Minimal eigenvalue of a matrix is larger than $-1$

Question

Let $A \in \mathbb{R}^{n\times n}$ be a matrix such that $I + A$ is invertible. Define $B:= A + A^T + A^T A$. Show that the minimal eigenvalue of $B$ satisfies $\lambda_{min} > -1 $.

I tried to use that $\lambda_{min} = \min_{||x|| = 1} x^T B x$, and once the computation done I get $$x^T B x \, \geq \, x^T A x + (Ax)^T x = 2 \cdot \langle Ax, x \rangle $$ But then I don't see how to get the result.

Any suggestion? Thanks in advance!

Answer 1

Note that $$\tag1 I+B=(I+A)^T(I+A).$$ Hence with $y:=(I+A)x$, $$x^T(I+B)x=y^Ty\ge 0 $$ so that $\lambda_\min\ge-1$. As the factors on the right of $(1)$ are invertible, $-1$ cannot be an eigenvalue of $B$.

Answer 2

Consider

$$I+B=(I+A)(I+A^T)=CC^T \ \text{with} \ C:=I+A$$

which is invertible (hypothesis).

$CC^T$, being symmetric definite positive (see remark below), has all its eigenvalues $> 0$.

Therefore, the eigenvalues of $B=CC^T-I$ are all $> -1$.

Remark : for $X \neq 0, X^TCC^TX=(C^TX)^T(C^TX)=\|C^TX\|^2 \geq 0$

and in fact $>0$ because $C$ is invertible.