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All eigenvalues of this \( A \) lie in a circle of radius \( R = 3 \) around one or more of the diagonal entries \( {d}_{1},{d}_{2},{d}_{3} \) : | \[ A = \left\lbrack \begin{matrix} {d}_{1} & 1 & 2 \\ 2 & {d}_{2} & 1 \\ - 1 & 2 & {d}_{3} \end{matrix}\right\rbrack \;\begin{array}{l} \left| {\lambda - {d}_{1}}\right| \leq 1 + 2 = {R}_{1} \\ \left| {\lambda - {d}_{2}}\right| \leq 2 + 1 = {R}_{2} \\ \left| {\lambda - {d}_{3}}\right| \leq 1 + 2 = {R}_{3} \end{array} \... | No |
Why is \( {AX} = {X\Lambda } \) ? | A multiplies its eigenvectors, which are the columns of \( X \) . The first column of \( {AX} \) is \( A{\mathbf{x}}_{1} \) . That is \( {\lambda }_{1}{\mathbf{x}}_{1} \) . Each column of \( X \) is multiplied by its eigenvalue :\n\n\[ A\text{ times }X\;{AX} = A\left\lbrack \begin{array}{lll} {x}_{1} & \cdots & {x}_{n}... | Yes |
Compute \( {A}^{k}{\mathbf{u}}_{0} \) for this faster Fibonacci: | \[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 1 & 0 \end{array}\right\rbrack \;\text{ has }\;{\lambda }_{1} = \mathbf{2}\;\text{ and }\;{\mathbf{x}}_{1} = \left\lbrack \begin{array}{l} 2 \\ 1 \end{array}\right\rbrack ,\;{\lambda }_{2} = - \mathbf{1}\;\text{ and }\;{\mathbf{x}}_{2} = \left\lbrack \begin{array}{r} 1 \\ ... | Yes |
Solve \( \frac{du}{dt} = A\mathbf{u} = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right\rbrack \mathbf{u}\; \) starting from \( \;\mathbf{u}\left( 0\right) = \left\lbrack \begin{array}{l} \mathbf{4} \\ \mathbf{2} \end{array}\right\rbrack \) . | This is a vector equation for \( \mathbf{u} \) . It contains two scalar equations for the components \( y \) and \( z \) . They are \ | No |
Example 2 Solve \( {du}/{dt} = {Au} \) knowing the eigenvalues \( \lambda = 1,2,3 \) of \( A \) : \( \begin{array}{l} \text{ Typical example } \\ \text{ Equation for }u \\ \text{ Initial condition }u\left( 0\right) \end{array}\;\frac{du}{dt} = \left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{arr... | The eigenvectors are \( {\mathbf{x}}_{1} = \left( {1,0,0}\right) \) and \( {\mathbf{x}}_{2} = \left( {1,1,0}\right) \) and \( {\mathbf{x}}_{3} = \left( {1,1,1}\right) \) . Step 1 The vector \( \mathbf{u}\left( 0\right) = \left( {9,7,4}\right) \) is \( \mathbf{2}{\mathbf{x}}_{1} + \mathbf{3}{\mathbf{x}}_{2} + \mathbf{4}... | Yes |
Example 3 Motion around a circle with \( {y}^{\prime \prime } + y = 0 \) and \( y = \cos t \) | This is our master equation with mass \( m = 1 \) and stiffness \( k = 1 \) and \( d = 0 \) : no damping. Substitute \( y = {e}^{\lambda t} \) into \( {y}^{\prime \prime } + y = 0 \) to reach \( {\lambda }^{2} + 1 = 0 \) . The roots are \( \lambda = i \) and \( \lambda = - i \) . Then half of \( {e}^{it} + {e}^{-{it}} ... | Yes |
When you substitute \( y = {e}^{\lambda t} \) into \( {y}^{\prime \prime } - 2{y}^{\prime } + y = 0 \), you get an equation with repeated roots: \( {\lambda }^{2} - {2\lambda } + 1 = 0 \) is \( {\left( \lambda - 1\right) }^{2} = 0 \) with \( \lambda = 1,1 \) . A differential equations course would propose \( {e}^{t} \)... | Linear algebra reduces \( {y}^{\prime \prime } - 2{y}^{\prime } + y = 0 \) to a vector equation for \( \mathbf{u} = \left( {y,{y}^{\prime }}\right) \) :\n\n\[ \frac{d}{dt}\left\lbrack \begin{matrix} y \\ {y}^{\prime } \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} {y}^{\prime } \\ 2{y}^{\prime } - y \end{matri... | Yes |
Use the infinite series to find \( {e}^{At} \) for \( A = \left\lbrack \begin{array}{rr} 0 & 1 \\ - 1 & 0 \end{array}\right\rbrack \) . Notice that \( {A}^{4} = I \) : | \[ {e}^{At} = I + {At} + \frac{1}{2}{\left( At\right) }^{2} + \frac{1}{6}{\left( At\right) }^{3} + \cdots = \left\lbrack \begin{matrix} 1 - \frac{1}{2}{t}^{2} + \cdots & t - \frac{1}{6}{t}^{3} + \cdots \\ - t + \frac{1}{6}{t}^{3} - \cdots & 1 - \frac{1}{2}{t}^{2} + \cdots \end{matrix}\right\rbrack . \] | Yes |
Example 6 Solve \( \frac{du}{dt} = {Au} = \left\lbrack \begin{array}{ll} 1 & 1 \\ 0 & 2 \end{array}\right\rbrack u \) starting from \( u\left( 0\right) = \left\lbrack \begin{array}{l} 2 \\ 1 \end{array}\right\rbrack \) at \( t = 0 \) . | Solution The eigenvalues 1 and 2 are on the diagonal of \( A \) (since \( A \) is triangular). The eigenvectors are \( \left( {1,0}\right) \) and \( \left( {1,1}\right) \) . The starting \( \mathbf{u}\left( 0\right) \) is \( {\mathbf{x}}_{1} + {\mathbf{x}}_{2} \) so \( {c}_{1} = {c}_{2} = 1 \) . Then \( \mathbf{u}\left... | Yes |
The eigenvectors of a 2 by 2 symmetric matrix have a special form: | Not widely known \( S = \left\lbrack \begin{array}{ll} a & b \\ b & c \end{array}\right\rbrack \) has \( {\mathbf{x}}_{1} = \left\lbrack \begin{matrix} b \\ {\lambda }_{1} - a \end{matrix}\right\rbrack \) and \( {\mathbf{x}}_{2} = \left\lbrack \begin{matrix} {\lambda }_{2} - c \\ b \end{matrix}\right\rbrack \) .\n\nThi... | No |
Example 4 This symmetric matrix has one positive eigenvalue and one positive pivot:\n\n\\[ \n\\text{Matching signs}\\;S = \\left\\lbrack \\begin{array}{ll} 1 & 3 \\\\ 3 & 1 \\end{array}\\right\\rbrack \\;\\begin{array}{l} \\text{ has pivots }1\\text{ and } - 8 \\\\ \\text{ eigenvalues }4\\text{ and } - 2. \\end{array}\... | Here is a proof that the pivots and eigenvalues have matching signs, when \\( S = {S}^{\\mathrm{T}} \\) .\n\nYou see it best when the pivots are divided out of the rows of \\( U \\) . Then \\( S \\) is \\( {LD}{L}^{\\mathrm{T}} \\) . The diagonal pivot matrix \\( D \\) goes between triangular matrices \\( L \\) and \\(... | Yes |
Test these symmetric matrices \( S \) and \( T \) for positive definiteness: | Solution The pivots of \( S \) are 2 and \( \frac{3}{2} \) and \( \frac{4}{3} \), all positive. Its upper left determinants are 2 and 3 and 4, all positive. The eigenvalues of \( S \) are \( 2 - \sqrt{2} \) and 2 and \( 2 + \sqrt{2} \), all positive. That completes tests \( \mathbf{1},\mathbf{2} \), and \( \mathbf{3} \... | Yes |
Find the axes of this tilted ellipse \( 5{x}^{2} + {8xy} + 5{y}^{2} = 1 \) . | Solution Start with the positive definite matrix that matches this equation:\n\n\[ \text{The equation is}\left\lbrack \begin{array}{ll} x & y \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 5 & 4 \\ 4 & 5 \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack = 1\text{. The matri... | Yes |
