Basis For Subspace Consisting of Matrices Commute With a Given Diagonal Matrix

Problem 287

Let $V$ be the vector space of all $3\times 3$ real matrices.
Let $A$ be the matrix given below and we define
\[W=\{M\in V \mid AM=MA\}.\] That is, $W$ consists of matrices that commute with $A$.
Then $W$ is a subspace of $V$.

Determine which matrices are in the subspace $W$ and find the dimension of $W$.

(a) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix},\] where $a, b, c$ are distinct real numbers.

(b) \[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix},\] where $a, b$ are distinct real numbers.

 
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Solution.

(a) Diagonal matrix with distinct diagonal entries

Let us first determine when a matrix $M$ commutes with $A$.
Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and suppose that $AM=MA$:
\[\begin{bmatrix}
a & 0 & 0 \\
0 & b &0 \\
0 & 0 & c
\end{bmatrix}
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}
\begin{bmatrix}
a & 0 & 0 \\
0 &b &0 \\
0 & 0 & c
\end{bmatrix}.\] Computing matrix products, we obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
ba_{2 1} & ba_{2 2} & ba_{2 3} \\
ca_{3 1} & ca_{3 2} &c a_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}b & a_{1 3}c \\
a_{2 1}a & a_{2 2}b & a_{2 3}c\\
a_{3 1}a & a_{3 2}b & a_{3 3}c
\end{bmatrix}. \tag{*}\] Compare the $(1,2)$ entries and we have $aa_{1 2}=ba_{1 2}$.
Since $a\neq b$, we must have $a_{1 2}=0$.

Similarly, comparing the off-diagonal entries and noting $a, b, c$ are distinct, we find that off diagonal entries $a_{i j} , i\neq j$ must be $0$.

Thus, $M$ commutes with $A$ if and only if
\[M=\begin{bmatrix}
a_{1 1} & 0 & 0 \\
0 & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix}.\]

Therefore, the subspace $W$ consists of all $3\times 3$ diagonal matrices:
\[W=\{W\in V\mid W \text{ is diagonal}\}.\] Then it is easy to see that the set $\{E_{1 1}, E_{2 2}, E_{3 3}\}$ is a basis for $W$, where $E_{i j}$ is the $3\times 3$ matrix whose $(i,j)$-entry is $1$ and the other entries are zero. Thus the dimension of $W$ is $3$.

(b) Diagonal matrix two diagonal entries are the same

Now consider the case
\[A=\begin{bmatrix}
a & 0 & 0 \\
0 &a &0 \\
0 & 0 & b
\end{bmatrix}.\] Let
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end{bmatrix}\] and compute $AM=MA$ as in part (a) (or you just need to replace $b, c$ in (*) by $a, b$, respectively) and obtain
\[\begin{bmatrix}
aa_{1 1} & aa_{1 2} & aa_{1 3} \\
aa_{2 1} & aa_{2 2} & aa_{2 3} \\
ba_{3 1} & ba_{3 2} & ba_{3 3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1 1}a & a_{1 2}a & a_{1 3}b \\
a_{2 1}a & a_{2 2}a & a_{2 3}b\\
a_{3 1}a & a_{3 2}a & a_{3 3}b
\end{bmatrix}. \] Comparing entries and noting $a\neq b$, we have
\[a_{1 3}=0, a_{2 3}=0, a_{3 1}=0, a_{3 2}=0.\]

Thus, $M$ commutes with $A$ is and only if
\[M=\begin{bmatrix}
a_{1 1} & a_{1 2} & 0 \\
a_{2 1} & a_{2 2} & 0 \\
0 & 0 & a_{3 3}
\end{bmatrix},\] and hence the subspace $W$ consists of such matrices.
From this, we see that the set
\[\{E_{1 1}, E_{1 2}, E_{2 1}, E_{2 2}, E_{3 3}\}\] is a basis for $W$, and we conclude that the dimension of $W$ is $5$.

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