Finding commuting matrices

The exercise says to find all the matrices that commute with

[ 1 0]

[-1 0]

I do

[ 1 0] _ [a b] ___ [a b] ___ [a b] _ [ 1 0] __ [a-b 0]

[-1 0] x [c d] = [-a -b] and [c d] x [-1 0] = [c-d 0].

(I inserted some underscores so the above reads better. The underscores don't denote anything)

Since the two resulting matrices must be equal I get a=a-b, b=0, -a=c-d and -b=0.

At this point I get stuck so I look at the solution which says "for equality it is necessary to have a=a-b, b=0, -a=c-d. thus those matrices that commute with the given matrix are all matrices of the form

[a 0]

[c c+a] where a and c can take any real values."

I don't understand this. How do you get from "a=a-b, b=0, -a=c-d" to

[a 0]

[c c+a] ?

Any ideas? The book (Linear Algebra: With Applications - Gareth Williams - Google Books) doesn't show how to find commuting matrices. I couldn't find any such instructions at the web.

Thanks

Re: Finding commuting matrices

Re: Finding commuting matrices

Re: Finding commuting matrices

You could, it would give you the same solution written in a slightly different way. Using Gusbob's method, you get

Doing it your way, you get

and

But those are not different sets of matrices.

Re: Finding commuting matrices

Thanks for the quick reply. I see all your "solution" matrices use only one variation of the condition. Is there a rule that dictates this? Can you write the solution as or ? This is what confuses me. Getting to the solution is rather trivial, but I'm not sure how should I write the solution as.

Re: Finding commuting matrices

Quote:

Originally Posted by

**jimtehma** Thanks for the quick reply. I see all your "solution" matrices use only one variation of the

condition. Is there a rule that dictates this?

Heuristically, you have 3 variables and 1 equation. Fixing one of the variables leaves the other two free. It doesn't matter which one you fix since you get the same information out of it anyways. Your first matrix is valid, though it is likely to be confusing to whomever is reading it. Since any of the two variables completely determine the other (solution space is 2 dimensions), your second matrix is not correct (because you have a,c,d free, essentially neglecting the condition -a = c-d).

Re: Finding commuting matrices

Thanks Gusbob. That clears things up.