# Is the Linear Transformation Between the Vector Space of 2 by 2 Matrices an Isomorphism? ## 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. Add to solve later

Contents

## 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.

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. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### Unit Vectors and Idempotent Matrices

A square matrix $A$ is called idempotent if $A^2=A$. (a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$....

Close