How to show that subspaces of finite-dimensional space are direct sum?

Question

Let $W$ be finite-dimensional vector space over the $\mathbb{R}$ and its subspaces are $U,V$. Let $X^*=\operatorname{Hom}(X,\mathbb{R})$ be a dual space to space $X$ which consists of all linear functionals on space $X$. We define a mapping $h: W^*\rightarrow U^*\times V^*$ with a formula $$h(\phi)=(\phi |_U, \phi |_V),$$ where $\phi |_U, \phi |_V$ are cuttings of the functional $\phi$ to subspaces $U$ and $V$ respectively.

Justify that if $h$ is an isomorphism then $W$ is a direct sum of $U$ and $V$.

---

We are obliged to show $W=U \oplus V$. So we have to show that

1. $U\cap V=\{0\},$
2. $U+V=W$.

We also know that $h$ is an isomorphism but I don't know how to use this fact in solution. Any advices and help will be much appreciated. Thanks.

Answer 1

Hints: Let $x \in U \cap V$. Show that there exists $\phi_1 \in U^{*}$ with $\phi_1(x)=1$ and $\phi_2 \in V^{*}$ with $\phi_2(x)=0$. Then there is no $\phi \in W^{*}$ with $h(\phi)=(\phi_1 ,\phi_2 )$ because we would the have $1=\phi_1(x)=\phi (x)=\phi_2(x)=0$. Hence $h$ is not surjective. This proves 1).

Suppose 2) is false. Then there exist $w \in W$ such that $w \notin U+V$. Let $\{x_1,x_2,...,x_k\}$ be basis for $U+V$. Then $\{w\}\cup \{x_1,x_2,...,x_k\}$ is linearly independent and it can be extended to basis for $W$. Now construct $\phi \in W^{*}$ such that $\phi(x_i)=0$ for all $i$ but $\phi (w) \neq 0$. [You can assign arbitrary values to basis elements]. It the follows that $h(\phi)=0$ but $\phi \neq 0$. Hence $h$ is not injective.

Answer 2

A roadmap of a proof is as follows.

$h$ being an isomorphism means it is surjective and injective (so $\operatorname{ker} h=0$). Moreover $\operatorname{ker} h = \{\phi: W\rightarrow \Bbb R \mid \phi \vert_U = 0 = \phi\vert_V\}$. Hence we can deduce $U+V = W$, as otherwise there would be a vector $w\notin U+W$, which can be used to define (by choosing a basis) a functional $\omega:W \rightarrow \Bbb R$ with the property that $\omega(w)=1, \omega\vert_{U+W} =0$.

Now assume there is a vector $w\in U \cap V$. Define a functional $\phi \in U^{*}$ with $\phi(w)=1$ and another one $\psi \in V^{*}$ with $\psi(w)=-1$. This contradicts surjectivity of $h$ as $(\phi,\psi)$ can’t possibly be obtained by restricting a functional on $W$. Hence $U \cap W = 0$.