Show that $\phi$ preserves invertibility.

Question

Let $A$ be the subalgebra of $M_4(\mathbb C)$ consisting of matrices of the form $\begin{bmatrix}a & b \\ 0 & c\end{bmatrix}$ where $a,b,c \in M_2(\mathbb C)$.

Now define $\phi:A \to A$ by $$\phi \bigg(\begin{bmatrix}a & b \\ 0 & c\end{bmatrix} \bigg) = \begin{bmatrix}a & b \\ 0 & c^t\end{bmatrix}.$$

To me it makes intuitive sense that $\phi$ preserves invertibility... However, I am not sure how I can prove it?

Answer 1

The determinant of a block triangular matrix is the product of the determinants of the matrices on the "diagonal". Therefore by the definition of $\phi$ we immediately see that $$\det \phi\left(\begin{bmatrix}a & b \\ 0 & c\end{bmatrix} \right) = \det a \det c^t= \det a \det c = \det\begin{bmatrix}a & b \\ 0 & c\end{bmatrix}.$$

Answer 2

Here is another way. It is not hard to see that $\begin{bmatrix} a&b\\ 0&c\end{bmatrix}$ is invertible if and only if both $a$ and $c$ are invertible; and, in that case, $$ \begin{bmatrix} a&b\\ 0&c\end{bmatrix}^{-1}=\begin{bmatrix} a^{-1}&-a^{-1}bc^{-1}\\ 0&c^{-1}\end{bmatrix}. $$