Science:Math Exam Resources/Courses/MATH307/December 2006/Question Section 101 08

MATH307 December 2006

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Question Section 101 08
Find the singular value decomposition $\displaystyle A=U\Sigma V^{T}$ for the matrix $A={\begin{pmatrix}0&2&0\\1&0&1\end{pmatrix}}$

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since the values in A are real, we know that $A^{*}A=A^{T}A$ and $AA^{*}=AA^{T}$. 1. Find the eigenvalues for $A^{T}A$ $A^{T}A$ = ${\begin{pmatrix}0&1\\2&0\\0&1\end{pmatrix}}$${\begin{pmatrix}0&2&0\\1&0&1\end{pmatrix}}$ = $A^{T}A-\lambda I$= ${\begin{pmatrix}1-\lambda &0&1\\0&4-\lambda &0\\1&0&1-\lambda \end{pmatrix}}$ $det(A^{T}A-\lambda I)=(1-\lambda )[(1-\lambda )(4-\lambda )]-(1)(4-\lambda )$ = $(4-\lambda )[(1-\lambda )(1-\lambda )-1]$ = $(4-\lambda )((\lambda ^{2}-2\lambda +1)-1)]$ = $(4-\lambda )[\lambda ^{2}-2\lambda ]$ = $(4-\lambda )(\lambda )(2-\lambda )$ $\therefore \lambda =4,2,0$ 2. Find the eigenvectors for each eigenvalue of $A^{T}A$. For $\lambda =4$ $A^{T}A-4I={\begin{pmatrix}1-4&0&1\\0&4-4&0\\1&0&1-4\end{pmatrix}}={\begin{pmatrix}-3&0&1\\0&0&0\\1&0&-3\end{pmatrix}}$ After row reducing, we have: ${\begin{pmatrix}1&0&-3\\-3&0&1\\0&0&0\end{pmatrix}}={\begin{pmatrix}1&0&-3\\0&0&-8\\0&0&0\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&0&1\\0&0&0\end{pmatrix}}$ From this matrix, we get ${\underline {x}}=\left\{{\begin{matrix}x_{1}=0\\x_{2}=x_{2}\\x_{3}=0\end{matrix}}\right.$ By letting $x_{2}=1$ (for simplicity), we get an eigenvector to be ${\begin{bmatrix}0\\1\\0\end{bmatrix}}$ For $\lambda =2$ $A^{T}A-2I={\begin{pmatrix}1-2&0&1\\0&4-2&0\\1&0&1-2\end{pmatrix}}={\begin{pmatrix}-1&0&1\\0&2&0\\1&0&-1\end{pmatrix}}$ After row reduce, we have: ${\begin{pmatrix}1&0&-1\\0&1&0\\0&0&0\end{pmatrix}}$ From this matrix, we get ${\underline {x}}$ = $\left\{{\begin{matrix}x_{1}=x_{3}\\x_{2}=0\\x_{3}=x_{3}\end{matrix}}\right.$ By letting $x_{3}=1$ (for simplicity), we get an eigenvector to be${\begin{bmatrix}1\\0\\1\end{bmatrix}}$ For $\lambda =0$ $A^{T}A={\begin{pmatrix}1&0&1\\0&4&0\\1&0&1\end{pmatrix}}$ From the above matrix, we get ${\underline {x}}=\left\{{\begin{matrix}x_{1}=-x_{3}\\x_{2}=0\\x_{3}=x_{3}\end{matrix}}\right.$ By letting $x_{3}=1$ (for simplicity), we get an eigenvector to be ${\begin{bmatrix}-1\\0\\1\end{bmatrix}}$ 3. Find the eigenvalues for $AA^{T}$. $AA^{T}$ = = Since we know that $A^{T}A$ and $AA^{T}$ have the same nonzero eigenvalues, the eigenvalues for $AA^{T}$ are 4 and 2. Check: $AA^{T}-\lambda I$ = $det(AA^{T}-\lambda I)=(4-\lambda )(2-\lambda )$ $\therefore \lambda =4,2$ 4. Find the eigenvectors for each eigenvalue of $AA^{T}$. For $\lambda =4$ $AA^{T}-4I$ = = From the above matrix, we get ${\underline {x}}=\left\{{\begin{matrix}x_{1}=x_{1}\\x_{2}=0\\\end{matrix}}\right.$ By letting $x_{2}=1$ (for simplicity), we get an eigenvector to be ${\begin{bmatrix}0\\1\end{bmatrix}}$ For $\lambda =2$ $AA^{T}-4I$ = = From the above matrix, we get ${\underline {x}}=\left\{{\begin{matrix}x_{1}=0\\x_{2}=x_{2}\\\end{matrix}}\right.$ By letting $x_{1}=1$ (for simplicity), we get an eigenvector to be ${\begin{bmatrix}1\\0\end{bmatrix}}$ 5. Combine information from the above sections to get Failed to parse (syntax error): {\displaystyle A = UΣV^T} . Create a matrix consisting of the eigenvectors of $A^{T}A$ as the columns. ${\begin{bmatrix}0&1&-1\\1&0&0\\0&1&1\end{bmatrix}}$ Normalize each column to get V. $V={\begin{bmatrix}0&{\frac {1}{\sqrt {2}}}&-{\frac {1}{\sqrt {2}}}\\1&0&0\\0&{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\end{bmatrix}}$ Create a matrix consisting of the eigenvectors of $AA^{T}$ as the columns. Note that the order of the eigenvectors in this matrix must match that of the matrix V. $U$ = ${\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ Create a 2x3 matrix which contains the singular values of A as the diagonal entries (i.e. 2 and ${\sqrt {2}}$). $\sum ={\begin{bmatrix}2&0&0\\0&{\sqrt {2}}&0\end{bmatrix}}$ $\therefore A=U\Sigma V^{T}=$ ${\begin{bmatrix}2&0&0\\0&{\sqrt {2}}&0\end{bmatrix}}$