Linear Algebra

Is the Linear Transformation Between the Vector Space of 2 by 2 Matrices an Isomorphism?

Linear Transformation problems and solutions

Problem 528

Let $V$ denote the vector space of all real $2\times 2$ matrices.
Suppose that the linear transformation from $V$ to $V$ is given as below.
\[T(A)=\begin{bmatrix}
2 & 3\\
5 & 7
\end{bmatrix}A-A\begin{bmatrix}
2 & 3\\
5 & 7
\end{bmatrix}.\] Prove or disprove that the linear transformation $T:V\to V$ is an isomorphism.

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Hint.

A linear transformation $T$ is an isomorphism if it is both surjective (onto) and injective (one to one).

Recall that a linear transformation $T$ is injective if and only if the kernel is trivial, that is, $\ker(T)=\{\mathbf{0}\}$.

Proof.

We claim that $T$ is not an isomorphism.

Recall that an isomorphism means that $T$ is surjective (onto) and injective (one to one).
Thus, to disprove that $T$ is a linear transformation, it suffices to prove that $T$ is not surjective or $T$ is not injective.

Let us prove that $T$ is not injective.
Equivalently, we show that the kernel $\ker(T)$ is nontrivial, that is, $\ker(T)\neq \{O\}$, where $O$ is the $2\times 2$ zero matrix.

If $A$ is in $\ker(T)$, then we have $O=T(A)$ and it follows that
\[\begin{bmatrix}
2 & 3\\
5 & 7
\end{bmatrix}A=A\begin{bmatrix}
2 & 3\\
5 & 7
\end{bmatrix}.\]

In other words, a matrix is in the kernel if it commutes with the matrix $\begin{bmatrix}
2 & 3\\
5& 7
\end{bmatrix}$.

So out goal is to find nonzero matrices that commute with $\begin{bmatrix}
2 & 3\\
5& 7
\end{bmatrix}$.

For example, the $2\times 2$ identity matrix $I$ will do.
Hence the kernel of $T$ contains a nonzero matrix, hence $T$ is not injective.
Thus, $T$ is not isomorphism.

Comments

Note that we did not have to determine the kernel.
What we needed is to show that $\ker(T)\neq \{O\}$.
We proved this by showing that $I\in \ker(T)$.

Another matrix in the kernel is $\begin{bmatrix}
2 & 3\\
5& 7
\end{bmatrix}$ itself.

Also, note that we proved that $T$ is not an isomorphism because $T$ is not injective.
We could have tried to prove that $T$ is not surjective but this is harder.

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