# Finding commuting matrices

• Mar 31st 2013, 07:00 PM
jimtehma
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
• Mar 31st 2013, 08:07 PM
Gusbob
Re: Finding commuting matrices
Quote:

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

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

[a 0]
[c c+a] ?

From $b=-b=0$, we $b=0$. Substituting this into $a=a-b$ gives $a=a$, so we can get rid of this condition, as it tells us absolutely nothing. The restriction $-a=c-d$ is the same as saying $d=c+a$. Since there are no other restrictions on $a$ and $c$, they can be anything.

To summarise, $a=a,b=0,c=c,d=a+c$. Putting this back into your original matrix $\begin{bmatrix}a&b\\ c& d\end{bmatrix}$ gives the desired matrix $\begin{bmatrix}a& 0\\ c& c+a\end{bmatrix}$
• Apr 1st 2013, 09:01 AM
jimtehma
Re: Finding commuting matrices
Thanks for the quick reply. I understand the steps. I have a followup question though!
Quote:

Originally Posted by Gusbob
The restriction $-a=c-d$ is the same as saying $d=c+a$. Since there are no other restrictions on $a$ and $c$, they can be anything.

Getting $d=c+a$ from $-a=c-d$ is rather trivial, but why don't we do the same for $c$ and $a$ and get $c=d-a$ and $a=d-c$ and use them in the answer matrix too? And thanks for the using the tex shortcuts. I saw them in the reply quote and used them in my post.
• Apr 1st 2013, 10:52 AM
HallsofIvy
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
$\begin{bmatrix}a & 0 \\ c & c+a\end{bmtatrix}$
Doing it your way, you get
$\begin{bmatrix} a & 0 \\ d- a & d \end{bmatrix}$
and
$\begin{bmatrix} d- c & 0 \\ c & d\end{bmatrix}$
But those are not different sets of matrices.
• Apr 1st 2013, 12:44 PM
jimtehma
Re: Finding commuting matrices
Thanks for the quick reply. I see all your "solution" matrices use only one variation of the $-a=c-d$ condition. Is there a rule that dictates this? Can you write the solution as $\begin{bmatrix} d-c & 0 \\ d-a & c+a \end{bmatrix}$ or $\begin{bmatrix}a & 0 \\c & d\end{bmatrix}$ ? This is what confuses me. Getting to the solution is rather trivial, but I'm not sure how should I write the solution as.
• Apr 1st 2013, 08:12 PM
Gusbob
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 $-a=c-d$ 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).
• Apr 2nd 2013, 09:37 AM
jimtehma
Re: Finding commuting matrices
Thanks Gusbob. That clears things up.