Is this set of linear maps valid?

Question

Notation: The set of all linear maps from a vector space $V$ to a vector space $W$ (over a field $mathbb{F}$) is denoted $mathcal{L}(V, W)$.

The question states:

Show that ${ T in mathcal{L}(mathbb{R}^5, mathbb{R}^4) : dim(null(T)) > 2}$ is not a subspace of $mathcal{L}(mathbb{R}^5, mathbb{R}^4)$.

If I understand correctly, $mathcal{L}(mathbb{R}^5, mathbb{R}^4)$ is the set of all linear maps from $mathbb{R}^5$ to $mathbb{R}^4$. So $dim(mathbb{R}^4) = 4$ and hence $range(T) = 4$ also.
But by the Fundamental Theorem of Linear Maps, could $T$ ever exist?
$$
dim(mathbb{R}^5) = dim(null(T)) + dim(range(T)) \
5 = dim(null(T)) + 4 \
1 = dim(null(T))
$$
And hence $dim(null(T)) not > 2$.
The answer to this questions seems to assume the linear mapping is from $mathbb{R}^5$ to a subspace of $mathbb{R}^4$, as it provides the following counter example:

Let $e_1, ldots, e_5$ be a basis of $mathbb{R}^5$ and $f_1, ldots, f_4$ be a basis of $mathbb{R}^4$. Define $S_1$ and $S_2$ by:
$$ S_1e_i = 0, S_1e_4 = f_1, S_1e_5 = f_2, i = 1, 2, 3 \
S_2e_i = 0,S_2e_3 = f_3, S_2e_5 = f_4, i = 1, 2, 4 $$
(goes onto show not closed under addition)

Have I midunderstood?

Kevin Aquino · Answer

Remember that a subset $U$ of a vector space $V$ is a subspace if and only if:

$0 in U$.
$U$ is closed under addition.
$U$ is closed under scalar multiplication.

In this case, the set you mentioned is not a subspace because it does not satisfy the second condition, as you can see in the example: the linear maps $S_{1}$ and $S_{2}$ belong to the set, but the linear map $S_{1} + S_{2}$ does not. Can you see why?

user810049 · Answer

I think you have misunderstood. $mathcal{L}(Bbb{R}^5, Bbb{R}^4)$ is the set of linear maps $T : Bbb{R}^5 to Bbb{R}^4$, which means linear maps whose domain is $Bbb{R}^5$ and whose codomain is $Bbb{R}^4$. This means that $Tv in Bbb{R}^4$, for any $v in Bbb{R}^5$. It does not mean that the map $T$ is surjective, i.e. for any $w in Bbb{R}^4$, there exists some $v in Bbb{R}^5$ such that $Tv = w$.
The range of a linear map is automatically a subspace of its codomain, and need not be the full codomain. For example, the $0$ map in $mathcal{L}(Bbb{R}^5, Bbb{R}^4)$ takes every vector in $Bbb{R}^5$ and maps it to $(0, 0, 0, 0) in Bbb{R}^4$. That is, it maps onto a (trivial) subspace of $Bbb{R}^4$.
The ranges of maps in this set, by the rank-nullity theorem, must have dimension strictly less than $3$. They cannot be surjective.

Is this set of linear maps valid?

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