Finding SVD from unit eigen values Suppose $A$ is a 2 by 2 symmetric matrix with unit eigenvectors $u_1$ and $u_2$. If its eigenvalues are $\lambda_1=3$ and $\lambda_2=-2$, what are the matrices $U,\Sigma,V^T $ in its SVD?
How to do this?
Is it something with the matrix beeing symmetric?


*

*David

 A: Since $A$ is symmetric, $A^*A=A^2$ and its singular values are simply the absolute values of its eigenvalues. So a SVD of $A$ is of the form $A=U\Sigma V^*$, with $U, V$ unitary and $\Sigma=\mbox{diag}(3,2)$. With the usual convention of ordering the singular values in descending order, only $\Sigma$ is unique. We will now see how to find some $U,V$ which work.
Since you are given an orthonormal basis $\{u_1,u_2\}$ of diagonalization, take $U$ to be the change of basis matrix from  $\{u_1,u_2\}$ to the canonical basis. That is the columns of $U$ are $u_1$ and $u_2$ expressed in the canonical basis.
$$U^*AU=\pmatrix{3&0\\0&-2}\quad\Rightarrow\quad U^*AU\pmatrix{1&0\\0&-1}=\pmatrix{3&0\\0&2}=\Sigma.$$
Therefore
$$
A=U\Sigma V^*\quad\mbox{with}\quad V=U\pmatrix{1&0\\0&-1}.
$$
Note: this works in general if you are given a symmetric (or normal, more generally) matrix with an orthonormal basis of diagonalization. Then the singular values are just the moduli of the eigenvalues, so it suffices to multiply $U$ by the appropriate diagonal unitary matrix to get $V$. If $A$ is not normal, this no longer works.
