# 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 $V$.

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## Hint.

Show that

- $\phi$ is a homomorphism
- $\phi$ is injective.
- $\phi$ 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.

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