Example 1 Don’t send \( A = \left\lbrack \begin{array}{llllll} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{array}\right\rbrack \; \) Send this \( A = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{... | 36 numbers become 12 numbers. With 300 by 300 pixels, 90,000 numbers become 600 . And if we define the all-ones vector \( \mathbf{x} \) in advance, we only have to send one number. That number would be the constant grayscale \( g \) that multiplies \( \mathbf{x}{\mathbf{x}}^{\mathrm{T}} \) to produce the matrix. | No |
Suppose the flag has two triangles of different colors. The lower left triangle has 1's and the upper right triangle has 0's. The main diagonal is included with the 1's. Here is the image matrix when \( n = 4 \) . It has full rank \( r = 4 \) so it is invertible : | Triangular\n\[ \nA = \left\lbrack \begin{array}{llll} \mathbf{1} & 0 & 0 & 0 \\ \mathbf{1} & \mathbf{1} & 0 & 0 \\ \mathbf{1} & \mathbf{1} & \mathbf{1} & 0 \\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \end{array}\right\rbrack \;\text{ and }\;{A}^{-1} = \left\lbrack \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 1 & 1 &... | Yes |
When is \( A = {U\sum }{V}^{\mathrm{T}} \) (singular values) the same as \( {X\Lambda }{X}^{-1} \) (eigenvalues)? | Solution \( A \) needs orthonormal eigenvectors to allow \( X = U = V \) . \( A \) also needs eigenvalues \( \lambda \geq 0 \) if \( \Lambda = \sum \) . So \( A \) must be a positive semidefinite (or definite) symmetric matrix. Only then will \( A = {X\Lambda }{X}^{-1} \) which is also \( {Q\Lambda }{Q}^{\mathrm{T}} \)... | Yes |
Example 2 If \( A = \mathbf{x}{\mathbf{y}}^{\mathrm{T}} \) (rank 1) with unit vectors \( \mathbf{x} \) and \( \mathbf{y} \), what is the SVD of \( A \) ? | Solution The reduced SVD in (2) is exactly \( \mathbf{x}{\mathbf{y}}^{\mathrm{T}} \), with rank \( r = 1 \) . It has \( {\mathbf{u}}_{1} = \mathbf{x} \) and \( {\mathbf{v}}_{1} = \mathbf{y} \) and \( {\sigma }_{1} = 1 \) . For the full SVD, complete \( {\mathbf{u}}_{1} = \mathbf{x} \) to an orthonormal basis of \( \mat... | Yes |
The SVD of \( A \) (centered data) shows the dominant direction in the scatter plot. | The second singular vector \( {\mathbf{u}}_{2} \) is perpendicular to \( {\mathbf{u}}_{1} \) . The second singular value \( {\sigma }_{2} \approx \sqrt{3} \) measures the spread across the dominant line. If the data points in \( A \) fell exactly on a line ( \( {\mathbf{u}}_{1} \) direction), then \( {\sigma }_{2} \) w... | No |
How do we know that \( \left| {\lambda }_{1}\right| \leq {\sigma }_{1} \) ? | The eigenvector for \( A\mathbf{x} = {\lambda }_{1}\mathbf{x} \) will give the ratio \( \parallel A\mathbf{x}\parallel /\parallel \mathbf{x}\parallel = \begin{Vmatrix}{{\lambda }_{1}\mathbf{x}}\end{Vmatrix}/\parallel \mathbf{x}\parallel \) which is \( \left| {\lambda }_{1}\right| \) . The maximum ratio \( {\sigma }_{1}... | Yes |
Example 2 The SVD example in Section 7.2 was \( A = \left\lbrack \begin{array}{ll} 3 & 0 \\ 4 & 5 \end{array}\right\rbrack = {U\sum }{V}^{\mathrm{T}} \) . Find the factors \( Q \) and \( S \) (rotation and stretch) in the polar decomposition \( A = {QS} \) . | Solution I will just copy the matrices \( U \) and \( \sum \) and \( V \) from Section 7.2:\n\n\[ Q = U{V}^{\mathrm{T}} = \frac{1}{\sqrt{20}}\left\lbrack \begin{array}{rr} 1 & - 3 \\ 3 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{rr} 1 & - 1 \\ - 1 & 1 \end{array}\right\rbrack = \frac{1}{\sqrt{20}}\left\lbra... | Yes |
Find the pseudoinverse of \( A = \left\lbrack \begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right\rbrack \) . | This matrix is not invertible. The rank is 1 . The only singular value is \( {\sigma }_{1} = 2 \) . That is inverted to \( 1/2 \) in \( {\sum }^{ + } \) (also rank 1).\n\n\[ \n{A}^{ + } = V{\sum }^{ + }{U}^{\mathrm{T}} = \frac{1}{\sqrt{2}}\left\lbrack \begin{array}{rr} 1 & 1 \\ 1 & - 1 \end{array}\right\rbrack \left\lb... | Yes |
Choose a fixed vector \( \mathbf{a} = \left( {1,3,4}\right) \), and let \( T\left( \mathbf{v}\right) \) be the dot product \( \mathbf{a} \cdot \mathbf{v} \): The input is \( \;\mathbf{v} = \left( {{v}_{1},{v}_{2},{v}_{3}}\right) \). | The output is \( T\left( \mathbf{v}\right) = \mathbf{a} \cdot \mathbf{v} = {v}_{1} + 3{v}_{2} + 4{v}_{3} \). Dot products are linear. The inputs \( v \) come from three-dimensional space, so \( \mathbf{V} = {\mathbf{R}}^{3} \). The outputs are just numbers, so the output space is \( \mathbf{W} = {\mathbf{R}}^{1} \). We... | Yes |
Example 2 The length \( T\left( \mathbf{v}\right) = \parallel \mathbf{v}\parallel \) is not linear. Requirement (a) for linearity would be \( \parallel \mathbf{v} + \mathbf{w}\parallel = \parallel \mathbf{v}\parallel + \parallel \mathbf{w}\parallel \) . Requirement (b) would be \( \parallel c\mathbf{v}\parallel = c\par... | Not (a): The sides of a triangle satisfy an inequality \( \parallel \mathbf{v} + \mathbf{w}\parallel \leq \parallel \mathbf{v}\parallel + \parallel \mathbf{w}\parallel \) .\n\nNot (b): The length \( \parallel - \mathbf{v}\parallel \) is \( \parallel \mathbf{v}\parallel \) and not \( - \parallel \mathbf{v}\parallel \) .... | Yes |
Is rotation linear? | Yes it is. We can rotate two vectors and add the results. The sum of rotations \( T\left( \mathbf{v}\right) + T\left( \mathbf{w}\right) \) is the same as the rotation \( T\left( {\mathbf{v} + \mathbf{w}}\right) \) of the sum. The whole plane is turning together, in this linear transformation. | Yes |
How do you find the derivative of \( u = 6 - 4x + 3x^2 \) ? | You start with the derivatives of \( 1, x \), and \( x^2 \). Those are the basis vectors. Their derivatives are 0,1, and \( 2x \). Then you use linearity for the derivative of any combination:\n\n\[ \frac{du}{dx} = 6 \text{ (derivative of } 1) - 4 \text{ (derivative of } x) + 3 \text{ (derivative of } x^2) = - 4 + 6x. ... | Yes |
