If real and imaginary part of a matrix are invertible then the matrix itself is invertible.

Question

Since invertibility means that the determinant is non-zero, we can rewrite the statement above as following:

Let $ M \in \mathbb{K}^{n \times n}$ then:

$$\det(\Re(M)) \neq 0 \wedge \det(\Im(M)) \neq 0 \Longrightarrow \det(M) \neq 0$$

I didn't manage to find a counterexample, but then I remembered that I could prove the statement by proving it's negation wrong:

Let $ M \in \mathbb{K}^{n \times n}$ then:

$$\det(\Re(M)) = 0 \vee \det(\Im(M)) = 0 \Longrightarrow \det(M) = 0$$

And this is obviously false, if you just consider real invertible matrices as a counterexample. Which means that the original statement is true.

Proving it this way feels a bit too easy. Is the proof correct? Are there other ways to prove this?

Answer 1

This is false. E.g. $X=\pmatrix{1&0\\ 0&-1}$ and $Y=\pmatrix{0&1\\ 1&0}$ are invertible, but $$ X+iY=\pmatrix{1&i\\ i&-1}=\pmatrix{1\\ i}\pmatrix{1& i} $$ is singular.