# A Linear Transformation $T: U\to V$ cannot be Injective if $\dim(U) > \dim(V)$

## Problem 541

Let $U$ and $V$ be finite dimensional vector spaces over a scalar field $\F$.
Consider a linear transformation $T:U\to V$.

Prove that if $\dim(U) > \dim(V)$, then $T$ cannot be injective (one-to-one).

Contents

## Hints.

You may use the folowing facts.

A linear transformation $T: U\to V$ is injective if and only if the nullity of $T$ is zero.

For the proof of this fact, see the post ↴
A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero

Rank-Nullity Theorem
For a linear transformation $T: U \to V$, we have
$\nullity(T)+\rk(T)=\dim(U).$

We give two proofs. The first one uses the rank-nullity theorem.
The second one avoids using the theorem.

## Proof 1.

By the rank-nullity theorem, we have
$\nullity(T)+\rk(T)=\dim(U).$ Note that the rank of $T$ is the dimension of the range $\calR(T)$, which is a subspace of the vector space of $V$.
Hence it yields that
$\rk(T)=\dim(\calR(T)) \leq \dim(V).$

It follows that
\begin{align*}
\nullity(T)=\dim(U)-\rk(T) > \dim(U)-\dim(V) > 0,
\end{align*}
where the last inequality follows from the assumption.

This implies that the nullity is nonzero, hence $T$ is not injective.
(See the hints above.)

## Proof 2.

In the second proof, we prove $T$ is injective without using the rank-nullity theorem.

Seeking a contradiction, assume that $T: U\to V$ is injective.
Let $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ be a basis of the vector space $U$, where $n=\dim(U)$.
Put $\mathbf{v}_1=T(\mathbf{u}_1), \dots, \mathbf{v}_n=T(\mathbf{u}_n)$.

We claim that the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent in $V$.
To see this, consider the linear combination
$c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n=\mathbf{0}_V,$ where $c_1, \dots, c_n$ are scalars in $\F$.
Then we have
\begin{align*}
\mathbf{0}_V&=c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n\\
&=c_1T(\mathbf{u}_1)+\cdots+c_n T(\mathbf{u}_n)\\
&=T(c_1\mathbf{u}_1+\cdots + c_n \mathbf{u}_n).
\end{align*}
Since $T$ is assumed to be injective, we must have
$c_1\mathbf{u}_1+\cdots + c_n \mathbf{u}_n=\mathbf{0}_U.$ Because $\mathbf{u}_1, \dots, \mathbf{u}_n$ are basis vectors, they are linearly independent.
It follows that
$c_1=\cdots=c_n=0,$ and hence the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent.

As the vector space $V$ contains $n$ linearly independent vectors, we see that
$\dim(V) \geq n=\dim(U),$ which contradicts the assumption that $\dim(U) > \dim(V)$.
Therefore, $T$ cannot be injective.

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##### A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero

Let $U$ and $V$ be vector spaces over a scalar field $\F$. Let $T: U \to V$ be a linear...

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