Science:Math Exam Resources/Courses/MATH307/December 2012/Question 06 (a)

MATH307 December 2012

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[hide] Question 06 (a)

Suppose that A is a matrix with singular value decomposition $A=U\Sigma V^{*}$ where ${\begin{aligned}U=&{\begin{bmatrix}1/{\sqrt {2}}&1/{\sqrt {2}}&0\\1/{\sqrt {2}}&-1/{\sqrt {2}}&0\\0&0&1\end{bmatrix}}\\\Sigma =&{\begin{bmatrix}2&0&0\\0&1&0\\0&0&1/2\end{bmatrix}}\\V=&{\begin{bmatrix}1/{\sqrt {3}}&0&2/{\sqrt {6}}\\1/{\sqrt {3}}&1/{\sqrt {2}}&-1/{\sqrt {6}}\\1/{\sqrt {3}}&-1/{\sqrt {2}}&-1/{\sqrt {6}}\end{bmatrix}}\end{aligned}}$ (a) Check all circles that apply. $U$ and $V$ are: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigcirc$, orthogonal $\bigcirc$ $\Sigma$ is: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigcirc$, orthogonal $\bigcirc$ $A^{*}A$ and $AA^{*}$ are: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigcirc$, orthogonal $\bigcirc$ $A$ is: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigcirc$, orthogonal $\bigcirc$

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[show] Hint 1
A matrix M is called hermitian if $M={\overline {M^{T}}}$ real symmetric if all entries of $M$ are real and $M=M^{T}$. unitary if $MM^{*}=M^{*}M=I$, where $M^{*}={\overline {M^{T}}}$. orthogonal if $MM^{T}=M^{T}M=I$.

[show] Hint 2
For real matrices the conditions hermitian and real symmetric are equivalent. Further, for real matrices the conditions unitary and orthogonal are equivalent.

[show] Hint 3
A square matrix is orthogonal if the columns (or rows) form an orthonormal basis.

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[show] Solution

Since all matrices are real, hermitian, real symmetric and symmetric are equivalent, and so are unitary and orthogonal. U and V are: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigotimes$, orthogonal $\bigotimes$ We easily see that $U$ is symmetric but $V$ is not. Hence, together, they are neither hermitian nor real symmetric. Since the columns of $U$ and $V$ are orthonormal, both matrices are unitary and orthogonal. This can be seen for $U$ since ${\begin{aligned}{\begin{bmatrix}1/{\sqrt {2}}\\1/{\sqrt {2}}\\0\end{bmatrix}}\cdot {\begin{bmatrix}1/{\sqrt {2}}\\1/{\sqrt {2}}\\0\end{bmatrix}}&=1\\{\begin{bmatrix}1/{\sqrt {2}}\\1/{\sqrt {2}}\\0\end{bmatrix}}\cdot {\begin{bmatrix}1/{\sqrt {2}}\\-1/{\sqrt {2}}\\0\end{bmatrix}}&=0\\{\begin{bmatrix}1/{\sqrt {2}}\\1/{\sqrt {2}}\\0\end{bmatrix}}\cdot {\begin{bmatrix}0\\0\\1\end{bmatrix}}&=0\\{\begin{bmatrix}1/{\sqrt {2}}\\-1/{\sqrt {2}}\\0\end{bmatrix}}\cdot {\begin{bmatrix}1/{\sqrt {2}}\\-1/{\sqrt {2}}\\0\end{bmatrix}}&=1\\{\begin{bmatrix}1/{\sqrt {2}}\\-1/{\sqrt {2}}\\0\end{bmatrix}}\cdot {\begin{bmatrix}0\\0\\1\end{bmatrix}}&=0\\{\begin{bmatrix}0\\0\\1\end{bmatrix}}\cdot {\begin{bmatrix}0\\0\\1\end{bmatrix}}&=1\end{aligned}}$ The corresponding vector multiplications give the same result for $V$. $\Sigma$ is: hermitian $\bigotimes$, real symmetric $\bigotimes$, unitary $\bigcirc$, orthogonal $\bigcirc$ Since $\Sigma$ is diagonal it is cleary symmetric. However, the first and the third column vectors don't have length 1 and hence $\Sigma$ is not unitary or orthogonal. $A^{*}A$ and $AA^{*}$ are: hermitian $\bigotimes$, real symmetric $\bigotimes$, unitary $\bigcirc$, orthogonal $\bigcirc$ Following the rules of transposition of matrix products we quickly check that $(A^{*}A)^{*}=A^{*}((A^{*})^{*})=A^{*}A$ and correspondingly $(AA^{*})^{*}=((A^{*})^{*})A^{*}=AA^{*}$ so that $A^{*}A$ and $AA^{*}$ are indeed unitary and hence also real symmetric. Next, observe that $\displaystyle A^{*}A=(U\Sigma V^{*})^{*}(U\Sigma V^{*})=V\Sigma ^{2}V^{-1}$ since, from the above, $U$ and $V$ are unitary. The equation above however describes a diagonalization of $A^{*}A$, and hence the eigenvalues of $A^{*}A$ are found on the diagonal of $\Sigma ^{2}$, namely $4,1,{\frac {1}{4}}$. However, all eigenvalues of a unitary matrix must have an absolute value of 1. Hence $A^{*}A$ is not unitary and not orthogonal. The equivalent argument also shows that $AA^{*}$ is not unitary and not orthogonal. $A$ is: hermitian $\bigcirc$, real symmetric $\bigcirc$, unitary $\bigcirc$, orthogonal $\bigcirc$ Since the largest singular value corresponds to the matrix norm of $A$ it follows that $\|A\|=2$, while unitary and orthogonal matrices have norm 1. To check if $A$ is symmetric we start by checking the entries $a^{2,1}$ and $a^{1,2}$ for equality. ${\begin{aligned}A&={\begin{bmatrix}1/{\sqrt {2}}&1/{\sqrt {2}}&0\\1/{\sqrt {2}}&-1/{\sqrt {2}}&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}2&0&0\\0&1&0\\0&0&1/2\end{bmatrix}}{\begin{bmatrix}1/{\sqrt {3}}&1/{\sqrt {3}}&1/{\sqrt {3}}\\0&1/{\sqrt {2}}&1/{\sqrt {2}}\\2/{\sqrt {6}}&-1/{\sqrt {6}}&-1/{\sqrt {6}}\end{bmatrix}}\\&={\begin{bmatrix}2/{\sqrt {2}}&1/{\sqrt {2}}&0\\2/{\sqrt {2}}&-1/{\sqrt {2}}&0\\0&0&1/2\end{bmatrix}}{\begin{bmatrix}1/{\sqrt {3}}&1/{\sqrt {3}}&1/{\sqrt {3}}\\0&1/{\sqrt {2}}&1/{\sqrt {2}}\\2/{\sqrt {6}}&-1/{\sqrt {6}}&-1/{\sqrt {6}}\end{bmatrix}}\\&={\begin{bmatrix}*&{\frac {2}{\sqrt {6}}}+{\frac {1}{2}}&*\\{\frac {2}{\sqrt {6}}}&*&*\\{}*&*&*\end{bmatrix}}\end{aligned}}$ We see that $a^{2,1}\neq a^{1,2}$ and hence $A$ is not hermitian or real symmetric.

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