If $A$ is positive definite, why are singular values of $A$ the same as eigenvalues of $A$ ?
Here's an answer (the accepted one).
But I feel like this answer only shows that "if $\lambda$ is an eigenvalue of $A$, then it is also a singular value of $A$".
How to show that if $\sigma$ is a singular value of $A$, then it is also an eigenvalue of $A$?
That is, starting from $A^T A v = \sigma^2 v$ $\implies $ $A^2 v = \sigma^2 v$.
How does the last expression imply $Av = \sigma v$?
I know $Av = \sigma v$ $\implies $ $A^2 v = \sigma^2 v$ but not so sure about the reverse implication?
Let $B=A^2$, $\beta=\sigma^2$ to simplify notation a bit. So you have $Bv=\beta v$. Apply $B$ again, you get $B^2v=\beta^2v$. Again, and again, etc., and you get $B^nv=\beta^nv$ for all $n$. Take linear combinations of this, and you get $$ p(B)v=p(\beta)v $$ for any polynomial $B$. Now choose $p$ to be a polynomial such that $p(\sigma_j^2)=\sigma_j$ for all $j$. Then $p(B)$ is the unique positive square root of $B$, namely $A$. So each $\sigma_j$ is an eigenvalue of $A$.