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). Add to solve later

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. 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 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...

Close