$L(\mathbb{R}^5, \mathbb{R}^4)$ 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.

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

2 Answers 2


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 $


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