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I am going through a proof and am struggling to understand a particular step. Let me setup the context.

Let $V$ be a finite dimensional vector space $V=\mathbb{C}^n$ with basis $e_1,\dots,e_n$ and Hermitian inner product $\Phi$. Let $U$ be a $\mathbb{C}$-subspace $U \subset V$. Define the orthogonal complement of $U$ to be

$U^{\perp} = \{w \in V\ \vert\ \Phi(u,w)=0\ \forall u \in U \}$.

I am going through a proof of the following lemma:

$U^{\perp}$ is a $\mathbb{C}$-vector space complementary to $U$, that is, $V=U\bigoplus U^{\perp}$.

The proof first shows that $U^{\perp}$ is in fact a $\mathbb{C}$-subspace of $V$, which I understand. I am confused by the final part of reasoning, however:

For a basis $e_1,\dots,e_m$ of $U$ and for any $v \in V$, set $$a_i=\frac{\Phi(e_i,v)}{\Phi(e_i,e_i)}.$$ Then setting $u = \sum_ia_ie_i$ and $w=v-u$ gives $\Phi(e_i,w)=0\ \forall i$, so that $w \in U^{\perp}$ and $v=u+w.$ Q.E.D.

Why does $\Phi(e_i,w)=0$?

Any help would be appreciated!

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    Is the basis orthogonal? – 2017-02-11
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    @Test123 the text doesn't specify – 2017-02-11
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    Can you provide the reference? What textbook/notes? – 2017-02-11

1 Answers 1

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You can consider that $\{e_i\}$ is an orthogonal basis w.r.t. $\Phi$ for $U$ i.e. $\Phi(e_i,e_j)=0$ for $i\neq j$.

$$\Phi(e_j,w)=\Phi(e_j,v)-\Phi(e_j,u)=\Phi(e_j,v)-\sum_i a_i\Phi(e_j,e_i)\\=\Phi(e_j,v)-\sum_i \frac{\Phi(e_i,v)}{\Phi(e_i,e_i)}\Phi(e_j,e_i)\\ =\Phi(e_j,v)-\Phi(e_j,v)=0 , \quad \forall j=1,\dots, m$$

since the only non-zero term of the sum is for $i=j$.

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    Thanks for the response, this makes sense. But why does the basis being orthogonal mean the hermitian product is 0 for $i \neq j$? – 2017-02-11
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    @Joseph Orthogonal basis w.r.t to the hermitian product means by definition that the hermitian product of two distinct basis elements is $0$ – 2017-02-11