Integration \( {T}^{ + } \) is also linear : \( {\int }_{0}^{x}\left( {D + {Ex}}\right) {dx} = {Dx} + \frac{1}{2}E{x}^{2} \) . | The input basis is now \( 1, x \) . The output basis is \( 1, x,{x}^{2} \) . The matrix \( {A}^{ + } \) for \( {T}^{ + } \) is 3 by 2 : \n\n\( \begin{array}{l} \text{ Input }v\;\text{ Multiplication }{A}^{ + }v = \left\lbrack \begin{array}{ll} 0 & 0 \\ 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right\rbrack \left\lbrack \beg... | Yes |
Example 6 Project every 3-dimensional vector onto the horizontal plane \( z = 1 \) . The vector \( \mathbf{v} = \left( {x, y, z}\right) \) is transformed to \( T\left( \mathbf{v}\right) = \left( {x, y,1}\right) \) . This transformation is not linear. Why not? | It doesn’t even transform \( \mathbf{v} = \mathbf{0} \) into \( T\left( \mathbf{v}\right) = \mathbf{0} \) . | Yes |
Suppose the input space \( \mathbf{V} = {\mathbf{R}}^{2} \) is also the output space \( \mathbf{W} = {\mathbf{R}}^{2} \). Suppose that \( T\left( \mathbf{v}\right) = \mathbf{v} \) is the identity transformation. You might expect its matrix to be \( I \), but that only happens when the input basis is the same as the out... | For this special case \( T\left( \mathbf{v}\right) = \mathbf{v} \), I will call the matrix \( B \) instead of \( A \). We are just changing basis from the \( \mathbf{v} \) ’s to the \( \mathbf{w} \) ’s. Each \( \mathbf{v} \) is a combination of \( {\mathbf{w}}_{1} \) and \( {\mathbf{w}}_{2} \). \n\n\[ \text{Input}\left... | Yes |
Example 5 \( S \) rotates the plane by \( \theta \) and \( T \) also rotates by \( \theta \) . Then \( {TS} \) rotates by \( {2\theta } \) . This transformation \( {T}^{2} \) corresponds to the rotation matrix \( {A}^{2} \) through \( {2\theta } \) : | By matching (transformation) \( {}^{2} \) with (matrix) \( {}^{2} \), we pick up the formulas for \( \cos {2\theta } \) and \( \sin {2\theta } \) . Multiply \( A \) times \( A \) :\n\n\[ \left\lbrack \begin{array}{rr} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array}\right\rbrack \left\lbrack \begin{... | Yes |
Example 6 \( S \) rotates by the angle \( \theta \) and \( T \) rotates by \( - \theta \) . Then \( {TS} = I \) leads to \( {AB} = I \) . | In this case \( T\left( {S\left( \mathbf{u}\right) }\right) \) is \( \mathbf{u} \) . We rotate forward and back. For the matrices to match, \( {ABx} \) must be \( \mathbf{x} \) . The two matrices are inverses. Check this by putting \( \cos \left( {-\theta }\right) = \cos \theta \) and \( \sin \left( {-\theta }\right) =... | Yes |
The projection matrix \( T \) projects every \( \mathbf{v} = \left( {x, y}\right) \) in \( {\mathbf{R}}^{2} \) onto the line \( y = - x \) . Using the standard basis, \( {\mathbf{v}}_{1} = \left( {1,0}\right) \) projects to \( T\left( {\mathbf{v}}_{1}\right) = \left( {\frac{1}{2}, - \frac{1}{2}}\right) \) . For \( {\ma... | Projection matrix \nStandard bases \nNot diagonal \n\( A = \left\lbrack \begin{array}{rr} \frac{1}{2} & - \frac{1}{2} \\ - \frac{1}{2} & \frac{1}{2} \end{array}\right\rbrack \) has \( {A}^{\mathrm{T}} = A \) and \( {A}^{2} = A. \) | Yes |
To construct the matrix \( A \) for the transformation \( T = \frac{d}{dx} \), we chose the input basis \( 1, x,{x}^{2},{x}^{3} \) and the output basis \( 1, x,{x}^{2} \) . The matrix \( A \) was simple but unfortunately it wasn't diagonal. But we can take each basis in the opposite order. | Now the input basis is \( {x}^{3},{x}^{2}, x,1 \) and the output basis is \( {x}^{2}, x,1 \) . The change of basis matrices \( {B}_{\text{in }} \) and \( {B}_{\text{out }} \) are permutations. The matrix for \( T\left( \mathbf{u}\right) = d\mathbf{u}/{dx} \) with the new bases is the diagonal singular value matrix \( {... | Yes |
This will be true for every matrix \( C = {BJ}{B}^{-1} \) that is similar to \( J \). | \[ \left( {{BJ}{B}^{-1} - {3I}}\right) {\mathbf{b}}_{4} = {BJ}{\mathbf{x}}_{4} - {3B}{\mathbf{x}}_{4} = B\left( {J - {3I}}\right) {\mathbf{x}}_{4} = B{\mathbf{x}}_{3} = {\mathbf{b}}_{3}. \] | Yes |
The same ideas work for a Fourier matrix \( F \) and a circulant matrix \( C \) of any size. Two by two matrices look trivial but they are very useful. Now eigenvalues of \( P \) have \( {\lambda }^{2} = 1 \) instead of \( {\lambda }^{4} = 1 \) and the complex number \( i \) is not needed: \( \lambda = \pm 1 \) . | Fourier matrix \( F \) from eigenvectors of \( P \) and \( C \)\n\( F = \left\lbrack \begin{array}{rr} \mathbf{1} & \mathbf{1} \\ \mathbf{1} & - \mathbf{1} \end{array}\right\rbrack \)\n\( P = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right\rbrack \)\nCirculant \( c_{0}I + c_{1}P \)\n\( C = \left\lbrack ... | Yes |
The double angle formula in trigonometry is \( \cos {2x} = 2{\cos }^{2}x - 1 \). | This tells us that \( {\cos }^{2}x = \frac{1}{2} + \frac{1}{2}\cos {2x} \). A very short Fourier series. So is \( {\sin }^{2}x = \frac{1}{2} - \frac{1}{2}\cos {2x} \). | Yes |
Find \( r \) and \( \theta \) for \( z = 1 + i \) and also for the conjugate \( \bar{z} = 1 - i \) . | Solution The absolute value is the same for \( z \) and \( \bar{z} \) . It is \( r = \sqrt{1 + 1} = \sqrt{2} \) :\n\n\[ \n{\left| z\right| }^{2} = {1}^{2} + {1}^{2} = 2\;\text{ and also }\;{\left| \bar{z}\right| }^{2} = {1}^{2} + {\left( -1\right) }^{2} = 2.\n\]\n\nThe distance from the center is \( r = \sqrt{2} \) . W... | Yes |
The inner product of \( \mathbf{u} = \left\lbrack \begin{array}{l} 1 \\ i \end{array}\right\rbrack \) with \( \mathbf{v} = \left\lbrack \begin{array}{l} i \\ 1 \end{array}\right\rbrack \) is \( \left\lbrack \begin{array}{ll} 1 & - i \end{array}\right\rbrack \left\lbrack \begin{array}{l} i \\ 1 \end{array}\right\rbrack ... | Example 1 is surprising. Those vectors \( \left( {1, i}\right) \) and \( \left( {i,1}\right) \) don’t look perpendicular. But they are. A zero inner product still means that the (complex) vectors are orthogonal. Similarly the vector \( \left( {1, i}\right) \) is orthogonal to the vector \( \left( {1, - i}\right) \) . T... | No |
If \( S = {S}^{\mathrm{H}} \) and \( z \) is any real or complex column vector, the number \( {z}^{\mathrm{H}}{Sz} \) is real. | Quick proof: \( {\mathbf{z}}^{\mathrm{H}}S\mathbf{z} \) is certainly 1 by 1 . Take its conjugate transpose:\n\n\[ \n{\left( {\mathbf{z}}^{\mathrm{H}}S\mathbf{z}\right) }^{\mathrm{H}} = {\mathbf{z}}^{\mathrm{H}}{S}^{\mathrm{H}}{\left( {\mathbf{z}}^{\mathrm{H}}\right) }^{\mathrm{H}}\text{which is}{\mathbf{z}}^{\mathrm{H}... | Yes |
