# A maximal vectorial subspace has a 1-dimensional complement.

I have seen this excercise on a book of Functional analysis and I am trying to prove it.

Let $$X$$ be a vectorial space over a field $$\Bbb{K} \, ( \Bbb{R}$$ or $$\Bbb{C} )$$ and $$H \subset X$$ a proper subspace. Then, the following are equivalent:

1. $$H$$ is maximal (that is, for every subspace $$H \subseteq M \subseteq X \rightarrow H=M \lor X=M$$ )
2. It exists a subspace $$L \subseteq X$$ such that $$\dim L = 1, H \cap L = \{ 0 \}$$ and $$X= H + L$$ .
3. $$\dim X/H = 1$$
4. It exists a linear functional $$\varphi : X \rightarrow \Bbb{K}$$ such that $$\ker \varphi = H$$

It is important to say that the book does not say anything about the dimension of $$X$$, in fact there is problem before this about infinite dimensional spaces. This is why I find this problem difficult, the finite case would be very simple.

I saw first that $$4 \Rightarrow 3$$ is easy to prove using the First Isomorhpy Theorem with $$\varphi$$ . In order to continue, I have tried to prove $$3 \Rightarrow 2$$ and $$1 \Rightarrow 2$$ by considering the linear function $$T: X \times H \rightarrow X, T(x,h) = x-h$$ and taking $$L := T( X \times H )$$ . This satisfies that $$H + L = X$$, but I do not know if it works because I am not seeing how to prove $$H \cap L = \{ 0 \}$$ and $$\dim L = 1$$ since the possibly infinite dimension of $$X$$ .

Also, I do not have very idea on how to "reach" $$4$$ from one of the others. I thought first on consider $$\varphi ( x ) = ||| x + H |||$$ where $$||| \cdot |||$$ is the norm on $$X/H$$ induced by a norm on $$X$$, but I am not sure about it, so any possible help would be appreciated.

• The $L$ in your attempt cannot work because it equals $X$. Apr 2 at 15:59
• This is a purely algebraic theorem. $\Bbb K$ does not need to be $\Bbb R$ or $\Bbb C$, and even if it is and if you have a norm on $X$, $H$ is not supposed to be closed for it. Moreover, the $\varphi$ in your attempt is $\ge0$ hence not linear. Apr 2 at 16:02
• Hint for $3\implies1$: for every subspace $H \subseteq M \subseteq X$, $\dim(X/H)=\dim( X/M)+\dim(M/H)$ Apr 2 at 16:06
• Hint for $1\implies2$: let $L$ be the line spanned by some $v\in X\setminus H$. Then, $M:=H\oplus L\supsetneq H$ hence $M=X$. Apr 2 at 16:09
• Hint for $2\implies4$: if $H\oplus\Bbb Kv=X$, define $\varphi:X\to\Bbb K$ by$$x-\varphi(x)v\in H.$$ Apr 2 at 16:13

$$1 \implies 2$$: Let $$v$$ be any nonzero vector in $$X \setminus H$$ and $$L = \mathbb{K}v$$. Then $$L$$ is one dimensional and $$L \cap H = 0$$. Also $$H \subsetneq H + L$$, which by (1) implies $$H + L = X$$.

$$2 \implies 3$$: The conditions $$H \cap L = 0$$ and $$H + L = X$$ amount to the stronger $$H \oplus L = X$$, i.e. every vector $$x \in X$$ has a unique decomposition as $$x = h + \ell$$ for $$h \in H$$, $$\ell \in L$$. (If you haven't seen direct sums before, try to prove this for yourself!) Thus every coset of $$H$$ contains a unique element of $$L$$, namely $$x + H \ni \ell$$ where the $$\ell$$ comes from the unique decomposition of $$x$$. Mapping $$x + H \mapsto \ell$$ defines a linear map $$\bar{f}: X/H \to L$$ which is in fact an isomorphism, with inverse sending $$\ell$$ back to its coset $$\ell + H$$.

$$3 \implies 4$$: Since it is one dimensional, choosing any nonzero vector as a basis vector gives an isomorphism $$X/H \cong \mathbb{K}$$. The quotient map $$\phi: X \to X/H \cong \mathbb{k}$$ is the desired functional.

$$4 \implies 1$$: Suppose $$H \subseteq M \subseteq X$$, and consider the image $$f(M) \subset \mathbb{K}$$. If $$f(M) = \{0\}$$, then $$M \subseteq \ker(f) = H$$, so $$M = H$$. Alternatively if $$f(M) = \mathbb{K}$$, then for any $$x \in X$$ there exists $$m \in M$$ with $$f(x) = f(m)$$, subsequently $$x - m \in \ker(f) = H \subset M$$, so $$x = (x-m) + m \in M$$ proving $$M = X$$.

$$3\implies 2$$: the condition $$\dim X/H=1$$ tells you that there exists nonzero $$x_0\in X$$ such that for any $$x$$ there exists a scalar $$\lambda$$ with $$x+H=\lambda x_0+H$$. The equality tells you that $$x\in\mathbb K x_0+H$$.

$$2\implies 1$$: suppose that $$H\subset M\subset X$$. By hypothesis any $$m\in M$$ can be written $$m=\lambda x_0+h$$ for some $$\lambda\in\mathbb K$$. If $$\lambda=0$$ then $$m\in H$$; and if $$\lambda\ne0$$, then $$x_0=\lambda^{-1}(m-h)\in M$$, so $$X=M$$.

$$1\implies 4$$: If $$H=X$$, take $$\varphi=0$$. Otherwise, let $$x_0\in X\setminus H$$. By hypothesis, $$M=\mathbb K x_0+H=X$$. That is, any $$x\in X$$ can be written as $$x=\lambda x_0+h$$, with $$\lambda\in\mathbb K$$ and $$h\in H$$. This $$\lambda$$ is unique, for if $$\lambda_1 x_0+h_1=\lambda_2 x_0+h_2$$, we have $$(\lambda_1-\lambda_2)x_0=h_2-h_1\in H$$, and so $$\lambda_1=\lambda_2$$. This allows us to define $$\varphi(\lambda x_0+h)=\lambda.$$

• The $$L$$ in your attempt cannot work because it equals $$X$$.
• This is a purely algebraic theorem. $$\Bbb K$$ does not need to be $$\Bbb R$$ or $$\Bbb C$$, and even if it is and if you have a norm on $$X$$, $$H$$ is not supposed to be closed for it. Moreover, the $$φ$$ in your attempt is $$≥0$$ hence not linear.
• For 3⟹1: for every subspace $$H⊆M⊆X$$, $$\dim(X/H)=\dim(X/M)+\dim(M/H)$$.
• For 1⟹2: let $$L$$ be the line spanned by some $$v∈X\setminus H$$. Then, $$M:=H⊕L⊋H$$ hence $$M=X$$.
• For 2⟹4: if $$H⊕\Bbb Kv=X$$, define $$φ:X→\Bbb K$$ by $$x−φ(x)v∈H.$$