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?
If a matrix is invertible, is its multiplication commutative?
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2@Pete:If the comment was meant this way, I think that it is formulated in a misleading way. – 2011-02-11
2 Answers
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)
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0Diagonal matrices are very good examples. A diagonal matrix with distinct diagonal entries commutes *only* with other diagonal matrices. – 2011-02-11
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 $P(A)$.
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0This is a good answer, +1. Not that it is much different than mine, but I don't know why some answers are never voted. – 2011-02-11