Eigenvalue and eigenvector of $A'A$

Question

Suppose that $\mathbf{A}\in\mathrm{R}^{m\times m}$ is a square but not necessarily symmetric matrix whose eigenvalues and eigenvectors are $\lambda_i$ and $\mathbf{x}_i,$ $i = 1,2,\cdots,m$.

1. Is there any relation between the eigenvalues of $\mathbf{A}$ and those of $\mathbf{A'A}$?

2. Is there any relation between the eigenvectors of $\mathbf{A}$ and those of $\mathbf{A'A}$?

Answer 1

If $A$ is normal (that is, if $A'A = AA'$), then the relation is "what you'd expect":

• $\lambda$ is an eigenvalue of $A\iff |\lambda|^2$ is an eigenvalue of $A'A$
• $A$ and $A'A$ have an identical set of orthonormal eigenvectors.

Also, $A$ is normal if and only if trace$(A'A) = \sum_{i=1}^k |\lambda_i(A)|^2$.

More generally, these are not true. However, we can say that $$ \sqrt{\lambda_1(A'A)} \leq |\lambda_i(A)| \leq \sqrt{\lambda_n(A'A)} $$ where $\lambda_1(X) \leq \cdots \leq \lambda_n(X)$ are the eigenvalues of $X$.

There is no general relation between the eigenvectors.