Singularity of a Symmetric Complex Matrix

Question

Let $B\in\mathbb{R}^{n\times n}$ be a real symmetric positive definite matrix. Note that since $B$ is positive definite, it is non-singular. Does there exist any real symmetric matrix $A\in\mathbb{R}^{n\times n}$ such that $$ \det(A+iB) = 0 $$ where $i$ is the imaginary unit?

Answer 1

The answer is no. One way to see this is as follows: the matrix $B$ has a positive-definite square root $B^{1/2}$. Note that $A + iB$ is invertible if and only if the matrix $$ B^{-1/2}(A + iB)B^{-1/2} = B^{-1/2}AB^{-1/2} + iI $$ is invertible. The eigenvalues of $B^{-1/2}AB^{-1/2} + iI$ have the form $\lambda + i$, where $\lambda$ is an eigenvalue of $B^{-1/2}AB^{-1/2}$. Because $B^{-1/2}AB^{-1/2}$ is (real and) symmetric, its eigenvalues are real. So, the eigenvalues of $B^{-1/2}AB^{-1/2} + iI$ are non-zero, which means that $B^{-1/2}AB^{-1/2} + iI$ is necessarily invertible.

The conclusion follows.