Is \( F \) unitary? | Yes. The squared length of every column is \( \frac{1}{3}\left( {1 + 1 + 1}\right) \) (unit vector). The first column is orthogonal to the second column because \( 1 + {e}^{{2\pi i}/3} + {e}^{{4\pi i}/3} = 0 \) . This is the sum of the three numbers marked in Figure 9.3.\n\nNotice the symmetry of the figure. If you rot... | Yes |
Suppose all \( {c}_{i} = c \) and \( {m}_{j} = m \) . Find the movements \( \mathbf{u} \) and tensions \( \mathbf{y} \) . | All springs are the same and all masses are the same. But all movements and elongations and tensions will not be the same. \( {K}^{-1} \) includes \( \frac{1}{c} \) because \( {A}^{\mathrm{T}}{CA} \) includes \( c \) :\n\n\[ \n\text{Movements}\;u = {K}^{-1}f = \frac{1}{4c}\left\lbrack \begin{array}{lll} 3 & 2 & 1 \\ 2 ... | Yes |
Example 2 If \( {c}_{1} = {c}_{2} = {c}_{3} = 1 \) and \( C = I \), this is the \( - 1,2, - 1 \) tridiagonal matrix \( {K}_{1} \) . The last entry of \( {K}_{1} \) is 1 instead of 2 because the spring at the bottom is free. Suppose all \( {m}_{j} = m \) : | \[ \text{Fixed-free}\;u = {K}_{1}^{-1}f = \frac{1}{c}\left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {mg} \\ {mg} \\ {mg} \end{array}\right\rbrack = \frac{mg}{c}\left\lbrack \begin{array}{l} 3 \\ 5 \\ 6 \end{array}\right\rbrack \text{.} \] Those... | Yes |
The fraction of rental cars in Denver starts at \( \frac{1}{50} = {.02} \) . The fraction outside Denver is .98. Every month, \( {80}\% \) of the Denver cars stay in Denver (and \( {20}\% \) leave). Also \( 5\% \) of the outside cars come in (95% stay outside). This means that the fractions \( {\mathbf{u}}_{0} = \left(... | First month \( \;A = \left\lbrack \begin{array}{ll} {.80} & {.05} \\ {.20} & {.95} \end{array}\right\rbrack \; \) leads to \( \;{\mathbf{u}}_{1} = A{\mathbf{u}}_{0} = A\left\lbrack \begin{array}{l} {.02} \\ {.98} \end{array}\right\rbrack = \left\lbrack \begin{array}{l} {.065} \\ {.935} \end{array}\right\rbrack \) . Not... | Yes |
Example 4 \( A = \left\lbrack \begin{array}{lll} {.2} & {.3} & {.4} \\ {.4} & {.4} & {.1} \\ {.5} & {.1} & {.3} \end{array}\right\rbrack \) has \( {\lambda }_{\max } = {.9} \) and \( {\left( I - A\right) }^{-1} = \frac{1}{93}\left\lbrack \begin{array}{lll} {41} & {25} & {27} \\ {33} & {36} & {24} \\ {34} & {23} & {36} ... | This economy is productive. \( A \) is small compared to \( I \), because \( {\lambda }_{\max } \) is .9 . To meet the demand \( \mathbf{y} \), start from \( \mathbf{p} = {\left( I - A\right) }^{-1}\mathbf{y} \) . Then \( A\mathbf{p} \) is consumed in production, leaving \( \mathbf{p} - A\mathbf{p} \) . This is \( \lef... | Yes |
Suppose the current corner is \( \mathbf{P} = \left( {4,0,0}\right) \), with the Ph.D. doing all the work (the cost is \( \$ {20} \) ). If the student works one hour, the cost of \( \mathbf{x} = \left( {3,1,0}\right) \) is down to \$18. The reduced cost is \( r = - 2 \) . If the machine works one hour, then \( \mathbf{... | Even in this small example, the first step may not go immediately to the best \( {\mathbf{x}}^{ * } \) . The method chooses the entering variable before it knows how much of that variable to include. We computed \( r \) when the entering variable changes from 0 to 1, but one unit may be too much or too little. The meth... | Yes |
Minimize the cost \( c \cdot x = 3{x}_{1} + {x}_{2} + 9{x}_{3} + {x}_{4} \). The constraints are \( x \geq 0 \) and two equations \( A\mathbf{x} = \mathbf{b} \): | \[ {x}_{1} + 2{x}_{3} + {x}_{4} = 4\;m = 2\text{ equations } \] \[ {x}_{2} + {x}_{3} - {x}_{4} = 2\;n = 4\;\text{ unknowns. } \] A starting corner is \( \mathbf{x} = \left( {4,2,0,0}\right) \) which costs \( \mathbf{c} \cdot \mathbf{x} = {14} \). It has \( m = 2 \) nonzeros and \( n - m = 2 \) zeros. The zeros are \( {... | Yes |
If \( \\mathbf{v} \) and \( \\mathbf{w} \) have finite length, how large can their dot product be? | The sum \( \\mathbf{v} \\cdot \\mathbf{w} = {v}_{1}{w}_{1} + {v}_{2}{w}_{2} + \\cdots \) also adds to a finite number. We can safely take dot products. The Schwarz inequality is still true:\n\n\[ \n\\text{Schwarz inequality}\\;\\left| {v \\cdot w}\\right| \\leq \\parallel v\\parallel \\parallel w\\parallel \\text{.} \n... | Yes |
The length of \( f\left( x\right) = \sin x \) comes from its inner product with itself: | \[ \left( {f, f}\right) = {\int }_{0}^{2\pi }{\left( \sin x\right) }^{2}{dx} = \pi \text{. The length of}\sin x\text{is}\sqrt{\pi }\text{.} \] | Yes |
If \( A \) is the identity matrix \( I \), the ratios are \( \parallel \mathbf{x}\parallel /\parallel \mathbf{x}\parallel \) . Therefore \( \parallel I\parallel = 1 \) . If \( A \) is an orthogonal matrix \( Q \) , lengths are again preserved: \( \parallel Q\mathbf{x}\parallel = \parallel \mathbf{x}\parallel \) . The r... | An orthogonal \( Q \) is good to compute with: errors don’t grow. | No |
The norm of a diagonal matrix is its largest entry (using absolute values): | \[ A = \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right\rbrack \text{ has norm }\parallel A\parallel = 3\text{. The eigenvector }\mathbf{x} = \left\lbrack \begin{array}{l} 0 \\ 1 \end{array}\right\rbrack \text{ has }A\mathbf{x} = 3\mathbf{x}. \] | Yes |
Example 3 When \( A \) is symmetric, \( c = \parallel A\parallel \begin{Vmatrix}{A}^{-1}\end{Vmatrix} \) comes from the eigenvalues: | \[ A = \left\lbrack \begin{array}{ll} 6 & 0 \\ 0 & 2 \end{array}\right\rbrack \text{ has norm }6.\;{A}^{-1} = \left\lbrack \begin{matrix} \frac{1}{6} & 0 \\ 0 & \frac{1}{2} \end{matrix}\right\rbrack \text{ has norm }\frac{1}{2}. \] This \( A \) is symmetric positive definite. Its norm is \( {\lambda }_{\max } = 6 \) . ... | Yes |
