# Proving that the subset of all linear transformations from $\mathbb{R}^5$ to $\mathbb{R}^4$ where $n(T) > 2$ is not a subspace

$$L(\mathbb{R}^5, \mathbb{R}^4)$$ is the set of all linear transformations between the two.

Let $$U = \{T \in L : n(T) > 2\}$$ ($$N(T)$$ is the null space of $$T$$ and $$n(T)$$ is the nullity)

I have to prove that $$U \nleq L$$.

I was thinking of finding $$T_1$$ and $$T_2$$ in $$U$$ (Assuming they have nullity 3) in a way that $$n(T_1 + T_2) \leq 2$$ and therefore showing that $$U$$ is not closed under addition.

This is what I wrote:

$$N(T_1) = \mathrm{Span}(v_1, v_2, v_3), \\ N(T_2) = \mathrm{Span}(w_1, w_2, w_3)$$

Where $$S_1 = \{v_1, v_2, v_3\}$$ and $$S_2 = \{w_1, w_2, w_3\}$$ are lineary independant. Then, I proved that $$S_1 \cap S_2$$ cannot be empty because that would mean $$6 \le\mathrm{Dim}(\mathbb{R}^5)$$ which is a contradiction. I thought maybe I could somehow show that $$T_1 + T_2$$ has nullity less than or equal to 2 if I could find how big $$S_1 \cap S_2$$ is.

But at this point, I don't know how to continue this idea or if it is even useful to do this.

• You should explain the notation in your question. What are $n(T)$ and $N(T)$? Commented Oct 20, 2020 at 12:17
• @PaulFrost I just edited it. n(T) is the nullity and N(T) is the null space.
– Zara
Commented Oct 20, 2020 at 12:20

Let $$T_1 : \mathbb R^5 \to \mathbb R^4, T_1(x_1,\ldots,x_5) = (x_1,x_2,0,0)$$ and $$T_2 : \mathbb R^5 \to \mathbb R^4, T_2(x_1,\ldots,x_5) = (0,0,x_3,x_4)$$.

Then $$N(T_1) = \{ (x_1,\ldots,x_5) \mid x_1 = x_2 = 0 \}$$, thus $$n(T_1) = 3$$. Similarly $$n(T_2) = 3$$. But $$(T_1 + T_2)(x_1,\ldots,x_5) = (x_1,x_2,x_3,x_4)$$, thus $$n(T_1 + T_2) = 1$$.

Take $$Range(T_1)= span \{x_1,x_2\}$$ , where , $$x_1,x_2$$ are linearly independent vectors in $$\mathbb{R}^4$$.

Now , take such $$x_3,x_4$$ in $$\mathbb{R}^4$$ that set $$\{x_1,x_2,x_3,x_4\}$$ is with vectors linearly independent to each other.

Take , $$Range(T_2)=span \{x_3,x_4\}$$

Clearly $$n(T_1)=3=n(T_2)$$

But now,as $$Range(T_{1}+T_{2})=span\{x_1,x_2,x_3,x_4\}$$ is of dimension $$4$$.

So, $$n(T_{1}+T_{2})=1 \lt 2$$