We have a matrix $A_{m \times n}$. I know that the two matrices $A^*A$ and $AA^*$ share nonzero eigenvalues. This comes from multiplying $A^*Ax=\lambda x$ on both sides by $A$ ($\lambda$ is an eigenvalue of $A^*A$, and $x$ is an eigenvector of $A^*A$). Then, we have:

$$AA^*(Ax)=\lambda (Ax)$$

So now $\lambda$ is also an eigenvalue of $AA^*$, but now the eigenvector is $Ax$ (for $x \notin KerA$, so that $Ax\neq0$).

All other $\lambda$ not common to both matrices $AA^*$ and $A^*A$ is equal to zero.

My question is: when will the multiplicity of the zero eigenvalues be the same for both matrices $(AA^*)_{m\times m}$ and $(A^*A)_{n\times n}$?

Intuition tells me that this will only happen when both matrices $AA^*$ and $A^*A$ have the same dimensions, which implies $A_{m \times n}$ being square $(m=n)$. Otherwise, one matrix will always have more eigenvalues than the other (or more repetitions of eigenvalues, so higher multiplicity), the extra eigenvalues being equal to zero.

Is this correct? and how would you prove this more rigorously?


1 Answer 1


Your intuition at the end of the post is correct.

One way to see this explicitly is to consider the SVD of $A$. Let $A=UDV^*$ where $U$ and $V$ are square orthogonal ($m\times m$ and $n\times n$ respectively), and $D$ is $m \times n$ rectangular diagonal with nonnegative diagonal entries in nondecreasing order. Noting that $A^*A=VD^* DV^*$ and $AA^* = UDD^*U^*$, you see that the eigenvalues of $A^*A$ and $AA^*$ are exactly the diagonal entries of $D^*D$ and $DD^*$ respectively. Walk through a few examples of rectangular diagonal $D$ to see how the diagonal entries of $D^*D$ and $DD^*$ differ. For example, if $D$ is a tall matrix ($m > n$), then $DD^*$ will have $m-n$ extra zeros on the diagonal compared to $D^*D$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .