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I'm trying to teach myself singular value decomposition, and part of that is finding eigenvalues and Eeigenvectors.

I have a matrix $$A=\begin{bmatrix}1 & 1\\ 0 & 1\\ -1 & 1\end{bmatrix}$$ So $$t(A)A =\begin{bmatrix}2 & 0\\ 0 & 3\end{bmatrix}$$

So my Eigenvalues are $$ \lambda_1 = 3,\qquad \lambda_2 = 2 $$

So now I need my eigenvectors.

To calculate the first one, I substitute $\lambda_1$ in $$\begin{bmatrix} \lambda-2 & 0\\ 0 & \lambda-3 \end{bmatrix}$$ that becomes $$ \begin{bmatrix} 3-2 & 0\\ 0 & 3-3\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} $$ So when I solve for the eigenvector by multiplying this by $[x_1; x_2]$ and setting it to $0$ I get \begin{cases} x_1 = 0 \\ x_2 = 0 \end{cases} But I cannot have a $[0 ; 0]$ eigenvector if I understand correctly, so I must be doing something wrong?

The result is supposed to be an eigenvector $[0 ; 1]$ but I fail to see how.

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When you multiply $\pmatrix{1 & 0\cr 0 & 0\cr} \pmatrix{x_1 \cr x_2\cr}$ you should get $\pmatrix{x_1 \cr 0\cr}$. So $x_1 = 0$, but there is no restriction on $x_2$. Thus $\pmatrix{0\cr x_2\cr}$ is an eigenvector, for any $x_2 \ne 0$.

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