Hi guys, can someone please explain this to me? How to get the eigenvalue/eigenvector pairs of this symmetric matrix? Thanks.
Efforts: Usually $Av_i = s_iu_i$ and for this question, I want to choose $v_i= [0, 1]^T$. Therefore isn't the first singular value be 1?
We know that the eigenvalues will be such that
$$\begin{bmatrix}0&1\\1&3/2\end{bmatrix}\begin{bmatrix}e_1\\e_2\end{bmatrix}=\lambda \begin{bmatrix}e_1\\e_2\end{bmatrix}\tag 1$$
Hence,
$$\left(\begin{bmatrix}0&1\\1&3/2\end{bmatrix}-\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}\right)\begin{bmatrix}e_1\\e_2\end{bmatrix}=0$$
which means that $\begin{bmatrix}e_1\\e_2\end{bmatrix} $ is in the null space of $\begin{bmatrix}-\lambda&1\\1&3/2-\lambda\end{bmatrix}$. The presence of a nullspace makes the matrix singular. Hence the determinant of $\begin{bmatrix}-\lambda&1\\1&3/2-\lambda\end{bmatrix}=0$.
$$\det\begin{bmatrix}-\lambda&1\\1&3/2-\lambda\end{bmatrix}=\lambda^2-3/2\lambda-1=0$$
which means that $\lambda_1=2$ and $\lambda_2=-1/2$.
Immediately we can go back to eq. (1) and see that:
$0\,e_1+1\,e_2=\lambda_1\, e_1$. Therefore $e_2=2\,e_1.$ And $e_2=-1/2\,e_1$
So we could propose as eigenvectors $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}-2\\1\end{bmatrix}$, but we want them to be unitary vectors, which is achieved by dividing each entry by $\sqrt 5$:
$$\text{evec}_1=\begin{bmatrix}\frac{1}{\sqrt 5}\\\frac{2}{\sqrt 5}\end{bmatrix}$$ for $\lambda_1= 2$,
and
$$\text{evec}_2=\begin{bmatrix}\frac{-2}{\sqrt 5}\\\frac{1}{\sqrt 5}\end{bmatrix}$$ for $\lambda_2= -1/2.$