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

Obviously $$\det(A^2)$$ is (casework), but is the above matrix non-negative? $$\det(A)\det(A) \geq 0$$ as $$\det(A) > 0$$ or $$\det(A) < 0$$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.

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

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

then

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

• Excellent solution and very compact. Especially step $\det(\overline{M})=\overline{\det(M)}$ is very fast. Commented Dec 11, 2018 at 15:16

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).

• With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+\alpha I$, where $\alpha\ge 0$, is non-negative. It's worth to notice that even in the case of negative $\alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs. Commented Dec 11, 2018 at 11:50