Example 4 Keep the same \( A \), with eigenvalues 6 and 2 . To make \( x \) small, choose \( b \) along the first eigenvector \( \left( {1,0}\right) \) . To make \( \Delta \mathbf{x} \) large, choose \( \Delta \mathbf{b} \) along the second eigenvector \( \left( {0,1}\right) \) . Then \( \mathbf{x} = \frac{1}{6}\mathbf... | This shows that the worst error allowed by the condition number \( \parallel A\parallel \begin{Vmatrix}{A}^{-1}\end{Vmatrix} \) can actually happen. Here is a useful rule of thumb, experimentally verified for Gaussian elimination: The computer can lose \( \log c \) decimal places to roundoff error. | Yes |
Find the variance \( {\sigma }^{2} \) of the ages of college freshmen. | Solution The probabilities of ages \( {x}_{i} = {17},{18},{19} \) were \( {p}_{i} = {0.2} \) and 0.5 and 0.3 . The expected value was \( m = \sum {\mathbf{p}}_{\mathbf{i}}{\mathbf{x}}_{\mathbf{i}} = \mathbf{{18}.1} \) . The variance uses those same probabilities :\n\n\[ \n{\sigma }^{2} = \left( {0.2}\right) {\left( \ma... | Yes |
The key question is the average \( {A}_{N} = \left( {{x}_{1} + \cdots + {x}_{N}}\right) /N \) . The independent \( {x}_{i} \) are \( \pm 1 \) and we are dividing their sum by \( N \) . The expected mean of \( {A}_{N} \) is still zero. The law of large numbers says that this sample average approaches zero with probabili... | \[ \text{By linearity}{\mathbf{\sigma }}_{\mathbf{N}}^{\mathbf{2}} = \frac{{\sigma }^{2}}{{N}^{2}} + \frac{{\sigma }^{2}}{{N}^{2}} + \cdots + \frac{{\sigma }^{2}}{{N}^{2}} = N\frac{{\sigma }^{2}}{{N}^{2}} = \frac{1}{N}\text{since}{\sigma }^{2} = 1\text{.} \] | Yes |
Example 3 Change outputs from 1 or -1 to \( x = 1 \) or \( x = 0 \) . Keep \( {p}_{1} = {p}_{0} = \frac{1}{2} \) . | The new mean value \( m = \frac{1}{2} \) falls halfway between 0 and 1. The variance moves to \( {\sigma }^{2} = \frac{1}{4} \) :\n\n\[ \mathbf{m} = \frac{1}{2}\left( 1\right) + \frac{1}{2}\left( 0\right) = \frac{\mathbf{1}}{\mathbf{2}}\text{ and }{\mathbf{\sigma }}^{\mathbf{2}} = \frac{1}{2}{\left( 1 - \frac{1}{2}\rig... | Yes |
Flip two coins separately. With 1 for heads and 0 for tails, the results can be \( \left( {1,1}\right) \) or \( \left( {1,0}\right) \) or \( \left( {0,1}\right) \) or \( \left( {0,0}\right) \) . Those four outcomes all have probability \( {p}_{11} = {p}_{10} = \) \( {p}_{01} = {p}_{00} = \frac{1}{4} \) . | Independent experiments have Prob of \( \left( {i, j}\right) = \left( {\text{Prob of }i}\right) \left( {\text{Prob of }j}\right) \). | No |
Example 3 Suppose that \( y \) is just \( - x \) . A coin flip has outputs \( x = 0 \) or 1 . The same flip has outputs \( y = 0 \) or -1 . The mean \( {m}_{x} \) is \( \frac{1}{2} \) for a fair coin, and \( {m}_{y} \) is \( - \frac{1}{2} \) . The covariance is \( {\sigma }_{xy} = - {\sigma }_{x}{\sigma }_{y} \) . The ... | In this case the correlation matrix \( R \) has determinant zero (singular and only semidefinite):\n\n\[ \text{ Correlation matrix }R = \left\lbrack \begin{matrix} 1 & {\rho }_{xy} \\ {\rho }_{xy} & 1 \end{matrix}\right\rbrack \]\n\n\[ R = \left\lbrack \begin{array}{rr} 1 & - 1 \\ - 1 & 1 \end{array}\right\rbrack \text... | Yes |
Example 4 Suppose the random variables \( x, y, z \) are independent. What matrix is \( R \) ? | Answer \( R \) is the identity matrix. All three correlations \( {\rho }_{xx},{\rho }_{yy},{\rho }_{zz} \) are 1 by definition. All three cross-correlations \( {\rho }_{xy},{\rho }_{xz},{\rho }_{yz} \) are zero by independence. | Yes |
Suppose a doctor measures your heart rate \( x \) three times \( \left( {m = 3, n = 1}\right) \) :\n\n\[ \begin{array}{l} x = {b}_{1} \\ x = {b}_{2} \\ x = {b}_{3} \end{array}\;\text{ is }\;{Ax} = b\;\text{ with }\;A = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \;\text{ and }\;V = \left\lbrack \... | The variances could be \( {\sigma }_{1}^{2} \cdot \;/9 \) and \( {\sigma }_{2}^{2} = 1/4 \) and \( {\sigma }_{3}^{2} = \mathbf{1} \) . You are getting more nervous as measurements are taken: ; is less reliable than \( {b}_{2} \) and \( {b}_{1} \) . All three measurements contain some information, so they all go into th... | Yes |
Example 1 Multiply \( A \) times \( x \) using the three rows of \( A \) . Then use the two columns : | \[ \text{ By rows }\;\left\lbrack \begin{array}{ll} 2 & 3 \\ 2 & 4 \\ 3 & 7 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right\rbrack \; = \left\lbrack \begin{array}{l} 2{x}_{1} + 3{x}_{2} \\ 2{x}_{1} + 4{x}_{2} \\ 3{x}_{1} + 7{x}_{2} \end{array}\right\rbrack \; = \begin{matrix}... | Yes |
Example 2 \( b = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \) is not in \( \mathbf{C}\left( A\right) .{Ax} = \left\lbrack \begin{array}{l} 2{x}_{1} + 3{x}_{2} \\ 2{x}_{1} + 4{x}_{2} \\ 3{x}_{1} + 7{x}_{2} \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbr... | The first two equations force \( {x}_{1} = \frac{1}{2} \) and \( {x}_{2} = 0 \) . Then equation 3 fails : \( 3\left( \frac{1}{2}\right) + 7\left( 0\right) = \mathbf{1.5} \) (not 1). This means that \( \mathbf{b} = \left( {1,1,1}\right) \) is not in the column space-the plane of \( {\mathbf{a}}_{1} \) and \( {\mathbf{a}... | Yes |
What are the column spaces of \( {A}_{2} = \left\lbrack \begin{array}{rrr} 2 & 3 & 5 \\ 2 & 4 & 6 \\ 3 & 7 & {10} \end{array}\right\rbrack \) and \( {A}_{3} = \left\lbrack \begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 1 \end{array}\right\rbrack \) ? | Solution. The column space of \( {A}_{2} \) is the same plane as before. The new column \( \left( {5,6,{10}}\right) \) is the sum of column 1 + column 2. So \( {a}_{3} = \) column 3 is already in the plane and adds nothing new. By including this \ | No |
Example 4 If \( A = \left\lbrack \begin{array}{lll} 1 & 3 & 8 \\ 1 & 2 & 6 \\ 0 & 1 & 2 \end{array}\right\rbrack \) then \( C = \left\lbrack \begin{array}{ll} 1 & 3 \\ 1 & 2 \\ 0 & 1 \end{array}\right\rbrack \;\begin{array}{l} n = 3\text{ columns in }A \\ r = 2\text{ columns in }C \end{array} \) | Column 3 of \( A \) is \( 2 \) (column 1) +2 (column 2). Leave it out of the basis in \( C \) . | No |
Example 5 If \( A = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right\rbrack \) then \( C = A \) . \( \;\begin{array}{l} n = 3\text{ columns in }A \\ r = 3\text{ columns in }C \end{array} \) | This matrix \( A \) is invertible. Its column space is all of \( {\mathbf{R}}^{3} \) . Keep all 3 columns. | Yes |
If \( A = \left\lbrack \begin{array}{lll} 1 & 2 & 5 \\ 1 & 2 & 5 \\ 1 & 2 & 5 \end{array}\right\rbrack \) then \( C = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \;\begin{array}{l} n = 3\text{ columns in }A \\ r = 1\text{ column in }C \end{array} \) | The number \( r \) is the “rank” of \( A. \) It is also the rank of \( C. \) It counts independent columns. Admittedly we could have moved from right to left in \( A \), starting with its last column. This would not change the final count \( r \) . Different basis, but always the same number of vectors. That number \( ... | No |
1 The column space \( \mathbf{C}\left( A\right) \) is the line through \( \mathbf{u} = \left\lbrack \begin{array}{l} 1 \\ 3 \end{array}\right\rbrack \) . Column 2 is on that line. | The column space \( \mathbf{C}\left( A\right) \) contains all combinations of the columns of \( A \) | No |
\[ {Bx} = \left\lbrack \begin{array}{lll} 1 & - 2 & - 2 \\ 3 & - 6 & - 6 \end{array}\right\rbrack \left\lbrack \begin{array}{l} a \\ b \\ c \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rbrack \text{ has solutions }{x}_{1} = \left\lbrack \begin{array}{l} 2 \\ 1 \\ 0 \end{array}\right... | In the textbook Introduction to Linear Algebra, those vectors \( {\mathbf{x}}_{1} \) and \( {\mathbf{x}}_{2} \) are called \ | No |
To project \( b = \\left( {3,3,3}\\right) \) on the \( {Q}_{1} \) line, multiply by \( {P}_{1} = {Q}_{1}{Q}_{1}^{\\mathrm{T}} \) . | \[ {P}_{1}b = \\frac{1}{9}\\left\\lbrack \\begin{array}{r} 2 \\\\ 2 \\\\ - 1 \\end{array}\\right\\rbrack \\;\\left\\lbrack \\begin{array}{lll} 2 & 2 & - 1 \\end{array}\\right\\rbrack \\;\\left\\lbrack \\begin{array}{l} 3 \\\\ 3 \\\\ 3 \\end{array}\\right\\rbrack = \\frac{1}{9}\\left\\lbrack \\begin{array}{r} 2 \\\\ 2 \... | Yes |
Example 1 \( S = \left\lbrack \begin{array}{ll} \mathbf{2} & \mathbf{1} \\ \mathbf{1} & \mathbf{2} \end{array}\right\rbrack \) has eigenvectors \( S\left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack = \mathbf{3}\left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack \) and \( S\left\lbrack \begin{array... | Then \( {\lambda }_{1} = 3 \) and \( {\lambda }_{2} = 1 \) . The powers \( {S}^{k} \) will grow like \( {3}^{k} \) . Those eigenvalues and eigenvectors have four properties to notice :\n\n(Trace of \( S \) ) The sum \( {\lambda }_{1} + {\lambda }_{2} = 3 + 1 \) equals the diagonal sum \( 2 + 2 = 4 \)\n\n(Determinant) T... | Yes |
Example 2 The rotation \( Q = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack \) has imaginary eigenvalues \( i \) and \( - i \) : | \( Q\left\lbrack \begin{array}{r} 1 \\ - i \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack \left\lbrack \begin{array}{r} 1 \\ - i \end{array}\right\rbrack = \left( i\right) \left\lbrack \begin{array}{r} 1 \\ - i \end{array}\right\rbrack \) and \( Q\left\lbrack \begin{... | Yes |
Example 3 Find the eigenvalues and eigenvectors of \( A = \left\lbrack \begin{array}{ll} 8 & 3 \\ 2 & 7 \end{array}\right\rbrack \) : not symmetric. | The determinant of \( A - {\lambda I} \) is \( \left| \begin{matrix} 8 - \lambda & 3 \\ 2 & 7 - \lambda \end{matrix}\right| = {\lambda }^{2} - {15\lambda } + {50} = \left( {\lambda - {10}}\right) \left( {\lambda - 5}\right) \). \n\n\( {\lambda }_{1} = {10} \) has \( \left\lbrack \begin{matrix} 8 - {10} & 3 \\ 2 & 7 - {... | Yes |
Example 4 If we divide Example 3 by 10, all eigenvalues are divided by 10. Then \( {\lambda }_{1} = 1 \) and \( {\lambda }_{2} = \frac{1}{2} \) . In this case \( A \) is a Markov matrix, with positive columns adding to 1 . | \[ A = \left\lbrack \begin{array}{ll} \mathbf{{0.8}} & \mathbf{{0.3}} \\ \mathbf{{0.2}} & \mathbf{{0.7}} \end{array}\right\rbrack \;\begin{array}{l} {A}^{k}v = {c}_{1}{\left( 1\right) }^{k}{x}_{1} + {c}_{2}{\left( \frac{1}{2}\right) }^{k}{x}_{2} \\ \text{ As }k\text{ increases,}{A}^{k}v\text{ approaches }{c}_{1}{x}_{1}... | Yes |
Example 1 Solve \( {Sx} = {\lambda Mx} \) when \( S = \left\lbrack \begin{array}{rr} 4 & - 2 \\ - 2 & 4 \end{array}\right\rbrack \) and \( M = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right\rbrack \) . | Solution Our eigenvalue problems are \( \left( {S - {\lambda M}}\right) \mathbf{x} = \mathbf{0} \) and \( \left( {H - {\lambda I}}\right) \mathbf{y} = \mathbf{0} \) . We will find the same \( \lambda \) ’s from both determinants : \( \det \left( {S - {\lambda M}}\right) = 0 \) and \( \det \left( {H - {\lambda I}}\right... | Yes |
Example 2 Suppose \( x = {b}_{1} \) and \( x = {b}_{2} \) are independent noisy measurements of the number \( x \) . We multiply those equations by their weights \( 1/\sigma \) .\n\nSolve \( {Ax} = \left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack x = \left\lbrack \begin{array}{l} {b}_{1} \\ {b}_{2} \end{ar... | The equations become \( x/{\sigma }_{1} = {b}_{1}/{\sigma }_{1} \) and \( x/{\sigma }_{2} = {b}_{2}/{\sigma }_{2} : \left\lbrack \begin{array}{l} 1/{\sigma }_{1} \\ 1/{\sigma }_{2} \end{array}\right\rbrack x = \left\lbrack \begin{array}{l} {b}_{1}/{\sigma }_{1} \\ {b}_{2}/{\sigma }_{2} \end{array}\right\rbrack \n\nWeig... | Yes |
Choose \( B \) as the \( r \) by \( r \) submatrix of \( A \) with the largest determinant. Then all entries of \( Y \) and \( Z \) have \( \left| {y}_{ij}\right| \leq 1 \) and \( \left| {z}_{ij}\right| \leq 1 \) . | Proof. Those \( {y}_{ij} \) and \( {z}_{ij} \) are the numbers in \( {C}_{m - r}{B}^{-1} \) and in \( {B}^{-1}{Z}_{n - r} \) . We start with the \( {z}_{ij} \) . Since \( B\left( {{B}^{-1}{Z}_{n - r}}\right) = {Z}_{n - r} \), we know that every column \( {z}_{j} \) of \( {B}^{-1}{Z}_{n - r} \) solves this system of lin... | Yes |
The matrix in Example 1 has this \( {B}_{\max } \) with maximum determinant \( = 2 \) . Then \( \left| {y}_{ij}\right| \leq 1 \) and \( \left| {z}_{ij}\right| \leq 1 \) : | \[ A = \left\lbrack \begin{array}{llll} \mathbf{1} & 2 & \mathbf{4} & 2 \\ \mathbf{0} & \mathbf{1} & \mathbf{2} & 1 \\ 1 & 3 & 6 & 3 \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} \mathbf{1} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} \\ 1 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} \mathbf{1} & ... | No |
