# Why do the eigenvalues of $A^TB$ equal the nonzero eigenvalues of $AB^T$? [duplicate]

Given two matrices, $A,B\in\mathbb{R}^{m\times n}$ where $m>n$, why do the eigenvalues of $A^TB$ equal the nonzero eigenvalues of $AB^T$? Essentially, why do the eigenvalues of the inner product between matrices equal the nonzero eigenvalues of the outer product between two matrices?

• @Stefan They may be related, but the question here is more directed and we have a better answer below. The question should not be flagged. – wyer33 Nov 8 '17 at 8:24
• you are right, the answer given here is better. – Stefan Nov 8 '17 at 8:28
• There are many similar questions, e.g. here, here and more. – A.Γ. Nov 8 '17 at 8:45
• @A.Γ.Right on that and the second link looks good. Thanks for the references. – wyer33 Nov 8 '17 at 8:51

If $\lambda\ne0$, and $x\ne 0$, such that $$AB^Tx=\lambda x,$$ then $$B^TAy=B^TA(B^Tx)=\lambda B^Tx=\lambda y.$$ Clearly, $y=B^Tx\ne 0$, otherwise, $AB^Tx=0$, as well, and hence $\lambda$ is an eigenvalue of $B^TA$.
When $A$ and $B$ are square matrices, $AB$ and $BA$ have the same characteristic polynomial and so the same eigenvalues.
When $A$ and $B$ are $m\times n$ and $n\times m$ ($m<n$), let $A'$ and $B'$ be the matrices got by appending extra zero rows to $A$ and $B$. Then $B'A'=BA$ and $A'B'$ is the diagonal sum of $AB$ with a zero matrix. So the eigenvalues of $AB$ are those of $BA$ plus $n-m$ extra zeros.
Applying to the problem in hand $A^TB$ and $BA^T$ have the same eigenvalues except for some zeros. But the eigenvalues of a matrix are not changed by transposition. So $(BA^T)^T=AB^T$ has the same eigenvalues as $BA^T$ etc.
• $A′$ and $B′$ be the matrices got by appending extra zero rows to $A$ and $B$? Isn't columns of $B$ correct? – C.F.G Aug 7 '20 at 18:55