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Specifically, let $V$ be a vector space and $U$ be a subspace, both over $\mathbb{C}$. Let $H(z_1, z_2)$ be some hermitian form. Define $W = \{w\in V\mid H(u, w) = 0\forall u\in U\}$.

It's fairly clear that $W$ is a vector space itself. However, I don't understand why it's true that $U\oplus W = V$ or why the choice of $H$ doesn't seem to matter. In the case where $H$ is just the standard dot product this seems intuitively obvious, but I don't know how to prove it.

Anon
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    The subspace $W$ highly depends on $H$. It can even intersect with $U$ nontrivially if it is not positive-definite. Worse, if $H$ is not nondegenerate, you don't even have $\dim_\mathbb{C}(U)+\dim_\mathbb{C}(W)=\dim_\mathbb{C}(V)$ (but you do have $\dim_\mathbb{C}(U)+\dim_\mathbb{C}(W)=\dim_\mathbb{C}(V)+\dim_\mathbb{C}\big(\ker(H)\big)$). So, depending on $H$, (1) the sum $U+W$ may not be direct, and (2) the sum $U+W$ may not equal $V$. – Batominovski Nov 16 '18 at 21:30
  • Then under what conditions does $U\oplus W = V$? In particular, why is $W$ referred to as the complement of $V$? – Anon Nov 16 '18 at 21:32
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    I do not know all possible criteria, but if $H$ is a positive-definite Hermitian form (well, or if $H$ is negative-definite), then it is guaranteed that $U\oplus W=V$ (at least, if $V$ is finite-dimensional). – Batominovski Nov 16 '18 at 21:33
  • So if we restrict $H$ to any positive-definite Hermitian form, then the specific choice of $H$ does not matter? Thanks! – Anon Nov 16 '18 at 21:34
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    The choice of $H$ still affects what $W$ can be. There are many possible $W$'s that fulfill the equation $U\oplus W=V$. – Batominovski Nov 16 '18 at 21:35
  • I thought that the complement of a subspace was unique? https://math.stackexchange.com/questions/1106649/a-proof-that-orthogonal-complement-is-unique – Anon Nov 16 '18 at 21:37
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    For a given Hermitian form $H$, the complement is unique. But if $H$ varies, the complement varies too. – Batominovski Nov 16 '18 at 21:37
  • I just realized. I do know all possible criteria. The equality $U\oplus W =V$ holds for all subspaces $U$ if and only if $H$ is a positive- or negative-definite Hermitian form. The proof is reduced to knowing the signature of $H$. – Batominovski Nov 16 '18 at 21:45

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Let $V = \mathbb{R}^2$ with the standard inner product $\langle(a, b), (c, d)\rangle_1 = ac + bd$, and let $L \subseteq V$ be the $x$-axis, $L = \{(a, 0) \mid a \in \mathbb{R}\}$. Then the orthgonal complement to $L$ are all those points $(c, d)$ satisfying $ac + 0d = 0$ for all $a \in \mathbb{R}$, i.e. the $y$-axis.

However, if we put a different inner product on $V$, say $\langle(a, b), (c, d)\rangle_2 = 2ac + ad + bc + 2bd$, (you need to check that this is an inner product, i.e. it is symmetric and positive-definite), then the orthogonal complement to $L$ are those points $(c, d)$ satisfying $2ac + ad = 0$ for all $a \in \mathbb{R}$, i.e. the line $2x + y = 0$.

In general, given a subspace $L \subseteq V$, the orthogonal complement of $L$ will depend on the choice of inner product. However, it will always be true (in the finite-dimensional case) that $V = L \oplus L^\perp$, by definiteness of the inner product. For $$L \cap L^\perp = \{v \in L \mid \langle v, u \rangle = 0 \text{ for all } u \in L\} \subseteq \{v \in L \mid \langle v, v \rangle = 0\} = \{0\}$$ (where we used the fact that $\langle v, v \rangle = 0$ iff $v = 0$). By picking an orthogonal basis of $L$ and extending to an orthogonal basis of $V$, it is easy to check that $\dim V = \dim L + \dim L^\perp$, and so $V = L + L^\perp$.

Joppy
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