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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?

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Though I guess $S$ has to be invertible in this case too, since it is changing a set of independent vectors to another set of independent vectors... But the question still persists (not just with respect to change of basis matrices I guess). –  LinAlgStudent Feb 11 '11 at 6:50
Only diagonal matrix commute with other matrix. –  Yuan Feb 11 '11 at 6:51
@Rasmus: Yuan said "only", so I would say that it is true. Rephrased: "If a matrix commutes with every invertible matrix, it is diagonal." Of course, it is better to replace "diagonal" by "scalar" since then the other implication also holds... –  Pete L. Clark Feb 11 '11 at 13:40
Note to the OP: your question is a perfectly fine one, but it's also a question you could probably have answered for yourself if you tried a few examples (in a sense I will not try to make precise here, "most" pairs of invertible matrices do not commute). The practice of testing one's questions out with actual examples is both useful and enjoyable -- in sister disciplines, it is called the "scientific method". I recommend it to you most highly. –  Pete L. Clark Feb 11 '11 at 13:47
@Pete:If the comment was meant this way, I think that it is formulated in a misleading way. –  Rasmus Feb 11 '11 at 15:05
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2 Answers

up vote 10 down vote accepted

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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Sorry, maybe I make a mistake. The multiplication of a scalar and the identity matrix commute with others. But I forgot the proof. –  Yuan Feb 11 '11 at 7:14
That is of course true since constants commute, and the identity can be put wherever we want. But a diagonal matrix and a constant multiple of the identity are very different. (I have definitely made that commuting mistake before too, it seems quite reasonable actually...) –  Eric Naslund Feb 11 '11 at 7:32
Diagonal matrices are very good examples. A diagonal matrix with distinct diagonal entries commutes only with other diagonal matrices. –  Jonas Meyer Feb 11 '11 at 13:50
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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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This is a good answer, +1. Not that it is much different than mine, but I don't know why some answers are never voted. –  Eric Naslund Feb 11 '11 at 7:58
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