On a particular matrix homomorphism

Question

I recently noticed that, for $2\times 2$ matrices, dividing one off-diagonal element by a constant while multiplying the other off-diagonal element by the same constant produces a matrix homomorphism.

i.e., if we call this operation $g$, then it's defined as

$$g\left(\begin{bmatrix}a & b\\c & d\end{bmatrix}\right) =\begin{bmatrix}a & (b\cdot\lambda)\\(c/\lambda) & d\end{bmatrix} $$

and then the homomorphism properties are:

$$ g(A) g(B) = g(AB) $$ (where juxtaposition indicates the usual matrix multiplication) and $$ g(A)+g(B) = g(A + B).$$

My question(s): Does this homomorphism have a name, any useful applications, and does it hint at any kind of deeper structure? Are there others like it?

Answer 1

You have $$ g \left( \begin{bmatrix}a & b\\c & d\end{bmatrix} \right) = \begin{bmatrix}\lambda & 0\\0 & 1\end{bmatrix} \begin{bmatrix}a & b\\c & d\end{bmatrix} \begin{bmatrix}\lambda & 0\\0 & 1\end{bmatrix}^{-1} = \begin{bmatrix}\lambda & 0\\0 & 1\end{bmatrix} \begin{bmatrix}a & b\\c & d\end{bmatrix} \begin{bmatrix}\lambda^{-1} & 0\\0 & 1\end{bmatrix}. $$

In fact, all automorphism of a matrix algebra over a field are inner (that is, of the form $x \mapsto t x t^{-1}$), see Skolem-Noether. See also this article for a more general situation.

Answer 2

Your homomorphism is an example of a similarity transform. A similarity transform is a transformation on matrices of the form $X\mapsto AXA^{-1}$ where $A$ is invertible. In your case $A = \begin{pmatrix}\lambda^{1/2} & 0 \\ 0 & \lambda^{-1/2}\end{pmatrix}$.