is $\det(A^2 + I)$ always non negative?

Solution 1:

I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so

$\overline{iA} = -iA; \tag 1$

then

$\det(I + A^2) = \det ((I + iA)(I - iA)) = \det(I + iA) \det(I - iA)$ $= \det(I + iA)\det(\overline{I + iA)} = \det(I + iA) \overline{\det(I + iA)} \ge 0. \tag 2$

Solution 2:

It is possible also to use eigenvalues to prove the claim. Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.

Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.

As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.

Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $\ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.

$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$ $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$

We have received once again a pair of conjugated complex numbers.

The product of conjugated complex numbers is non-negative number ( $cc^*=\vert c\vert^2$) and the whole product is non-negative.

Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $\pm i \ \ $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).