$$A=\begin{pmatrix}-3 & -1 \\ -3& 1\end{pmatrix}$$
- Find the singular values of $σ_1$ and $σ_2$,
- Find unit vectors $v_1$ and $v_2$ such that $||Av_1|| =σ_1$ and $||Av_2|| =σ_2$
I figured out $σ_1$ and $σ_2$ are $\sqrt{18}$ and $\sqrt{2}$ but I can't figure out the second part.
When we have $$A^TA = \begin{pmatrix}18 & 0 \\ 0& 2\end{pmatrix}$$ I thought we were supposed to take each of our lambda values ($18$ and $2$) and subtract them from the diagonals of the $A^TA$ matrix.
For 18 we get $$\begin{pmatrix}18-18 &0 \\ 0& 2-18\end{pmatrix}$$ = $$\begin{pmatrix}0 &0 \\ 0& -16\end{pmatrix}$$ = $$\begin{pmatrix}0 &1 \\ 0& 0\end{pmatrix}$$ So for v1 it should be $$V1=\begin{pmatrix}1 \\ 0\end{pmatrix}$$.I get v2 as $$V2=\begin{pmatrix}0 \\ 1\end{pmatrix}$$. However, that answer comes back as incorrect. Not sure what I am doing wrong here, I think I am getting caught up in what exactly I am trying to find? Please for help!
EDIT: OK I am not sure how to do ||Av1|| and ||Av2||. For ||Av1|| am I just multiplying A*v1? $$A=\begin{pmatrix}-3 & -1 \\ -3& 1\end{pmatrix}$$ * $$V1=\begin{pmatrix}1 \\ 0\end{pmatrix}$$
Which equals $$\begin{pmatrix}-3 \\ -3\end{pmatrix}$$ Normalize it and get $$\begin{pmatrix}-1 \\ -1\end{pmatrix}$$
But I am getting this answer as incorrect so I'm not sure what I'm supposed to be doing.
Hint. This seems like a careless mistake. Have you actually calculate the values of $\|Av_1\|$ and $\|Av_2\|$?