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

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MATH307 December 2006

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Question Section 101 08
Find the singular value decomposition $A=U\mathrm{\Sigma }{V}^T$ for the matrix $A=\left(\begin{array}{ccc}0& 2& 0\\ 1& 0& 1\end{array}\right)$

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Solution

Since the values in A are real, we know that $A^{*}A=A^{T}A$ and $A{A}^{*}=A{A}^T$. 1. Find the eigenvalues for $A^{T}A$ $A^{T}A$ = $\left(\begin{array}{cc}0& 1\\ 2& 0\\ 0& 1\end{array}\right)$$\left(\begin{array}{ccc}0& 2& 0\\ 1& 0& 1\end{array}\right)$ = $A^{T}A-\lambda I$= $\left(\begin{array}{ccc}1-\lambda & 0& 1\\ 0& 4-\lambda & 0\\ 1& 0& 1-\lambda \end{array}\right)$ $det(A^{T}A-\lambda I)=(1-\lambda )[(1-\lambda )(4-\lambda )]-(1)(4-\lambda )$ = $(4-\lambda )[(1-\lambda )(1-\lambda )-1]$ = $(4-\lambda )(({\lambda }^2-2\lambda +1)-1)]$ = $(4-\lambda )[{\lambda }^2-2\lambda ]$ = $(4-\lambda )(\lambda )(2-\lambda )$ $\therefore \lambda =4,2,0$ 2. Find the eigenvectors for each eigenvalue of $A^{T}A$. For $\lambda =4$ $A^{T}A-4I=\left(\begin{array}{ccc}1-4& 0& 1\\ 0& 4-4& 0\\ 1& 0& 1-4\end{array}\right)=\left(\begin{array}{ccc}-3& 0& 1\\ 0& 0& 0\\ 1& 0& -3\end{array}\right)$ After row reducing, we have: $\left(\begin{array}{ccc}1& 0& -3\\ -3& 0& 1\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}1& 0& -3\\ 0& 0& -8\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$ From this matrix, we get $\underset{\_}{x}=\{\begin{array}{c}{x}_1=0\\ x_2=x_2\\ x_3=0\end{array}$ By letting $x_2=1$ (for simplicity), we get an eigenvector to be $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ For $\lambda =2$ $A^{T}A-2I=\left(\begin{array}{ccc}1-2& 0& 1\\ 0& 4-2& 0\\ 1& 0& 1-2\end{array}\right)=\left(\begin{array}{ccc}-1& 0& 1\\ 0& 2& 0\\ 1& 0& -1\end{array}\right)$ After row reduce, we have: $\left(\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$ From this matrix, we get $\underset{\_}{x}$ = $\{\begin{array}{c}{x}_1=x_3\\ x_2=0\\ x_3=x_3\end{array}$ By letting $x_3=1$ (for simplicity), we get an eigenvector to be$\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$ For $\lambda =0$ $A^{T}A=\left(\begin{array}{ccc}1& 0& 1\\ 0& 4& 0\\ 1& 0& 1\end{array}\right)$ From the above matrix, we get $\underset{\_}{x}=\{\begin{array}{c}{x}_1=-x_3\\ x_2=0\\ x_3=x_3\end{array}$ By letting $x_3=1$ (for simplicity), we get an eigenvector to be $\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$ 3. Find the eigenvalues for $A{A}^T$. $A{A}^T$ = = Since we know that $A^{T}A$ and $A{A}^T$ have the same nonzero eigenvalues, the eigenvalues for $A{A}^T$ are 4 and 2. Check: $A{A}^T-\lambda I$ = $det(A{A}^T-\lambda I)=(4-\lambda )(2-\lambda )$ $\therefore \lambda =4,2$ 4. Find the eigenvectors for each eigenvalue of $A{A}^T$. For $\lambda =4$ $A{A}^T-4I$ = = From the above matrix, we get $\underset{\_}{x}=\{\begin{array}{c}{x}_1=x_1\\ x_2=0\\ \end{array}$ By letting $x_2=1$ (for simplicity), we get an eigenvector to be $\left[\begin{array}{c}0\\ 1\end{array}\right]$ For $\lambda =2$ $A{A}^T-4I$ = = From the above matrix, we get $\underset{\_}{x}=\{\begin{array}{c}{x}_1=0\\ x_2=x_2\\ \end{array}$ By letting $x_1=1$ (for simplicity), we get an eigenvector to be $\left[\begin{array}{c}1\\ 0\end{array}\right]$ 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. $\left[\begin{array}{ccc}0& 1& -1\\ 1& 0& 0\\ 0& 1& 1\end{array}\right]$ Normalize each column to get V. $V=\left[\begin{array}{ccc}0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ 1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$ Create a matrix consisting of the eigenvectors of $A{A}^T$ as the columns. Note that the order of the eigenvectors in this matrix must match that of the matrix V. $U$ = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ Create a 2x3 matrix which contains the singular values of A as the diagonal entries (i.e. 2 and $\sqrt{2}$). $\sum =\left[\begin{array}{ccc}2& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right]$ $\therefore A=U\mathrm{\Sigma }{V}^T=$ $\left[\begin{array}{ccc}2& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right]$

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