Suppose $V$ and $W$ are finite-dimensional with $2\le\dim(V)\le\dim(W)$. Show that $K=\{T\in\mathcal{L}(V,W): T\ \text{not injective}\}$ is not a subspace of $\mathcal{L}(V,W)$. My attempt is to pass by fundamental theorem of linear maps. I would like to know if my solution holds, please.
For me, to show that $K$ is not a subspace of $\mathcal{L}(V,W)$ is equivalent to show that $\forall S\in K$ $S$ is not linear. Is it correct?
Suppose by absurd that $S\in K$ is linear. So, by fundamental theorem of linear map we have that $\dim(V)=\dim(\text{range} \ S)+\dim(\text{null}\ S)$
But, $\dim(\text{range} \ S)+\dim(\text{null}\ S)\ge\dim(V)+\dim(\text{null} \ S)\ge\dim(V)+1\neq \dim(V)$ as $\dim(\text{null} \ S)\ \neq 0$ because $S$ is not injective. Thus, each element $S$ of $K$ is not a linear application. We conclude that $K$ is not a subspace of $\mathcal{L}(V,W)$.
Best Answer
As Bargabbiati wrote in his comment, your proof is not valid.
Hint: Let $P \in \mathcal{L}(V,W)$ be a projection (this means $P^2=P$) with $0 \ne P \ne I.$
Then $P \in K.$ Now let $Q:= I-P$.
Show that $Q \in K$, but $P+Q \notin K.$