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In an exam question I was asked to calculate the eigenvalues of $A^\top A$, where $A = (a_1\ a_2\ a_3); a_1=(0\ 2\ 1)^\top; a_2=(1\ -1\ 1)^\top; a_3=(1\ 1\ -1)^\top;$ and $A^\top$ stands for the transpose of $A$.

By mistake I calculated $AA^\top$ instead of $A^\top A$, and I got a diagonal matrix whose diagonal entries are 2, 3, 6, which are equal to the eigenvalues of $A^\top A$.

My question is whether this is a coincidence; that is, whether the eigenvalues of $A^\top A$ are equal to those of $AA^\top$.

Pteromys
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  • It's no coincidence. In general $MM^T$ and $M^TM$ share the same non-zero eigenvalues. Since $A$ is a square matrix, the eigenvalues are exactly the same. – Git Gud Jun 16 '13 at 15:12
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    For two matrices $A,B\in M_n(K)$, $\chi_{AB}=\chi_{BA}$ (the characteristic polynomials are the same). Sorry for the self-promotion, but here is how you can prove this general result, and even more in the non square case. So not only the eigenvalues are the same, but they have the same multiplicities. – Julien Jun 16 '13 at 16:11

3 Answers3

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In general, note that if $A \in \mathbb{C}^{m \times n}$, then $A^TA \in \mathbb{C}^{n \times n}$, has $n$ eigenvalues, and $AA^T \in \mathbb{C}^{m \times m}$ has $m$ eigenvalues. However, the non-zero eigenvalues of both the matrices are same and positive. This can be seen by using the singular value decomposition of $A$. If $A$ is of rank $r$, i.e., only $r$ non-zero eigenvalues, we then have $$A = U_{m \times r} \Sigma_{r \times r} V^T_{r \times n}$$ This gives us $$A^TA = V_{n \times r} \Sigma^2_{r \times r} V^T_{r \times n}$$ while $$AA^T = U_{m \times r} \Sigma^2_{r \times r} U^T_{r \times m}$$ The above gives the corresponding eigenvalue decomposition of $A^TA$ and $AA^T$.

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Eigen Values are same, Here is the Simple Proof.

Let $$B=A^T A$$ and $$ C=AA^T$$ Let the Eigen value of B be $\lambda$ and Corresponding Eigen Vector be $X$, Then $$ BX=\lambda X \implies A^TAX=\lambda X$$ Pre Multiplying with $A$ we get

$$ AA^TAX=\lambda AX \implies CY=\lambda Y$$ So $\lambda$ is an Eigen Value of $C$ and note that $Y=AX$ is an Eigen Vector of $C$.

Ekaveera Gouribhatla
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    This proof is incorrect. If $AX=0$, then your equation says $C0=\lambda0$, which does not imply that $\lambda$ is an eigenvalue of $C$. – pre-kidney Apr 13 '15 at 00:55
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It is no coincidence. For any two square matrices $A$ and $B$, $AB$ is similar to $BA$. Similar matrices always have the same eigenvalues (but different eigenvectors).

rurouniwallace
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  • It should be for any two square matrices $A$ and $B$. – Milind Hegde Jun 16 '13 at 15:15
  • This is wrong: it holds if $A$ or $B$ is nonsingular, but if both are singular, there are counterexamples, i.e., $AB$ and $BA$ need not be similar. See the second answer here (the one with the most upvotes). – Vedran Šego Jun 16 '13 at 15:28