Linear Transformation $T(X)=AX-XA$ and Determinant of Matrix Representation

Linear algebra problems and solutions

Problem 330

Let $V$ be the vector space of all $n\times n$ real matrices.
Let us fix a matrix $A\in V$.
Define a map $T: V\to V$ by
\[ T(X)=AX-XA\] for each $X\in V$.

(a) Prove that $T:V\to V$ is a linear transformation.

(b) Let $B$ be a basis of $V$. Let $P$ be the matrix representation of $T$ with respect to $B$. Find the determinant of $P$.

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(a) Prove that $T:V\to V$ is a linear transformation.

To prove $T$ is a linear transformation, we need to show the following properties.

  1. For any $X, Y\in V$, we have $T(X+Y)=T(X)+T(Y)$.
  2. For any $X\in V, r\in \R$, we have $T(rX)=rT(X)$.

To check condition 1, let $X, Y \in V$. Then we have
T(X+Y)&=A(X+Y)-(X+Y)A && \text{by definition of $T$}\\
&=T(X)+T(Y) && \text{by definition of $T$}.
Hence condition 1 is met.

To verify condition 2, let $X\in V, r\in \R$.
Then we have
T(rX)&=A(rX)-(rX)A && \text{by definition of $T$}\\
&=rAX-rXA && \text{$r$ is a scalar}\\
&=rT(X) && \text{by definition of $T$}.
So condition 2 is also met, hence $T$ is a linear transformation.

(b) Find the determinant of the matrix representation of $T$.

Let $B$ be a basis of the vector space $V$ and let $P$ be the matrix of the linear transformation $T$ with respect to $B$. We prove that the determinant of $P$ is zero.

Let $I$ be the $n\times n$ identity matrix. Then we have
where $O$ is the $n\times n$ zero matrix.
Since $T(O)=O$, this implies that the linear transformation $T$ is not injective, hence $P$ is a singular matrix.

Let us explain the details.
Let $v=[I]_B \in \R^{n^2}$ be the coordinate vector of $I$ with respect to the basis $B$.
Then since $I\neq O$, the vector $v$ is not zero.
Then $T(I)=O$ implies that
\[Pv=0\in \R^{n^2}.\] As the nonzero vector $v$ is a solution of the matrix equation $Px=0$, the matrix $P$ is singular.

Since $P$ is singular, the determinant of $P$ is zero.

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