As a part of an exercise I have to prove the following:

Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.

What I need to prove is the following:

$AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$

In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.

How can I prove this?

Thanks and regards.

  • 1
    $\begingroup$ In fact, if $n\geq m$ then $\text{det}(\lambda I_n - AA^T )=\lambda^{n-m}\text{det}(\lambda I_m - A^TA ).$ $\endgroup$ Dec 31, 2014 at 22:00
  • $\begingroup$ how can we prove this with counting geometric multiplicities? $\endgroup$ May 31, 2020 at 14:19

4 Answers 4


Let $\lambda$ be an eigenvalue of $A^TA$, i.e. $$A^T A x = \lambda x$$ for some $x \neq 0$. We can multiply $A$ from the left and get $$A A^T (Ax) = \lambda (Ax).$$

What can you conclude from this?

  • 4
    $\begingroup$ what happens if $x \in ker(A)?$ not that it can happen. $\endgroup$
    – abel
    Dec 31, 2014 at 23:47
  • 5
    $\begingroup$ @abel If $Ax=0$, then $\lambda x=0$ and $\lambda=0$. So, if we assume $\lambda\ne0$… $\endgroup$
    – egreg
    Jan 1, 2015 at 0:18
  • 2
    $\begingroup$ @egreg, i know; but that needed to be part of the answer. egret you must like linear. this is the second time we meet, the other one at eigenvalues 1, -1, and 4. $\endgroup$
    – abel
    Jan 1, 2015 at 0:21
  • 4
    $\begingroup$ Further, same proof holds for any $A$ and $B$ as mentioned by @abel. Given $ABx = \lambda x$, multiplying by $B$, yields $BABx = B \lambda x$ $\Rightarrow$ $BA(Bx) = \lambda (Bx)$. So, $AB$ and $BA$ share the eigenvalues. $\endgroup$
    – kamalbanga
    Sep 5, 2018 at 12:03

in fact, nonzero eigenvalues $AB$ and $BA$ are the same for any rectangular matrices $A$ and $B$. this follows from the fact that $trace((AB)^k) = trace((BA)^k)$ and the coefficients of the characteristic polynomials of a square matrix $A$ are a function of $trace(A^k).$

  • $\begingroup$ So we can we say that the non-zero eigenvalues of $AB$ and $BA$ also have the same algrebraic multiplicity. $\endgroup$
    – Curtis74
    Nov 23, 2019 at 15:00

One proof that comes to mind is to use Sylvester's determinant theorem.

In particular: $$ \mu \neq 0 \text{ is an eigenvalue of }A^TA \implies\\ \det(A^TA - \mu I) = 0 \implies\\ \det(I + (-1/\mu)A^TA) = 0 \implies\\ \det(I + A(-1/\mu)A^T) = 0 \implies\\ \det(AA^T - \mu I) = 0 \implies\\ \mu \neq 0 \text{ is an eigenvalue of }AA^T $$


SVD is definitely an overkill, but maybe it would be helpful to you (as it is for me) to draw the matrices that the decomposition gives us:


Recall that:

  • The columns of $V$ (right-singular vectors) are eigenvectors of $A^TA$.
  • The columns of $U$ (left-singular vectors) are eigenvectors of $AA^T$.
  • $V^T=V^{-1}$ and $U^T=U^{-1}$.

With some simple operations you can get:

derived from SVD 1

And also:

derived from SVD 2

Write down both of these equations, but for the $i^\text{th}$ column.
From there, the solution is not far.


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