Science:Math Exam Resources/Courses/MATH307/December 2012/Question 06 (f)/Solution 1

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

$\|{\hat {A}}\|=\|{\hat {\Sigma }}-\Sigma \|=\left\|{\begin{bmatrix}0&0&0\\0&0&0\\0&0&-1/2\end{bmatrix}}\right\|=1/2$