Can zero be any dimension?

Question

When I am working on the problem:

Suppose $V$ and $W$ are finite-dimensional with $\text{dim }V \geq \text{dim }W \geq 2$. Show that $\{T \in \mathcal{L}(V,W): T \text{ is not surjective}\}$ is not a subspace of $\mathcal{L}(V,W)$.

So, I naturally want to check if $0 \in \{T \in \mathcal{L}(V,W): T \text{ is not surjective}\}$. Am I correct?

My first thought was the zero map is surjective because it maps any dimension, say 2, 3, or 4 to zero in 2,3,or 4 dimension. So, range of $T$ is equal to $W$. I think there are some problem in this statement that only compare the dimension to determine whether T and W are equal. But I am not sure.

If range of $T$ is equal to $W$, then it is surjective. So it violates the additive identity properties of suspaces?

Answer 1

The range of the zero map $\{0\}$ and therefore it is surjective when and only when $W=\{0\}$. So, since you are assuming that $\dim W\geqslant 2$, the zero map is not surjective.

Concerning the question from the title of your question: $\dim\{0\}=0$.

Answer 2

The zero map from $\mathbb{R}^{\dim V}$ to $\mathbb{R}^{\dim W}$ is not surjective, unless $\dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.

Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.