If a matrix is invertible, is its multiplication commutative?

Question

The question is prompted by change of basis problems -- the book keeps multiplying the bases by matrix $S$ from the left in order to keep subscripts nice and obviously matching, but in examples bases are multiplied by $S$ (the change of basis matrix) from whatever side. So is matrix multiplication commutative if at least one matrix is invertible?

Answer 1

Definitely not. Yuan's comment is also not correct, diagonal matrices do not necessarily commute with non-diagonal matrices. Consider $$\left[\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right]\left[\begin{array}{cc} a & 0\\ 0 & b\end{array}\right]=\left[\begin{array}{cc} a & b\\ 0 & b\end{array}\right] $$

Changing the order I get $$ \left[\begin{array}{cc} a & 0\\ 0 & b\end{array}\right]\left[\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right]=\left[\begin{array}{cc} a & a\\ 0 & b\end{array}\right] $$ Which is different for $a\neq b$.

Hope that helps. (Sometimes change of basis matrices can go on different sides for different reasons, but without seeing the exact text you are talking about I can't comment)

Answer 2

In general, two matrices (invertible or not) do not commute. For example $$\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right) = \left(\begin{array}{cc} 2 & 1\\ 1 & 1\end{array}\right) $$ $$ \left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right) = \left(\begin{array}{cc} 1 & 1\\ 1 & 2\end{array}\right)$$

Also, to change a basis you usually need to conjugate and not just multiply from the left (or just right).

What you do know is that a matrix A commutes with $A^n$ for all $n$ (negative too if it is invertible, and $A^0 = I$), so for every polynomial P (or Laurent polynomial if A is invertible) you have that A commutes with $P(A)$.

Answer 3

If we ignore zero matrices, for commmutivity to hold it must also be part of the same automorphism group as the other matrix. Consider the following:

$AB=BA$

=> $B^{-1}AB=B^{-1}BA$

=> $A=B^{-1}AB$

Thus invertibility is required, but it is not sufficient.