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Find all $2\times 2$ matrices that commute with

$$\left( \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right)$$

My progress:

I know that a square matrix commutes with itself, the identity matrix of that order, the null matrix of that order and any scalar matrix of that order.

The answer has been given as:

$$\left( \begin{array}{cc} m & 3n \\ n & m+2n \end{array} \right)$$

I don't understand how they're getting that form. Can someone please explain?

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5 Answers 5

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Let's call your matrix $$A = \left( \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right)$$

We want a matrix $X_{2\times 2} = \begin{pmatrix} a & b\\ c&d\end{pmatrix}$ such that $AX = XA$.

$$AX = \begin{pmatrix} 2a + 3c & 2b+3d\\ a + 4c&b+4d\end{pmatrix}$$

$$XA = \begin{pmatrix} 2a + b&3a + 4b\\2c+d & 3c+4d\end{pmatrix}$$

Now you have a system of equations in four variables:

$$2a + 3c = 2a + b \implies b = 3c$$

$$2b+3d = 3a + 4b$$

$$a + 4c = 2c + d$$

$$b+4d= 3c+4d$$

Solve the system of equations. (Note, if you do Gaussian Elimination, you'll have two of the four rows all zero.)

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You need $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) \left(\begin{array}{cc}2&3\\1&4\end{array}\right)= \left(\begin{array}{cc}2&3\\1&4\end{array}\right)\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Multiply out the matrices; that will give you four equations that connect $a,b,c$ and $d$. Then solve those equations.

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If you write down the unknown matrix as $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Then you want $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Write out the left and right sides as matrices with entries like 2a + 1b, etc. Set them equal. You get four equations in 4 unknowns. Solve, and you get the answer above.

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A somewhat more insightful approach is to think in terms of the eigendecomposition of the given matrix. Namely, if $A$ is a 2x2 diagonalisable non-degenerate matrix (which yours is), then $[A,B]=0$ iff $B$ is also diagonalisable in the same eigenbasis.

Case in point, your matrix decomposes as $$A = \begin{pmatrix}2&3\\1&4\end{pmatrix} = \frac14\begin{pmatrix}1&3\\-1 & 1 \end{pmatrix} \begin{pmatrix}5&0\\0&1\end{pmatrix} \begin{pmatrix}1&-3\\1 & 1 \end{pmatrix}.$$ It immediately follows that $[A,B]=0$ iff $B$ decomposes as $$B = \frac14\begin{pmatrix}1&3\\-1 & 1 \end{pmatrix} \begin{pmatrix}a&0\\0&b\end{pmatrix} \begin{pmatrix}1&-3\\1 & 1 \end{pmatrix} =\frac14\begin{pmatrix}a+3b & 3a-3b \\ a-b & 3a+b\end{pmatrix}$$ for any $a,b$. You can then get the parametrisation via $n,m$ by a simple change of variables.

See e.g. this post for the generalisation of these statements about finding commutants from eigendecompositions.

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If B be the required matrix then both AB, and BA should be defined and also AB = BA

∴ Order of B is 2×2 and let is be $\begin{bmatrix}p & q\\r & s\end{bmatrix}$

AB = $\begin{bmatrix}2 & 3\\1 & 4\end{bmatrix}$×$\begin{bmatrix}p & q\\r & s\end{bmatrix}$=$\begin{bmatrix}2p+3r & 2q+3s\\p+4r & q+4s\end{bmatrix}$

BA = $\begin{bmatrix}p & q\\r & s\end{bmatrix}$×$\begin{bmatrix}2 & 3\\1 & 4\end{bmatrix}$=$\begin{bmatrix}2p+q & 3p+4q\\2r+s & 3r+4s\end{bmatrix}$

2p+3r = 2p+q >>> ∴ r = q/3

2q+3s = 3p+4q >>> ∴ s = p+(2q/3)

p+4r = 2r+s >>> p+4r = 2r+(p+2q/3) >>> ∴ r = q/3

q+4s = 3r+4s >>> ∴ r = q/3

=> Let p=m & r=n

∴ q = 3n   &   s = m+2n

$\begin{bmatrix}m & 3n\\n & m+2n\end{bmatrix}$

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