If a matrix is invertible, is its multiplication commutative? 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?
 A: 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 $P(A)$.
A: 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)
