# Isomorphism of the Endomorphism and the Tensor Product of a Vector Space

## Problem 80

Let $V$ be a finite dimensional vector space over a field $K$ and let $\End (V)$ be the vector space of linear transformations from $V$ to $V$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$ be a basis for $V$.
Show that the map $\phi:\End (V) \to V^{\oplus n}$ defined by $f\mapsto (f(\mathbf{v}_1), \dots, f(\mathbf{v}_n))$ is an isomorphism.
Here $V^{\oplus n}=V\oplus \dots \oplus V$, the direct sum of $n$ copies of $V$.

## Hint.

Show that

1. $\phi$ is a homomorphism
2. $\phi$ is injective.
3. $\phi$ is surjective.

You may want to consider the matrix representation of a linear transformation $f$ with respect to the given basis $\mathbf{v}_1, \dots, \mathbf{v}_n$.

## Proof.

### We first show that $\phi$ is a linear transformation.

Let $f, g\in \End(V)$. Then
\begin{align*}
\phi(f+g)&=((f+g)(\mathbf{v}_1), \dots, (f+g)(\mathbf{v}_n))\\
&=(f(\mathbf{v}_1)+g(\mathbf{v}_1), \dots, f(\mathbf{v}_n)+g(\mathbf{v}_n))\\
&=(f(\mathbf{v}_1), \dots, f(\mathbf{v}_n))+(g(\mathbf{v}_1), \dots, g(\mathbf{v}_n))\\
&=\phi(f)+\phi(g).
\end{align*}
For $f\in \End(V)$, $c\in K$, we have
\begin{align*}
\phi(cf)&= (cf(\mathbf{v}_1), \dots, cf(\mathbf{v}_n))\\
&=c(f(\mathbf{v}_1), \dots, f(\mathbf{v}_n))\\
&=c\phi(f).
\end{align*}
Therefore, $\phi$ is a linear transformation.

We prove that $\phi$ is bijection.

### To prove $\phi$ is injection,

suppose that $\phi(f)=\mathbf{0}$ for $f \in \End(V)$. That is, we have $f(\mathbf{v}_1)=\dots=f(\mathbf{v} _n)=0$.
For any vector $\mathbf{v}\in V$, we have a linear combination
$\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n$ for $c_1,\dots, c_n \in K$. Then
\begin{align*}
f(\mathbf{v})&=f(c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n)\\
&=c_1f(\mathbf{v}_1)+\cdots+c_n f(\mathbf{v}_n)\\
&=c_1 0+\dots +c_n 0=0.
\end{align*}
Thus $f(\mathbf{v})=0$ for any $\mathbf{v} \in V$, hence $f=0$ and $\phi$ is injective.

### To show that $\phi$ is surjective,

let $(\mathbf{u}_1, \dots, \mathbf{u}_n)\in V^{\oplus n}$ be an arbitrary vector in $V^{\oplus n}$.
Using the basis $\mathbf{v}_1, \dots, \mathbf{v}_n$, we have linear combinations
$\mathbf{u}_i=c_{1i}\mathbf{v}_1+\cdots+c_{ni} \mathbf{v}_n$ for some $c_{1i}, \dots, c_{ni} \in K$ for $1\leq i \leq n$.
Let $f \in \End(V)$ be a linear transformation whose matrix representation with respect to the basis $B:=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is given by
$[f]_B=\begin{bmatrix} c_{11} & c_{12} & \dots & c_{1n} \\ c_{21} &c_{22} & \dots & c_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ c_{n1} & c_{n2} & \dots & c_{nn} \end{bmatrix}.$ Since $[f]_B=[\,[f(\mathbf{v}_1)]_B \dots, [f(\mathbf{v}_n)]_B\,]$, we have
$[f(\mathbf{v}_i)]_B=\begin{bmatrix} c_{1i} \\ c_{2i} \\ \vdots \\ c_{ni} \end{bmatrix}.$ Namely, we have
$f(\mathbf{v}_i)=c_{1i}\mathbf{v}_1+\cdots+c_{ni} \mathbf{v}_n=\mathbf{u}_i.$ Therefore $\phi(f)= (\mathbf{u}_1, \dots, \mathbf{u}_n)\in V^{\oplus n}$ and $\phi$ is surjective.

In conclusion, $\phi$ is a bijective homomorphism from $\End(V)$ to $V^{\oplus n}$. Thus $\End(V)$ and $V^{\oplus n}$ are isomorphic.

Let $V$ be the set of all $n \times n$ diagonal matrices whose traces are zero. That is, \begin{equation*} V:=\left\{...