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.

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 $\dim(U) > \dim(V)$

3 Responses

1. 08/17/2017

[…] 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 […]

2. 08/22/2017

[…] 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 […]

3. 10/26/2017

[…] 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 […]

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