Consider $X$ a $n\times d$ real matrix. By the Singular value decomposition we have that
$X = USV^{t}$ where $V$ is an orthogonal matrix of size $d\times d$, $S$ the matrix containing the singular values of size $n\times d$ and $U$ a matrix with orthonormal column of size $n\times n$. My aim is to prove that the eigenvalues of $X^{t}X$ are proportional to the square singular values of $X$. For this, I need to consider an eigendecomposition of $X^{t}X$ wich is possible since this matrix is symmetric. We have $X^{T}X = (USV^{t})^{t}(USV^{t})=VS^{t}SV^{t}$, and we know that $S^{t}S$ is a diagonal matrix and that $V^{t}=V^{-1}$.
Now if I consider an eigendecomposition of $X^{t}X$ given by $PDP^{-1}$, we have the equality $VS^{t}SV^{-1} = PDP^{-1} $.
But I don't see how I can continue, nothing is telling me that $V=P$ at my knowledge ?
Thank you a lot for your help
Hopefully this hint helps with your next step.
From $ VS^tSV^{−1}=PDP^{−1}$, we could move the P matrices over to get $ P^{-1}VS^tSV^{−1}P=D$. As $S^tS$ is diagonal and $(P^{-1}V)^{-1} = V^{-1}P$, does this tell us anything about the relationship between the eigenvalues of $D$ and $S^tS$?
Similarly, what can find out about the eigenvalues of $S^tS$, if we move the $V$ matrices over to get $ S^tS=V^{−1}PDP^{−1}V$?