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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?

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

2 Answers 2

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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.

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  • $\begingroup$ Maps in this set, by the rank-nullity theorem, must have a range with dimension $\textbf{at most}$ $3$. $\endgroup$ Jul 22, 2020 at 2:13
  • $\begingroup$ @KevinLópezAquino Yes, good note, thank you. $\endgroup$
    – user810049
    Jul 22, 2020 at 4:08
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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?

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