# Basis for $\operatorname{null}T$ given basis for $\operatorname{range}T$

I was trying to solve exercise 24 of chapter 3.B of "Linear Algebra done right", by Sheldon Axler. It states:

Suppose $$W$$ is finite-dimensional and $$T_1, T_2 \in \mathcal{L}(V, W)$$. Prove that $$\operatorname{null} T_1 \subset \operatorname{null} T_2$$ if and only if there exists $$S \in \mathcal{L}(W,W)$$ such that $$T_2 = ST_1$$.

I found a "solution" to this exercise at https://linearalgebras.com/3b.html. It starts by proving one direction of the implication and states at the beginning: "Suppose $$\operatorname{null} T_1 \subset \operatorname{null}T_2$$. [...] Let $$Tv_1, \cdots, Tv_n$$ be a basis for $$\operatorname{range}T$$, then the list $$v_1, \cdots, v_n$$ is linearly independent. Let $$K = \operatorname{span}(v_1, \cdots,v_m)$$, then $$V = K \oplus \operatorname{null} T$$."

I think the last result can be showed like this: given the list $$n_1, \cdots, n_k$$ is a basis for $$\operatorname{null} T$$, then we can extend this list to a basis $$n_1, \cdots, n_k, v_1, \cdots, v_n$$. Since the list $$v_1, \cdots, v_n$$ is linearly independent and each $$v_i$$ is linearly independent from all $$n_i$$ (otherwise $$T(v_i) = 0$$), by the rank-nullity theorem, $$\dim V = k + n$$, so $$n_1,\cdots, n_k, v_1, \cdots, v_n$$ indeed forms a basis of $$V$$ and $$V = K \oplus \operatorname{null} T$$. Is this reasoning correct?

I also wanted to consider a similar statement. Let $$Tv_1, \cdots, Tv_n$$ be a basis for $$\operatorname{range}T$$, if we extend $$v_1, \cdots, v_n$$ to a basis $$v_1, \cdots, v_n, v_{n+1},\cdots, v_m$$ of $$V$$ then the list $$v_{n+1},\cdots, v_m$$ is a basis for $$\operatorname{null} T$$. This would mean that if $$V = K \oplus G$$ then $$G = \operatorname{null} T$$. Is this statement true?

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Aug 16, 2019 at 18:06
• Why do you want to prove this? You will have some difficulty since it is not true. Aug 16, 2019 at 22:06
• Thank you for your feedback. I have added context and reformulated the question. Is it acceptable now?
– JoJo
Aug 17, 2019 at 8:11
• Very good! @Joca Aug 17, 2019 at 9:02

This is not true.

Let $$T : \mathbb{R}^{3} \longrightarrow \mathbb{R}^{4}$$ be a linear transformation such that:

1) $$T(1,0,0) = (1,0,0,0)$$ and $$T(0,1,0) = (0, 1, 0, 0)$$

Let $$A = \lbrace T(1,0,0), T(0,1,0) \rbrace \subset \mathbb{R}^{4}$$ be a base for $$\text{Range}(T)$$.

2) $$B =\lbrace (1,0,0), (0,1,0) \rbrace \subset \mathbb{R}^{3}$$ is linearly independent.

Completing $$B$$ to a base of $$\mathbb{R}^{3}$$, for example : $$B_{1} = \lbrace (1,0,0), (0,1,0), (0,0,1) \rbrace$$

we can do it : $$T(0,0,1) = (1,1,0,0)$$ and this case $$T(0,0,1) \not \in Null(T)$$.

Therefore $$\lbrace (1,0,0) \rbrace$$ is not a base for $$\text{Null}(T)$$.

• But then, $T(1, 0, 0), T(0, 1, 0)$ cannot be a base for $range \ T$ since $T(0,0,1) \in range T$ but $T(0,0,1) \notin span(T(1,0,0), T(0,1,0))$
– JoJo
Aug 17, 2019 at 8:05
• @Joca. You are right. I edited my answer. Thank you very much. Aug 17, 2019 at 14:47
• That's right, thank you! I was trying to hard to prove it was true and didn't consider counter examples.
– JoJo
Aug 17, 2019 at 16:16
• @Joca. Normal. Sometimes we forget to even think about counterexamples. Success !! Aug 17, 2019 at 22:13