Prove a invertible square matrix has imaginary numbers

inverselinear algebramatricesproof-writing

Let $C$ be an invertible square matrix. Assume that $C$ is a complex number matrix, and that at least one entry is not entirely a real number (so including the complex number $i$). Prove or disprove that the inverse of $C$, $C^{-1}$ has at least one entry that is also not entirely a real number. I did some rough work, and this seems to be true. We can calculate the inverse by using the adjunct or adjoint of the matrix. When you calculate that, you calculate for cofactors. When you calculate for cofactors, you are also calculating for determinants. This means if there's a complex number involved, it is always going to be there somewhere. Now, how do I write this as a formal proof?

Best Answer

Suppose that all entries of an invertible matrix $A$ are real numbers. Then its determinant is real (and not $0$) and all entries of the adjugate matrix of $A$ are real numbers (its entries are determinants of real matrices). So, $A^{-1}$ is a real matrix too.

Now, apply this to $C^{-1}$: if it was a real matrix, $C$ would be a real matrix too.

By the way, this works for every field.