Injective Linear Transformation $K[x]_{\leq 4}\rightarrow V$

Question

When $V$ is the vector space of all $2\times2$ matrices, why is no injective linear transformation $T: K[x]_{\leq 4}\longrightarrow V$? ($K$ is a field)

Answer 1

Consider a basis $B$ of $k_{\leq4}[X]$. Suppose exists $f:k_{\leq4}[X]\to V$ injective.

If we define $f$ over $B$ we are done.

Claim: if there is $p,q\in k_{\leq4}[X]$ l.i. such that $f(p)=\alpha f(q)$ then $f$ is not injective.

In fact, $0=f(p)-\alpha f(q)=f(p-\alpha q)$ and we have $0\neq p-\alpha q \in \ker f$ thus $dim(\ker f)\geq 1$ and $f$ is not injective.

This means that if $f$ is injective and we have a basis $B=\{v_i\}$ (in particular, l.i. vectors), $\{f(v_i)\}$ is a l.i. set.

In your case, this is impossible because $dim(k_{\leq4}[X])=5\not\leq 4=dim(V)$.

Answer 2

Well, you have that $dim(K[x]_{\leq 4}) = 5$ while $dim(V) = 4$. So you have that $dim(ker(T)) \geq 1$ so $T$ cannot be injective.