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

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

Well I could go many ways just by defining a eigenvalue for $$AA^T$$ (or $$A^TA$$). But I don´t know how to find an statement about the sign of the eigenvalues. I tried to suppouse that there exists an eigenvalue $$\lambda<0$$ and from there get a contradiccion, but I feel is not the right way.

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 $$A A^T$$:

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

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 $$\langle x,x\rangle \gt 0$$.

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 $$A^TA$$.