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

Linear algebra problems and solutions

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.

 
LoadingAdd to solve later
Sponsored Links

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 $5$.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Prove a Given Subset is a Subspace  and Find a Basis and DimensionProve a Given Subset is a Subspace and Find a Basis and Dimension Let \[A=\begin{bmatrix} 4 & 1\\ 3& 2 \end{bmatrix}\] and consider the following subset $V$ of the 2-dimensional vector space $\R^2$. \[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\] (a) Prove that the subset $V$ is a subspace of $\R^2$. (b) Find a basis for […]
  • Basis and Dimension of the Subspace of All Polynomials of Degree 4 or Less Satisfying Some Conditions.Basis and Dimension of the Subspace of All Polynomials of Degree 4 or Less Satisfying Some Conditions. Let $P_4$ be the vector space consisting of all polynomials of degree $4$ or less with real number coefficients. Let $W$ be the subspace of $P_2$ by \[W=\{ p(x)\in P_4 \mid p(1)+p(-1)=0 \text{ and } p(2)+p(-2)=0 \}.\] Find a basis of the subspace $W$ and determine the dimension of […]
  • Vector Space of Polynomials and Coordinate VectorsVector Space of Polynomials and Coordinate Vectors Let $P_2$ be the vector space of all polynomials of degree two or less. Consider the subset in $P_2$ \[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align*} &p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\ &p_3(x)=2x^2, &p_4(x)=2x^2+x+1. \end{align*} (a) Use the basis […]
  • Dimension of the Sum of Two SubspacesDimension of the Sum of Two Subspaces Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$. Then prove that \[\dim(U+V) \leq \dim(U)+\dim(V).\]   Definition (The sum of subspaces). Recall that the sum of subspaces $U$ and $V$ is \[U+V=\{\mathbf{x}+\mathbf{y} \mid […]
  • The Subset Consisting of the Zero Vector is a Subspace and its Dimension is ZeroThe Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$. Then prove that $V$ is a subspace of $\R^n$.   Proof. To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]
  • Symmetric Matrices and the Product of Two MatricesSymmetric Matrices and the Product of Two Matrices Let $A$ and $B$ be $n \times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$.   Hint. A matrix $A$ is called symmetric if $A=A^{\trans}$. In […]
  • Linear Properties of Matrix Multiplication and the Null Space of a MatrixLinear Properties of Matrix Multiplication and the Null Space of a Matrix Let $A$ be an $m \times n$ matrix. Let $\calN(A)$ be the null space of $A$. Suppose that $\mathbf{u} \in \calN(A)$ and $\mathbf{v} \in \calN(A)$. Let $\mathbf{w}=3\mathbf{u}-5\mathbf{v}$. Then find $A\mathbf{w}$.   Hint. Recall that the null space of an […]
  • A Matrix Commuting With a Diagonal Matrix with Distinct Entries is DiagonalA Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal Let \[D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 &d_2 & \dots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \dots & d_n \end{bmatrix}\] be a diagonal matrix with distinct diagonal entries: $d_i\neq d_j$ if $i\neq j$. Let $A=(a_{ij})$ be an $n\times n$ matrix […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Problems and solutions in Linear Algebra
Determine Whether a Set of Functions $f(x)$ such that $f(x)=f(1-x)$ is a Subspace

Let $V$ be the vector space over $\R$ of all real valued function on the interval $[0, 1]$ and let...

Close