Proving that $\det \left( I_n + A^{2} \right) \geq 0$ Suppose I have an $n \times n$  real-valued matrix, how can one prove that $\det \left( I_n + A^{2} \right) \geq 0$?
It is straightforward to prove that $\det \left( A^2 \right)$ is positive based on the multiplicative property of determinants, but I am unsure of how to handle the identity matrix.
 A: Here's a proof not passing through the complex numbers. Let's assume that $A$ is diagonal, since if we write $A=P^{-1}DP$ for a diagonal $D$, then
$$\det\left(I+P^{-1}D^2P\right)=\det\left(P^{-1}P+P^{-1}D^2P\right)=\det\left[P^{-1}\left(I+D^2\right)P\right]=\det\left(I+D^2\right).$$
Now the diagonal entries of $A^2$ are all nonnegative, so $I+A^2$ has all positive entries on the diagonal. Since the determinant of a diagonal matrix is the product of all the terms on the diagonal, it follows that the determinant of $I+A^2$ is positive.
A: The determinant is equal to the product of the (complex) eigenvalues. Note that the eigenvalues of $I+A^2$ are the $1+\lambda_i^2$ where the $\lambda_i$ are the complex eigenvalues of $A$.
If $\lambda_i$ is real, then $1+\lambda_i^2 \geq 0$
If $\lambda_i$ is a (non-real) complex number, then since the original matrix is real-valued it means that the complex conjugated  $\bar{\lambda_i}$ is also an eigenvalue and the product $(1+\lambda_i^2)(1+\bar{\lambda_i}^2)=(1+\lambda_i^2)(\overline{1+\lambda_i^2})=|1+\lambda_i^2|^2$
is still positive.
Therefore the complete product is positive, hence the given determinant is positive.
