Prove that $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$

Question

As I am self learning Linear Algebra Done Right, I would like to make sure I learn it correctly. Please help on the following the proof.

Suppose $\varphi \in \mathcal{L}(V, \mathbb{F})$. Suupose $u \in V$ is not in null $\varphi$. Prove that $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$

My approach:

Suppose $\varphi \in \mathcal{L}(V, \mathbb{F})$, and $u \in V$ is not in null $\varphi$. Let $U = \{au:a \in \mathbb{F}\}$ .

Let $w_1,...,w_m$ be the basis of null $\varphi$. Extend $w_1,...,w_m$ to be a basis of $V$ as $w_1,...,w_m, v_1,...,v_n$. Then $v_1,...,v_n$ is the basis of $\{au:a \in \mathbb{F}\}$ [I cannot convince myself here. Seems right.Not sure]

For any $v \in V$, $v = a_1m_1 + ... + a_mw_m + b_1v_1+...+b_nv_n = \text{null }\varphi + U$.

Suppose $v \in \text{null }\varphi \cap U$, then $v = a_1m_1 + ... + a_mw_m = b_1v_1+...+b_nv_n $. $a_1m_1 + ... + a_mw_m - b_1v_1 -...-b_nv_n = 0$. Since $w_1,...,w_m, v_1,...,v_n$ is a basis of $V$, $a_1 = ... = a_m = b_1 = ... = b_n = 0$ which implies $v = 0$.

Therefore $V = \text{null }\varphi \oplus \{au:a \in \mathbb{F}\}$.

Answer 1

As the author-@Sheldon Axler-has mentioned, there is no assumption that $V$ is finite-dimensional. Thus you could not use bases.

I'll give you a complete answer now. Based on the following theorem:

Direct sum of two subspaces: Suppose $U$ and $W$ are subspaces of $V$. Then $U+W$ is a direct sum if and only if $U \cap W = {0}$.

It is obvious that $\text{null} \varphi \cap \{au: a \in F\} = 0$. (You can verify it yourself.)

What is left to do is to prove that for any vector $v \in V$, $v$ can be written as $v=k+au$ where $k\in \text{null} \varphi$. That is, we wanna prove $V = \text{null} \varphi + \{au: a \in F\}$.

Because $u$ is not in $\text{null} \varphi$, let $T(u)=f_1\neq 0$. For an arbitrary $v$, let $T(v)=f_0$ ($f_0$ is any number in $F$). If $T(v) = f_0=0$, then $v$ is in $\text{null} \varphi$. If $T(v) = f_0 \neq 0$, then $T(v)= mT(u)= T(mu)$, where $m=\frac{f_0}{f_1}$. So $T(v-mu) = 0$ and $v=k+mu$ where $k\in \text{null} \varphi$. In both cases, $v$ can be written as $v=k+au$.

Proof completed.

Be careful! You can't use the following theorems in the book, which use basis in their proofs or require space is finite dimensional.

Every subspace of V is part of a direct sum equal to V: Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = U \oplus W$.

or

Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V,W)$. Then range $T$ is finite-dimensional and $\text{dim} V = \text{dim} T + \text{dim range} T$.