What is the inverse of \( M = I - \left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right\rbrack \) ? In this case \( \mathbf{u} = \mathbf{v} = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \) . | Solution Here \( {\mathbf{v}}^{\mathrm{T}}\mathbf{u} = 3 \) and \( {M}^{-1} = I + \frac{\mathbf{u}{\mathbf{v}}^{\mathrm{T}}}{1 - 3} \) . So \( {M}^{-1} \) equals \( I - \frac{1}{2}\left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right\rbrack \) . | Yes |
If \( M = I - \left\lbrack \begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right\rbrack = I - U{V}^{\mathrm{T}} \) then \( {M}^{-1} = \left\lbrack \begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right\rbrack \) | That came from writing the first displayed matrix as \( U{V}^{\mathrm{T}} \) and reversing to \( {V}^{\mathrm{T}}U \) :\n\n\[ U{V}^{\mathrm{T}} = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right\rbrack \left\lbrack \begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right\rbrack \text{ and }{V... | Yes |
ADI solves \( {AX} - {XB} = C \) | \[ {X}_{j + 1/2}\left( {B - {p}_{j}I}\right) = C - \left( {A - {p}_{j}I}\right) {X}_{j} \] \n\nMatrices \( {X}_{j + 1/2} \) and \( {X}_{j + 1} \) \n\[ \left( {A - {q}_{j}I}\right) {X}_{j + 1}\; = C - {X}_{j + 1/2}\left( {B - {q}_{j}I}\right) \] | Yes |
Minimize \( f\left( \mathbf{x}\right) \) for \(\mathbf{x}\) in \( K \). | ADMM rewrites “ \( x \) in \( K \) ” as a minimization of \( g \) . It connects \( x \) to \( z \) by a constraint.\n\n\[ \text{Minimize}f\left( \mathbf{x}\right) + g\left( \mathbf{z}\right) \text{subject to}\mathbf{x} - \mathbf{z} = \mathbf{0}.\ \]\n\n\( g \) is the indicator function of the set \( K : g\left( z\right... | Yes |
An important \( {\ell }^{1} \) problem has an exact solution. \( f\left( \mathbf{x}\right) = \lambda \parallel \mathbf{x}{\parallel }_{1} = \lambda \left| {x}_{1}\right| + \cdots + \lambda \left| {x}_{n}\right| \) splits into \( n \) scalar functions \( \lambda {\left| x\right| }_{i}. \) Separation by ADMM leads to the... | \[ {f}_{i}\left( {x}_{i}\right) = \lambda \left| {x}_{i}\right| + \frac{1}{2}\rho {\left( {x}_{i} - {v}_{i}\right) }^{2}\text{ with }{v}_{i} = {z}_{i} - {u}_{i} \] The solution \( {x}_{i}^{ * } \) is the \ | No |
Nonnegative Matrix Factorization \( A \approx {CR} \) with \( {C}_{ij} \geq 0 \) and \( {R}_{ij} \geq 0 \) | ADMM begins with an alternating minimization-our favorite way to factor a matrix.\n\n<table><tr><td>Find \( C \geq 0 \)</td><td>Minimize \( \parallel A - {CR}{\left| \right| }_{F}^{2} \) with \( R \geq 0 \) fixed</td></tr><tr><td>Find \( R \geq 0 \)</td><td>Minimize \( \parallel A - {CR}{\parallel }_{F}^{2} \) with \( ... | Yes |
Example 4 LASSO aims for a sparse solution to \( {Ax} = b \) by including an \( {\ell }^{1} \) penalty : | \[ \text{LASSO Minimize}\frac{1}{2}\parallel {Ax} - b{\parallel }^{2} + \lambda \parallel x{\parallel }_{1} \] (25) Immediately that problem splits into \( f\left( \mathbf{x}\right) + g\left( \mathbf{z}\right) \) with the constraint \( \mathbf{x} - \mathbf{z} = \mathbf{0} \) . The subproblem for \( \mathbf{x} \) is lea... | Yes |
The transform of \( f = \left( {\mathbf{1},\mathbf{0},\ldots ,\mathbf{0}}\right) \) is \( c = \frac{\mathbf{1}}{N}\left( {\mathbf{1},\mathbf{1},\ldots ,\mathbf{1}}\right) . \) | That vector \( f \) with one spike is a discrete delta function. It is concentrated at one point. Its transform \( c \) spreads out over all frequencies. Multiplying \( {\Omega f} \) picks out the zeroth column of \( \Omega \) . Therefore \( c \) shows the same Fourier coefficients \( 1/N \) from all frequencies. Here ... | Yes |
Example 3 A shift in the delta vector to \( f = \left( {0,1,0,\ldots ,0}\right) \) produces a “modulation” in its transform. This shifted \( f \) picks out the next column \( \left( {1,\omega ,\ldots ,{\omega }^{N - 1}}\right) \) of \( F \) : | \[ c = \frac{1}{N}{\Omega }_{N}f = \frac{1}{N}\left\lbrack \begin{matrix} 1 \\ \omega \\ \vdots \\ {\omega }^{N - 1} \end{matrix}\right\rbrack \text{ and }f = {Fc} = \frac{1}{N}\left\lbrack \begin{matrix} 1 & 1 & \cdots & 1 \\ 1 & w & \cdots & {w}^{N - 1} \\ \cdot & \cdot & \cdots & \cdot \\ 1 & {w}^{N - 1} & \cdots & ... | Yes |
With \( N = 4 \) there are \( {N}^{2} = {16} \) pixels in a square. The block matrix \( {\Omega }_{4 \times 4} \) for the \( {2D} \) transform is 16 by 16 : | \[ {\Omega }_{4 \times 4} = {\Omega }_{4} \otimes {\Omega }_{4} = \left\lbrack \begin{array}{rrrr} {\Omega }_{4} & {\Omega }_{4} & {\Omega }_{4} & {\Omega }_{4} \\ {\Omega }_{4} & - i\;{\Omega }_{4} & {\left( -i\right) }^{2}\;{\Omega }_{4} & {\left( -i\right) }^{3}\;{\Omega }_{4} \\ {\Omega }_{4} & {\left( -i\right) }^... | Yes |
Example 1 A 20-node graph has two built-in clusters \( P \) and \( Q \) (to find from \( z \) ). The MATLAB code creates edges within \( P \) and within \( Q \), with probability 0.7 . Edges between nodes in \( P \) and \( Q \) have smaller probability 0.1 . All edges have weights \( {w}_{ij} = 1 \), so \( C = I \) . \... | With \( G = {A}^{\mathrm{T}}A \), the eigenvalue command \( \left\lbrack {V, E}\right\rbrack = \mathbf{{eig}}\left( {G, D}\right) \) solves \( {A}^{\mathrm{T}}A\mathbf{x} = {\lambda D}\mathbf{x} \) . Sorting the \( \lambda \) ’s leads to \( {\lambda }_{2} \) and its Fiedler vector \( z \) . Des Higham’s third graph sho... | Yes |
Find the variance \( {\sigma }^{2} \) of the ages of college freshmen. | Solution The probabilities of ages \( {x}_{i} = {17},{18},{19} \) were \( {p}_{i} = {0.2} \) and 0.5 and 0.3 . The expected value was \( m = \sum {p}_{i}{x}_{i} = {18.1} \) . The variance uses those same probabilities :\n\n\[ \n{\sigma }^{2} = \left( {0.2}\right) {\left( {17} - {18.1}\right) }^{2} + \left( {0.5}\right)... | Yes |
Change outputs from 1 or -1 to \( x = 1 \) or \( x = 0 \) . Keep \( {p}_{1} = {p}_{0} = \frac{1}{2} \) . | The new mean value \( m = \frac{1}{2}\; \) falls halfway between 0 and 1. The variance moves to \( {\sigma }^{2} = \frac{1}{4} \) :\n\n\[ \mathbf{m} = \frac{1}{2}\left( 1\right) + \frac{1}{2}\left( 0\right) = \frac{\mathbf{1}}{\mathbf{2}}\text{ and }{\mathbf{\sigma }}^{\mathbf{2}} = \frac{1}{2}{\left( 1 - \frac{1}{2}\r... | Yes |
