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

## Problem 540

Let $U$ and $V$ be vector spaces over a scalar field $\F$.

Let $T: U \to V$ be a linear transformation.

Prove that $T$ is injective (one-to-one) if and only if the nullity of $T$ is zero.

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## Definition (Injective, One-to-One Linear Transformation).

A linear transformation is said to be **injective** or **one-to-one** if provided that for all $\mathbf{u}_1$ and $\mathbf{u}_1$ in $U$, whenever $T(\mathbf{u}_1)=T(\mathbf{u}_2)$, then we have $\mathbf{u}_1=\mathbf{u}_2$.

## Proof.

### $(\implies)$: If $T$ is injective, then the nullity is zero.

Suppose that $T$ is injective.

Our objective is to show that the null space $\calN(T)=\{\mathbf{0}_U\}$.

Since $T$ is a linear transformation, it sends the zero vector $\mathbf{0}_U$ of $U$ to the zero vector $\mathbf{0}_V$ of $V$.

In fact, we have

\begin{align*}

T(\mathbf{0}_U)&=T(\mathbf{0}_U-\mathbf{0}_U)\\

&=T(\mathbf{0}_U+(-1)\mathbf{0}_U)\\

&=T(\mathbf{0}_U)+(-1)T(\mathbf{0}_U) &&\text{by linearity of $T$}\\

&=T(\mathbf{0}_U)-T(\mathbf{0}_U)=\mathbf{0}_V.

\end{align*}

Hence $\mathbf{0}_U\in \calN(T)$.

On the other hand, if $\mathbf{u}\in \calN(T)$, then we have

\[T(\mathbf{u})=\mathbf{0}_V=T(\mathbf{0}_U).\]
Since $T$ is injective, it yields that $\mathbf{u}=\mathbf{0}_U$.

Therefore we obtain $\calN(T)=\{\mathbf{0}_U\}$, and the nullity of $T$ is zero.

(Recall that the nullity of $T$ is the dimension of $\calN(T)$.)

### $(\impliedby)$: If the nullity is zero, then $T$ is injective.

Next, suppose that the nullity of $T$ is zero.

This is equivalent to the condition $\calN(T)=\{\mathbf{0}_U\}$.

Our goal is to show that $T: U \to V$ is injective.

Suppose that $T(\mathbf{u}_1)=T(\mathbf{u}_2)$ for some $\mathbf{u}_1, \mathbf{u}_2\in U$.

Then we have

\begin{align*}

\mathbf{0}_V&=T(\mathbf{u}_1)-T(\mathbf{u}_2)\\

&=T(\mathbf{u}_1)+(-1)T(\mathbf{u}_2)\\

&=T(\mathbf{u}_1+(-1)\mathbf{u}_2) && \text{by linearity of $T$}\\

&=T(\mathbf{u}_1-\mathbf{u}_2)

\end{align*}

It follows that the vector $\mathbf{u}_1-\mathbf{u}_2$ is in the null space $\calN(T)=\{\mathbf{0}_U\}$.

Thus, we have $\mathbf{u}_1-\mathbf{u}_2=\mathbf{0}_U$, or $\mathbf{u}_1=\mathbf{u}_2$.

So the linear transformation $T$ is injective.

## Related Question.

**Problem**.

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.

The proof of this problem is given in the post ↴

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

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## 3 Responses

[…] 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 […]

[…] to show that the null space of $T$ is trivial: $calN(T)={mathbf{0}}$. (See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for a proof of this […]

[…] that is, $T(mathbf{x})=mathbf{0}$ implies that $mathbf{x}=mathbf{0}$. (See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for the proof of this […]