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Find all $2\times 2$ matrices that commute with $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.

Solution and ask:

$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$

$$AX = \left( \begin{array}{cc} 1\cdot a + 1\cdot c & 1\cdot b + 1\cdot d \\ 0\cdot a + 0\cdot c & 0\cdot b + 0\cdot d \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right)$$

$$XA = \left( \begin{array}{cc} a\cdot 1 + b\cdot 0 & a\cdot 1 + b\cdot 0 \\ c\cdot 1+d\cdot 0 & c\cdot 1+d\cdot 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$

Since $AX = XA$, we obtain $$\left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$

so that $a+c = a$, $b+d = a$, $0 = c$ and $0 = c$.

Is the calculation correct?

What is the value of $a,b,c,d$ ?

$c = 0, a = ?, b = ?, d = ?$

Yuval Gat
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LSM
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    Your calculations seem right to me. I think the main thing to note is that there doesn't have to necessarily be a single specific value for $a, b, c, d$ since you're finding all matrices that commute (not just one). Granted, you have $c=0$ and $b+d=a$. So really your matrices would be any in the form $$X=\begin{bmatrix} b+d & b\ 0 & d \end{bmatrix}$$

    This basically means what $b,d$ are is up to the choice o the reader - they're arbitrary, they're variables, you could have multiple applicable matrices. So long as the form above is matched, you should get commuting.

     – PrincessEev Nov 04 '18 at 16:21
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    If it helps cement the idea to you, you could try showing $AX=XA$ by using $X$ of the form I specified above (keeping $b,d$ arbitrary). Sort of verifying the identity you found, in other words. It's a nice way to double-check an answer. – PrincessEev Nov 04 '18 at 16:22

3 Answers3

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Your calculations are correct. To summarize, you have found that $c=0$ and that $a=b+d$. This does not determine $a$, $b$ and $d$ uniquely; it only shows that such a matrix must be of the form $$X = \left( \begin{array}{cc} b+d & b \\ 0 & d \end{array} \right).$$ Now the question that remains is; does every matrix of this form commute with $A$?

Servaes
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@Servaes answer is exactly right, but let me add one more thing: the set of all such matrices $X$ forms a subspace of the space of $2 \times 2$ matrices.

One way to see this is to add two of them and see that the resulting matrix has the same form; the other is a little more abstract: suppose that $AX = XA$ and $AY = YA$. Then $A(X+Y) = AX + AY = XA + YA = (X+Y) A$, so the set of such matrices is closed under addition. It's also closed under scalar multiplication (which you can probably show for yourself).

When you come across a subspace like this, it's not a bad idea to ask yourself, "What's the dimension of the subspace? What's a basis for it?"

You can then choose a different matrix $A$, and ask yourself whether the dimension changes when $A$ changes, and if so, what's the relationship?

I know you're not likely to pursue this if this happens to just be a homework problem and you've got others to do as well, but I offer it as a possible general approach as you advance in mathematics.

John Hughes
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Here is a synthetic solution. Let $V$ denote the vector space of $2\times2$ matrices. Let $T:V\to V$ be the linear operator $TX=AX-XA$. The kernel of $T$ consists of eigenvectors of $T$ with eigenvalue $0$. Observe that $A$ is diagonalizable with eigenvalues $0$ and $1$ (well, this is obvious since $A$ is triangular, so the eigenvalues are the diagonal entries, and since they are distinct, $A$ must be diagonalizable).

From this thread, we know that the eigenspace of $T$ with eigenvalue $0$ is spanned by $u_0^t v_0$ and $u_1^t v_1$, where $u_\lambda$ is an eigenvector of $A^t$ with eigenvalue $\lambda\in\{0,1\}$, and $v_\lambda$ is an eigenvector of $A$ with eigenvalue $\lambda\in\{0,1\}$. We can take $$u_0=\begin{pmatrix}0\\1\end{pmatrix},\ u_1=\begin{pmatrix}1\\1\end{pmatrix},\ v_0=\begin{pmatrix}-1\\1\end{pmatrix},\ v_1=\begin{pmatrix}1\\0\end{pmatrix}.$$ So, $$u_0^tv_0=\begin{pmatrix}0&-1\\0&1\end{pmatrix},\ u_1^tv_1=\begin{pmatrix}1&1\\0&0\end{pmatrix}.$$ Thus, every $X\in \ker T$ must be of the form $$a\ u_0^tv_0+b\ u_1^tv_1=\begin{pmatrix}b&-a+b\\0&a\end{pmatrix}.$$ This gives the same answer as Servaes' solution.