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

MATH307 December 2012

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Question 06 (f)

Suppose that A is a matrix with singular value decomposition $A=U\Sigma V^{*}$ where ${\begin{aligned}U=&{\begin{bmatrix}1/{\sqrt {2}}&1/{\sqrt {2}}&0\\1/{\sqrt {2}}&-1/{\sqrt {2}}&0\\0&0&1\end{bmatrix}}\\\Sigma =&{\begin{bmatrix}2&0&0\\0&1&0\\0&0&1/2\end{bmatrix}}\\V=&{\begin{bmatrix}1/{\sqrt {3}}&0&2/{\sqrt {6}}\\1/{\sqrt {3}}&1/{\sqrt {2}}&-1/{\sqrt {6}}\\1/{\sqrt {3}}&-1/{\sqrt {2}}&-1/{\sqrt {6}}\end{bmatrix}}\end{aligned}}$ (f) Let ${\hat {A}}=U{\begin{bmatrix}2&0&0\\0&1&0\\0&0&0\end{bmatrix}}V^{*}$ where U and V are the matrices given above. Then ${\hat {A}}$ is a matrix with a non-trivial null space. What is a basis for $N({\hat {A}})$? What is the norm $\|{\hat {A}}-A\|$?

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
To find the nullspace $N({\hat {A}})$, look for vectors x such that $0={\hat {A}}x=U{\hat {\Sigma }}V^{*}x$

Hint 2
Since U is unitary, the only way that $U{\hat {\Sigma }}V^{*}x=0$ is that ${\hat {\Sigma }}V^{*}x=0$. Find the nullspace of ${\hat {\Sigma }}$ and relate the result to $V^{*}x$. Solve for x.

Hint 3
To calculate $\|{\hat {A}}-A\|$ use that ${\hat {A}}-A=U{\hat {\Sigma }}V^{*}-U\Sigma V^{*}=U({\hat {\Sigma }}-\Sigma )V^{*}$.

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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. Let's start by calculating the nullspace of ${\hat {A}}$, that is, we are looking for vectors x such that $0={\hat {A}}x=U{\begin{bmatrix}2&0&0\\0&1&0\\0&0&0\end{bmatrix}}V^{*}x=U{\hat {\Sigma }}V^{*}x$ Since U is unitary, Uy = 0 implies y = 0. Hence, for ${\hat {A}}x=0$ is only possible if and only if ${\hat {\Sigma }}V^{*}x=0.$ Let's abbreviate $\displaystyle V^{*}x=z$. Then, in order to find the nullspace of ${\hat {A}}$ we are looking for vectors z such that ${\hat {\Sigma }}z=0.$ This equation can quickly be solved: $z={\begin{bmatrix}0\\0\\t\end{bmatrix}}=te_{3}$, for any real value of t. Since $\displaystyle z=V^{*}x$ we solve for x and find $x=(V^{*})^{-1}z=Vz=tVe_{3}=t{\begin{bmatrix}2/{\sqrt {6}}\\-1/{\sqrt {6}}\\-1/{\sqrt {6}}\end{bmatrix}}$ Therefore, $N({\hat {A}})=t{\begin{bmatrix}2/{\sqrt {6}}\\-1/{\sqrt {6}}\\-1/{\sqrt {6}}\end{bmatrix}}$ Next, observe that ${\hat {A}}-A=U{\hat {\Sigma }}V^{*}-U\Sigma V^{*}=U({\hat {\Sigma }}-\Sigma )V^{*}$ Again following the logic of part (c) it holds that $\|{\hat {A}}\|=\|{\hat {\Sigma }}-\Sigma \|=\left\|{\begin{bmatrix}0&0&0\\0&0&0\\0&0&-1/2\end{bmatrix}}\right\|=1/2$

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