x}_{1} + 4{x}_{2} = 2 \) and \( 5{x}_{1} + 6{x}_{2} = 4 \) needs three determinants: | \[ \det A = \left| \begin{array}{ll} 3 & 4 \\ 5 & 6 \end{array}\right| \;\det {B}_{1} = \left| \begin{array}{ll} 2 & 4 \\ 4 & 6 \end{array}\right| \;\det {B}_{2} = \left| \begin{array}{ll} 3 & 2 \\ 5 & 4 \end{array}\right| \] Those determinants of \( A,{B}_{1},{B}_{2} \) are -2 and -4 and 2 . All ratios divide by \( \d... | Yes |
For \( n = 2 \), find the columns of \( {A}^{-1} = \left\lbrack \begin{array}{ll} \mathbf{x} & \mathbf{y} \end{array}\right\rbrack \) by solving \( A{A}^{-1} = I \) : | \[ \begin{array}{l} \text{ Columns of }{A}^{-1}\left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right\rbrack = \left\lbrack \begin{array}{l} \mathbf{1} \\ \mathbf{0} \end{array}\right\rbrack \;\left\lbrack \begin{array}{ll} a & b \\ c & ... | Yes |
Property 2 The cross product \( u \times v \) is perpendicular to \( u \) (and also to \( v \) ). | The direct proof is to watch terms cancel, producing a zero dot product:\n\n\[ \mathbf{u} \cdot \left( {\mathbf{u} \times \mathbf{v}}\right) = {u}_{1}\left( {{u}_{2}{v}_{3} - {u}_{3}{v}_{2}}\right) + {u}_{2}\left( {{u}_{3}{v}_{1} - {u}_{1}{v}_{3}}\right) + {u}_{3}\left( {{u}_{1}{v}_{2} - {u}_{2}{v}_{1}}\right) = 0. \] | Yes |
Example 7 \( \;\mathbf{u} = \left( {3,2,0}\right) \) and \( \mathbf{v} = \left( {1,4,0}\right) \) are in the \( {xy} \) plane, \( \mathbf{u} \times \mathbf{v} \) goes up the \( z \) axis: | \[ \mathbf{u} \times \mathbf{v} = \left| \begin{array}{lll} i & j & k \\ 3 & 2 & 0 \\ 1 & 4 & 0 \end{array}\right| = {10k}\text{. The cross product is}\mathbf{u} \times \mathbf{v} = \left( {\mathbf{0},\mathbf{0},\mathbf{{10}}}\right) \text{.} \] | Yes |
Example 8 The cross product of \( \mathbf{u} = \left( {1,1,1}\right) \) and \( \mathbf{v} = \left( {1,1,2}\right) \) is \( \left( {1, - 1,0}\right) \) | \[ \left| \begin{array}{lll} i & j & k \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array}\right| = i\left| \begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right| - j\left| \begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right| + k\left| \begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right| = i - j. \] This vector \( \left( {1, - 1,0}\right... | Yes |
The reflection matrix \( R = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right\rbrack \) has eigenvalues 1 and -1. | The eigenvector \( \left( {1,1}\right) \) is unchanged by \( R \) . The second eigenvector is \( \left( {1, - 1}\right) \) -its signs are reversed by \( R \) . A matrix with no negative entries can still have a negative eigenvalue! The eigenvectors for \( R \) are the same as for \( P \), because reflection \( = 2 \) (... | Yes |
Example 4 \( A = \left\lbrack \begin{array}{ll} \mathbf{1} & \mathbf{2} \\ \mathbf{2} & \mathbf{4} \end{array}\right\rbrack \) is already singular (zero determinant). Find its \( \lambda \) ’s and \( x \) ’s. | When \( A \) is singular, \( \lambda = 0 \) is one of the eigenvalues. The equation \( {Ax} = {0x} \) has solutions. They are the eigenvectors for \( \lambda = 0 \) . But \( \det \left( {A - {\lambda I}}\right) = 0 \) is the way to find all \( \lambda \) ’s and \( \mathbf{x} \) ’s. Always subtract \( {\lambda I} \) fro... | No |
Example 5 The \( {90}^{ \circ } \) rotation \( Q = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack \) has no real eigenvectors. Its eigenvalues are \( {\lambda }_{1} = i \) and \( {\lambda }_{2} = - i \) . Then \( {\lambda }_{1} + {\lambda }_{2} = \) trace \( = 0 \) and \( {\lambda }_{1}{\lambd... | After a rotation, no real vector \( {Qx} \) stays in the same direction as \( \mathbf{x}(\mathbf{x} = \mathbf{0} \) is useless). There cannot be an eigenvector, unless we go to imaginary numbers. Which we do.\n\nTo see how \( i = \sqrt{-1} \) can help, look at \( {Q}^{2} \) which is \( - I \) . If \( Q \) is rotation t... | Yes |
Every eigenvalue \( \lambda \) of this \( A \) falls into one or both of the Gershgorin circles: The centers are \( a \) and \( d \), the radii are \( {R}_{1} = \left| b\right| \) and \( {R}_{2} = \left| c\right| \) . | \[ A = \left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \;\begin{array}{ll} \text{ First circle: } & \left| {\lambda - a}\right| \leq \left| b\right| \\ \text{ Second circle: } & \left| {\lambda - d}\right| \leq \left| c\right| \end{array} \] | Yes |
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