# Computing Singular Value question

Good afternoon.

I'm studying for my finals of this year, currently studying for the exam "Numerical Linear Algebra". I'm trying to solve some of the questions the teacher asked the past years (for preparation), but I can't seem to figure out a few, namely this one. We've got this setup:

• $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},{-}\dfrac{1}{\sqrt{2}},0\Bigr]$

$Ax=[2,0,0]$

• $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\Bigr]$

$Ax=[0,-3,0]$

• $x=[0,0,1]$

$Ax=[0,0,6]$

The questions were as follows: 1. Determine the SVD of A (WITHOUT COMPUTING THE ACTUAL MATRIX) 2. Determine the condition number of A 3. Determine the rank and the determinant

Questions 2 and 3 are quite logical (condition number = largest SVD divided by the smallest, rank = number of non zero SV's and the determinant is the product of the SV's, if I'm not mistaken), but I can't seem to find a logical answer or explanation to question 1. So I'm hoping anyone here has a pointer in the right direction? I'm not asking for the asnwer (since I have to be able to replicate it on the exam, most likely), but some hints would be appreciated.

Thanks.

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Since the three given vectors are orthonormal and their images are orthogonal, these are the right singular vectors. Thus, the singular value decomposition of $A$ is $$A \;=\; \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6\end{bmatrix}\begin{bmatrix}-1/\sqrt{2} & -1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{bmatrix}.$$ The rows of the last matrix are the given vectors, the diagonal entries of the middle matrix are the norms of the images of these vectors, and the columns of the first matrix are the normalizations of these images.