Show that a set of matrices that commute is a subspace?

Question

I am having problems with this question, would appreciate any help.

Let $$A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix}$$

Show that the set of 2-by-2 matrices that commute with $A$ (i.e., matrices B for which $AB = BA$) is a subspace of the vector space of 2-by-2 matrices, and find the dimension of and a basis for this subspace.

Right now, I am trying to show that dimension of this matrix is 4, dimension of matrices that do not commute is 2, thus dimension of matrices that do commute is 2. However, I'm not sure how I should go about proving that the dimension of matrices which do not commute is 2. Any idea how should I go about doing this?

The set of matrices which do not commute is not a subspace. Indeed, it does not contan zero, for example! — Mariano Suárez-Álvarez, Feb 06 '15 at 23:48
Let $B = \begin{bmatrix} a & b \ c & d \ \end{bmatrix}$ and find out what $AB = BA$ means in terms of the $a, b, c, d$. — Arthur, Feb 06 '15 at 23:48
Dimension is not complementary. If the complement of a $2$ dimensional subspace were somehow a subspace, its dimension would be 4. — Matt Samuel, Feb 06 '15 at 23:48
@MattSamuel, that does not make any more sense than claiming that «if the complement were a kitten, it would be a very cute one» :-) — Mariano Suárez-Álvarez, Feb 06 '15 at 23:50
Ok, what do you think is the best way to solve this problem then? Is my method appropriate? — thbcm, Feb 07 '15 at 00:03

score 2 · Accepted Answer · answered Feb 07 '15 at 00:35

Since the matrices that do not commute is not a subspace, we need to take an approach different from the one you outline in your last paragraph. A better approach is the one given by @Arthur in his comment to the question. We want to find all matrices $B$ where $AB=BA$. We let

$$B = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$

We then want

$$\begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix} \begin{bmatrix}a & b \\c & d \\\end{bmatrix} =\begin{bmatrix}a & b \\c & d \\\end{bmatrix} \begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix}$$

$$\begin{bmatrix}a+2c & b+2d \\2a+c & 2b+d \\\end{bmatrix} =\begin{bmatrix}a+2b & 2a+b \\c+2d & 2c+d \\\end{bmatrix}$$

Then the upper left corners give us

$$a+2c=a+2b$$ $$b=c$$

The upper right corners give us

$$b+2d=2a+b$$ $$a=d$$

This means that

$$B = \begin{bmatrix}a & b \\b & a \\\end{bmatrix}$$

Checking, we get

$$AB=\begin{bmatrix}a+2b & 2a+b \\2a+b & a+2b \\\end{bmatrix}=BA$$

so that answer is correct. We see that

$$B = a\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}+b\begin{bmatrix}0 & 1 \\1 & 0 \\\end{bmatrix}$$

Therefore we conclude that the matrices that commute with $A$ are a two-dimensional subspace of the vector space of $2$-by-$2$ matrices, with the basis

$$\left\{\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}, \begin{bmatrix}0 & 1 \\1 & 0 \\\end{bmatrix}\right\}$$

Thanks for the complete explanation! – thbcm Feb 07 '15 at 01:08 — thbcm, Feb 07 '15 at 01:08

Show that a set of matrices that commute is a subspace?

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