I was looking though this answer and had a question regarding the following statement:
Then he shows that if $v$ is a unit eigenvector of $A^TA$ with eigenvalue $\sigma^2$, then $u = \frac{1}{\sigma}Av$ is a unit eigenvector of $AA^T$ with eigenvalue $\sigma^2$.
Let $u_i = \frac{1}{\sigma_i}Av_i$, where $\sigma_i^2$ and $v_i$ are the non-zero eigenvalue and its corresponding unit length eigenvector of $A^TA = Q\Sigma Q^T$. In this case $Q$ would contain $v_i$ in its ith column and $\Sigma$ containing the $\sigma_i^2$ along its diagonals in the ith row and column (with the zero eigenvalues at the end). Also $Q^TQ=I$ as the eigenvectors are orthonormal.
I'm wondering if the my logic is correct:
$$AA^Tu_i = AA^T\frac{1}{\sigma_i}Av_i=\frac{1}{\sigma_i}AQ\Sigma Q^Tv_i=\frac{1}{\sigma_i}A\sigma_i^2v_i=\sigma_i^2\frac{1}{\sigma_i}Av_i=\sigma_i^2 u_i$$
Also $u_i^Tu_i = \frac{1}{\sigma_i^2}v_i^TA^TAv_i = \frac{1}{\sigma_i^2}v_i^TQ\Sigma Q^Tv_i = 1$. Does this show that $u_i$ is an unit eigenvector of $AA^T$ with eigenvalue $\sigma_i^2$?
Best Answer
You are doing it too complicated. You have $$ AA^Tu=\frac1\sigma\,A(A^TAv)=\frac1\sigma\,A(\sigma^2v)=\sigma^2\,\frac1\sigma\,Av=\sigma^2u. $$ And, to check that $u$ is a unit vector, $$ u^Tu=\frac1{\sigma^2}(Av)^TAv=\frac1{\sigma^2}\,v^TA^TAv=v^Tv=1. $$ There is not need to use the singular value decomposition.