# Is this set of linear maps valid?

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?

• $T(\mathbb{R}^5)\subseteq\mathbb{R}^4$ only imply $\mbox{range}(T)\leq 4$, which in turn imply $\dim\mbox{null}(T)\geq1$. Commented Jul 22, 2020 at 1:50

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.

• Maps in this set, by the rank-nullity theorem, must have a range with dimension $\textbf{at most}$ $3$. Commented Jul 22, 2020 at 2:13
• @KevinLópezAquino Yes, good note, thank you. Commented Jul 22, 2020 at 4:08

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

1. $$0 \in U$$.
2. $$U$$ is closed under addition.
3. $$U$$ 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?