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I know this should be really simple, but for some reason I can;t figure it out. I need to find all matrices which commute with the following 2x2 matrix A:

$$ B=\begin{bmatrix} 1 & -1\\ 5 & -4 \end{bmatrix} $$

I've tried using the definition of two commuting matrices, AB=BA and a generic matrix A where $ A=\begin{bmatrix} a & b\\ c & d \end{bmatrix} $ to generate a system of linear equations which I then solve to obtain a, b, c, and d such that A and B commute as follows:

$$ AB = \begin{bmatrix} a+5b & -a-4b\\ c+5d & -c-4d \end{bmatrix}=BA=\begin{bmatrix} a-c & b-d\\ 5a-4c & 5b-4d \end{bmatrix}$$ This gives the equations: $$ c+ 5b=0 $$ $$ 5b-d+a=0 $$ $$ 5c-5a+5d=0 $$ $$5b+c=0 $$

This can then be converted to a matrix and reduced to row echelon form:

$$ \begin{bmatrix} 1 & 5 & 0 & -1 & 0 \\ 0 & 5 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Introducing a parameter s for the variable c and a parameter t for the variable d, it seems this system can be solved to give: $ a=s+t $, $b =\frac {-s} {5} $, $ c=s $, and $ d=t $.

Substituting these values into AB or BA yields the following matrix which, by rights should represent all matrices which commute with c: $$ \begin{bmatrix} t & \frac {-s} {5} -t \\ s+5t & -s-4t \end{bmatrix} $$

However, the given solution is: $$ \begin{bmatrix} a & b \\ -5b & 5b+a \end{bmatrix} $$

Our solutions seem quite similar (except of course for different variable names), but I jsut can;t for the life of me figure out where I've gone wrong, any help would be greatly appreciated!

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  • $\begingroup$ Try substitution and see if they agree. $\endgroup$ – Henricus V. Jan 12 '16 at 5:35
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You've done everything just fine. You just need to perform the transformation $(a,b) \to (t,\frac{-s}{5}-t)$.

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  • $\begingroup$ Oh wow, I feel stupid now. Let a=t, $b=\frac{-s}{5}$ then s+5t=-5b and -s-4t=5b+a, thats so simple! One quick question though, is there any reason for doing this, or any way I should know that this is required? Thank you so much by the way. $\endgroup$ – CoffeeCrow Jan 12 '16 at 5:44
  • $\begingroup$ When one gets a solution to a problem in linear algebra which is a subspace, rather than a point (such as in this instance, where row reduction yielded a system of equations) then there is no canonical way to express it. The explicit form for the solution one gets depends on the basis for the solution subspace which is singled out during algebraic manipulation. At the end of the day, though, it doesn't matter what basis you use to describe the solution space, so if you get two solutions which look different, a natural thing to check is whether they can be related by a linear transformation. $\endgroup$ – Archaick Jan 12 '16 at 5:50
  • $\begingroup$ Don't feel stupid, everyone's been there, I don't care how smart they are. It was a good question! $\endgroup$ – Archaick Jan 12 '16 at 5:50
  • $\begingroup$ Okay, I think that mostly makes sense! we haven't really covered subspaces and such yet, which I guess explains why I had such a hard time! Thank you again for all the help, I really appreciate it. $\endgroup$ – CoffeeCrow Jan 12 '16 at 6:14
  • $\begingroup$ Don't mention it. Linear algebra takes a while to get used to, but given how well you handled this problem, it seems you'll get a good handle on it really quickly. $\endgroup$ – Archaick Jan 12 '16 at 6:16

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