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

MATH307 December 2012

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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 $\bigcirc$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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 $MM^{T}=M^{T}M=I$.

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

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

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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 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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