The singular value decomposition of a complex-values $n\times m$ matrix $A$ is defined as
$$A = U \Sigma V^*$$
where $U$ and $V$ are unitary $n \times n$ matrices and $m \times m$ matrices respectively, and $Σ$ is an $n \times m$ rectangular diagonal matrix, whose diagonal elements are the singular values $\sigma _I$ of $A$. $V^*$ denotes the conjugate transpose of $V$. The singular value decomposition s a generalization of the eigen decomposition to arbitrary $n\times m$ matrices
I have attempted to solve the following question below. I am not sure if my proof is enough and I am looking for some guidance! thanks!
Verify that the left singular-vectors (the left columns of $U$) are the eigenvectors $AA^*,$that the right-singular vectors (the columns of $V$) are the eigenvectors of $A^*A$ and that the non-zero singular values of $A$ are the square roots of the non-zero eigenvalues of $AA^* and A^*A$
Here are my workings below!
$$A = U \Sigma V^T$$ $$AA^T = (U\Sigma V^T)(U\Sigma V^T)^T = U \Sigma V^TV\Sigma U^T = U\Sigma ^2U^T$$
The, the left singular vectors of $A$ (i.e columns of $U$) are the eigen vectors $AA^T$
The singular values of $A$ (i.e diagonal elements of $\Sigma$) are square-roots of the eigen values of $AA^T$
Next,
$$A^TA = (U \Sigma V^T)^T(U \Sigma V^T) = V \Sigma U^T U \Sigma V^T = V \Sigma ^2 V^T$$
Thus the right singular vectors of $A$(i.e columns of $V$) are the eigenvectors of $A^TA$.
The singular values of $A$ (i.e diagonal elements of $\Sigma$) are square roots of the eigenvalues of $A^TA$