Flip two coins separately. With 1 for heads and 0 for tails, the results can be \( \left( {1,1}\right) \) or \( \left( {1,0}\right) \) or \( \left( {0,1}\right) \) or \( \left( {0,0}\right) \) . Those four outcomes all have probability \( {\left( \frac{1}{2}\right) }^{2} = \frac{1}{4} \) . | For independent experiments we multiply probabilities :\n\n\[ \n{p}_{ij} = \text{Probability of}\left( {i, j}\right) = \left( {\text{Probability of}i}\right) \text{times}\left( {\text{Probability of}j}\right) \text{.} \n\] | Yes |
Example 2 Glue the coins together, facing the same way. The only possibilities are \( \left( {1,1}\right) \) and \( \left( {0,0}\right) \) . Those have probabilities \( \frac{1}{2} \) and \( \frac{1}{2} \) . The probabilities \( {p}_{10} \) and \( {p}_{01} \) are zero. \( \left( {1,0}\right) \) and \( \left( {0,1}\righ... | Joint probability matrices\n\[ \mathbf{P} = \left\lbrack \begin{array}{ll} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right\rbrack . \] | Yes |
The glued coins show perfect correlation. Heads on one means heads on the other. The covariance \( {\sigma }_{12} \) moves from 0 to \( {\sigma }_{1} \) times \( {\sigma }_{2} \) . This is the largest possible value of \( {\sigma }_{12} \) . Here it is \( \left( \frac{1}{2}\right) \left( \frac{1}{2}\right) = {\sigma }_... | \[ \text{Means} = \frac{1}{2}\;{\sigma }_{12} = \frac{1}{2}\left( {1 - \frac{1}{2}}\right) \left( {1 - \frac{1}{2}}\right) + 0 + 0 + \frac{1}{2}\left( {0 - \frac{1}{2}}\right) \left( {0 - \frac{1}{2}}\right) = \frac{1}{4} \] | No |
Example 4 Suppose that \( y \) is just \( - x \) . A coin flip has outputs \( x = 0 \) or 1 . The same flip has outputs \( y = 0 \) or -1 . The mean \( {m}_{x} \) is \( \frac{1}{2} \) for a fair coin, and \( {m}_{y} \) is \( - \frac{1}{2} \) . The covariance of \( x \) and \( y \) is \( {\sigma }_{xy} = - {\sigma }_{x}... | In this case the correlation matrix \( R \) has determinant zero (singular and only semidefinite):\n\n\[ \text{ Correlation matrix }R = \left\lbrack \begin{matrix} 1 & {\rho }_{xy} \\ {\rho }_{xy} & 1 \end{matrix}\right\rbrack \]\n\n\[ R = \left\lbrack \begin{array}{rr} 1 & - 1 \\ - 1 & 1 \end{array}\right\rbrack \text... | Yes |
Example 5 Suppose the random variables \( x, y, z \) are independent. What matrix is \( R \) ? | Answer \( R \) is the identity matrix. All three correlations \( {\rho }_{xx},{\rho }_{yy},{\rho }_{zz} \) are 1 by definition. All three cross-correlations \( {\rho }_{xy},{\rho }_{xz},{\rho }_{yz} \) are zero by independence. | Yes |
Suppose a doctor measures your heart rate \( x \) three times \( \left( {m = 3, n = 1}\right) \) :\n\n\[ \begin{array}{l} x = {b}_{1} \\ x = {b}_{2} \\ x = {b}_{3} \end{array}\;\text{ is }\;A\mathbf{x} = \mathbf{b}\;\text{ with }\;A = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \;\text{ and }\;V ... | The variances could be \( {\sigma }_{1}^{2} = 1/9 \) and \( {\sigma }_{2}^{2} = 1/4 \) and \( {\sigma }_{3}^{2} = \mathbf{1} \) . The weights are 3 then 2 then 1. You are getting more nervous as measurements are taken: \( {b}_{3} \) is less reliable than \( {b}_{2} \) and \( {b}_{1} \) . All three measurements contain ... | Yes |
The transition matrix can be \( P = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right\rbrack = \) switching matrix. | This means : The system changes its state at every time step. State 1 at time \( n \) leads to State 2 at time \( n + 1 \) . If the initial probabilities were in \( {\mathbf{y}}_{0} = \left( {\frac{1}{3},\frac{2}{3}}\right) \) then \( {\mathbf{y}}_{1} = P{\mathbf{y}}_{0} = \left( {\frac{2}{3},\frac{1}{3}}\right) \) . T... | Yes |
Example 3 \( \;{P}_{3} = \left\lbrack \begin{array}{ll} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{array}\right\rbrack \) has \( \lambda = 1,\frac{1}{2} \) with \( {v}_{1} = \left\lbrack \begin{array}{l} 1 \\ 0 \end{array}\right\rbrack \) and \( {v}_{2} = \left\lbrack \begin{array}{r} 1 \\ - 1 \end{array}\right\rbrack \) ... | Even with that zero in \( {P}_{3} \), all columns of \( {\left( {P}_{3}\right) }^{n} \) approach \( {\mathbf{v}}_{1} = \) first eigenvector:\n\n\[ \n{\left( {\mathbf{P}}_{\mathbf{3}}\right) }^{n} = {\left\lbrack \begin{array}{ll} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{array}\right\rbrack }^{n} = \left\lbrack \begin{ma... | Yes |
For a constant vector \( \mathbf{a} = \left( {{a}_{1},\ldots ,{a}_{n}}\right), F\left( \mathbf{x}\right) = {\mathbf{a}}^{\mathrm{T}}\mathbf{x} \) has gradient \( \mathbf{\nabla }F = \mathbf{a} \) . | The partial derivatives of \( F = {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n} \) are the numbers \( \partial F/\partial {x}_{k} = {a}_{k} \) . | Yes |
For a symmetric matrix \( S \), the gradient of \( F\left( x\right) = {x}^{\mathrm{T}}{Sx} \) is \( \nabla F = {2Sx} \) . | To see this, write out the function \( F\left( {{x}_{1},{x}_{2}}\right) \) when \( n = 2 \) . The matrix \( S \) is 2 by 2 :\n\n\[ F = \left\lbrack \begin{array}{ll} {x}_{1} & {x}_{2} \end{array}\right\rbrack \left\lbrack \begin{array}{ll} a & b \\ b & c \end{array}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\... | Yes |
For a positive definite symmetric \( S \), the minimum of a quadratic \( F\left( x\right) = \frac{1}{2}{x}^{\mathrm{T}}{Sx} - {a}^{\mathrm{T}}x \) is the negative number \( {F}_{\min } = - \frac{1}{2}{a}^{\mathrm{T}}{Sa} \) at \( {x}^{ * } = {S}^{-1}a \). | The minimum occurs where first derivatives of \( F \) are zero :\n\n\[ \nabla F = \left\lbrack \begin{matrix} \partial F/\partial {x}_{1} \\ \vdots \\ \partial F/\partial {x}_{n} \end{matrix}\right\rbrack = {Sx} - a = 0\text{ at }{x}^{ * } = {S}^{-1}a = \arg \min F. \] | Yes |
The logarithm of the determinant is a most remarkable function : | \[ \mathbf{L}\left( \mathbf{X}\right) = \log \left( {\det \mathbf{X}}\right) \text{has partial derivatives}\frac{\partial L}{\partial {x}_{ij}} = \frac{{C}_{ij}}{\det X} = j, i\text{entry of}{\mathbf{X}}^{-1}\text{.} \] The chain rule for \( L = \log F \) is \( \left( {\partial L/\partial F}\right) \left( {\partial F/\... | Yes |
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