Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective

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Theorem

Let $K$ be a field.

Let $V$ be a finite dimensional vector space over $K$.

Let $f: V \to V$ be a linear transformation on $V$.


Then $f$ is an injection if and only if $f$ is a surjection.


Proof

Let $n = \dim V$.

From Vector Space has Basis, there exists a basis:

$\BB = \set {e_1, \ldots, e_n}$

for $V$.

Then from Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image, $f \sqbrk \BB$ is a generating set for $f \sqbrk V$.


Sufficient Condition

Suppose that $f$ is a surjection.

Then $f \sqbrk V = V$.

So $f \sqbrk \BB$ is a generating set for $f \sqbrk V$ of size $n$.

From Sufficient Conditions for Basis of Finite Dimensional Vector Space, $f \sqbrk \BB$ is therefore a basis for $V$.

In particular, $\map f {e_1}, \map f {e_2}, \ldots, \map f {e_n}$ are linearly independent.

Now take $x \in V$ such that $\map f x = \mathbf 0_V$ and write:

$\ds x = \sum_{i \mathop = 1}^n \alpha_i e_i$

for coefficients $\alpha_1, \ldots, \alpha_n \in K$.

Since $f$ is linear, we have:

$\ds \mathbf 0_V = \map f x = \sum_{i \mathop = 1}^n \alpha_i \map f {e_i}$

Since $\map f {e_1}, \map f {e_2}, \ldots, \map f {e_n}$ are linearly independent, we have $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$.

Hence $x = 0$.

So we have $\ker f = \set {\mathbf 0_V}$.

From Linear Transformation is Injective iff Kernel Contains Only Zero, $f$ is injective.

$\Box$


Necessary Condition

Suppose that $f$ is injective.

From Linear Transformation is Injective iff Kernel Contains Only Zero, $\ker f = \set {\mathbf 0_V}$.

So if $\alpha_1, \alpha_2, \ldots, \alpha_n \in K$ are such that:

\(\ds \sum_{i \mathop = 1}^n \alpha_i \map f {e_i}\) \(=\) \(\ds f \paren {\sum_{i \mathop = 1}^n \alpha_i e_i}\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \mathbf 0_V\)

Then we have:

$\ds \sum_{i \mathop = 1}^n \alpha_i e_i = \mathbf 0_V$

and hence $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$ since $e_1, e_2, \ldots, e_n$ is linearly independent.

So $f \sqbrk \BB$ is a linearly independent set of size $n$.

So $\map \span {f \sqbrk \BB} = f \sqbrk V$ is a subspace of $V$ of dimension $n$.

From Dimension of Proper Subspace is Less Than its Superspace, we have $f \sqbrk V = V$.

So $f$ is surjective.

$\blacksquare$


Sources

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