# 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).

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Contents

## Hints.

You may use the folowing facts.

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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