Linear Map to Lower-Dimensional Space is Not Injective

Linear map to a lower-dimensional space is not injective

Suppose $V$ and $W$ are finite-dimensional vector spaces such that $\dim V>\dim W$. Then, no linear map from $V$ to $W$ is injective.

Proof. Let $T\in \mathcal{L}(V,W)$. Then

$$\begin{array}{rl}\dim \text{null }T& =\dim V-\dim \text{range }T\\ & \ge \dim V-\dim W\\ & >0\end{array}$$

where the first line above comes from the fundamental theorem of linear maps and the second line follows from the fact that $\text{range }T$ is a subspace of $W$ (see dimension of a subspace).

This inequality states that $\dim \text{null }T>0$. This means that $\text{null }T$ contains vectors other than $0$. Thus, $T$ is not injective, as injectivity implies null space is 0 and vice versa. $■$

For example, let’s say we have a linear map $T:{\mathbb{F}}^4\to {\mathbb{F}}^3$ such that

$$T(z_1,z_2,z_3,z_4)=(\sqrt{7}{z}_1+\pi z_2+z_4,97z_1+3z_2+2z_3,z_2+6z_3+7z_4)$$

Because $\dim {\mathbb{F}}^4>\dim {\mathbb{F}}^3$, we can immediately say that $T$ is not injective without any calculations.