Show that the eigenvalues of $AA^T$ and $A^TA$ are non-negative.

Question

Let $A\in \mathbb{R^{m\times n}}$. Show that the eigenvalues of $AA^T$ and $A^TA$ are non-negative.

I could just apply the definition of an eigenvalue for $AA^T$ (or $A^TA$), but I don´t know how to determine the sign of the eigenvalue. Here is what I tried: suppose that $\lambda<0$ is an eigenvalue and proceed via contradiction. However, I feel that this might not be correct.

Answer 1

If $\langle,\rangle$ is the standard scalar product, $\langle A(x),y\rangle=\langle x,A^T(y)\rangle$, we deduce that if $AA^T(x)=cx$, $\langle AA^T(x),x\rangle=\langle A^T(x),A^T(x)\rangle=c\langle x,x\rangle$. We conclude that $c\geq 0$ since $\langle A^T(x),A^T(x)\rangle\geq 0$ and $\langle x,x\rangle \gt 0$.

Answer 2

Let $\lambda$ be an eigenvalue of $A^T A$ and $v$ an associated unit eigenvector.

(We can obtain a unit eigenvector from a non-unit one by scaling).

Then we have

$\lambda = ||v||^2 \lambda$

$= v^T \lambda v$

$= v^T A^T A v$

$= (Av)^T(Av)$

$= ||Av||^2 \ge 0$

The same proof will work for $A A^T$:

$ \lambda =$ ... $= ||A^T v||^2 \ge 0$

Answer 3

Given

$A \in \Bbb R^{n \times m}, \tag 1$

we have

$A^T \in \Bbb R^{m \times n}, \tag 2$

whence

$AA^T \in R^{n \times n}; \tag 3$

we observe that

$(AA^T) = (A^T)^TA^T = AA^T, \tag 4$

that is, $AA^T$ is a symmetric matrix operating on $\Bbb R^n$,

$AA^T: \Bbb R^n \to \Bbb R^n, \tag 5$

thus if $\mu$ is an eigenvalue of $AA^T$,

$\exists 0 \ne x \in \Bbb R^n, AA^Tx = \mu x, \tag 6$

then

$\mu \in \Bbb R; \tag 7$

now for any $p \in \Bbb N$ we let

$\langle \cdot, \cdot \rangle: \Bbb R^p \times \Bbb R^p \to \Bbb R \tag 8$

denote the standard inner product on $\Bbb R^p$; then

$\mu \langle x, x \rangle_n = \langle x, \mu x \rangle_n = \langle x, AA^Tx \rangle_n = \langle A^Tx, A^Tx \rangle_m \ge 0; \tag 9$

since

$\langle x, x \rangle_n > 0, \tag{10}$

this forces

$\mu = \dfrac{\langle A^Tx, A^Tx \rangle_m}{\langle x, x \rangle_n} \ge 0. \tag{11}$

By interchanging the roles of $A$ and $A^T$ in the above, it is easily seen that virutally the same argument yields

$\mu = \dfrac{\langle Ax, Ax \rangle_n}{\langle x, x \rangle_m} \ge 0, \tag{12}$

where $\mu$ is now an eigenvalue of $A^TA$.