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Let $V$ be an inner product space over $F$. Let $W$ be a subspace of $V$. I saw that Axler's Linear Algebra Done Right claimed that if $W$ is finite-dimensional, then $V=W\oplus W^\perp$. It is trivial that $W\cap W^\perp=\{0\}$ is always true. So let us focus on $V=W+W^\perp$. As you can see in the picture, the proof for $V=W+W^\perp$ used the fact that $W$ is finite-dimensional. However, does $V=W+W^\perp$ hold when $W$ is infinite-dimensional? If so, how to prove it?

Update: I also wonder that if $W$ is finite-dimensional, must $W^\perp$ be finite-dimensional or infinite-dimensional?

reference: enter image description here

Eric
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    $W$ has to be a closed subspace of $V$ for $W \oplus W^\perp = V$ to hold. Finite-dimensional spaces are automatically closed so the result holds for them. – mechanodroid Sep 12 '17 at 14:34
  • @mechanodroid Thanks. What does closed mean for a subspace? I haven't found it in my linear algebra textbooks. – Eric Sep 12 '17 at 15:00
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    Topologically closed, so that means that you can take limits without leaving the subspace. This is automatic in the finite dimensional case but not in the general case. For instance, in the example below the polynomial space is not closed because polynomials can converge to any continuous function, not just polynomials. – Zach Boyd Sep 12 '17 at 18:55
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    @Eric The inner product on $V$ induces a norm $|x| = \sqrt{\langle x, x\rangle}$ which behaves similarly as the absolute value on $\mathbb{R}$. A sequence $(x_n)_{n=1}^\infty$ of vectors in $V$ is said to converge to a vector $x \in V$ if:

    $$\forall \varepsilon > 0,, \exists n_0 \in \mathbb{N} \text{ such that } n \geq n_0 \implies |x_n - x| < \varepsilon$$

    A set $S \subseteq V$ is said to be closed if it contains all limits of its convergent sequences, that is if $(x_n){n=1}^\infty$ is a sequence of vectors in $S$ such that $\lim{n\to\infty} x_n = x$ then it must be $x \in S$.

     – mechanodroid Sep 12 '17 at 19:29
  • @mechanodroid Very very clear, I get it! – Eric Sep 13 '17 at 02:50
  • One more question, if $V$ is infinite-dimensional and $W\leq V$ is finite-dimensional, then $V=W+W^\perp$ must hold right? – Eric Sep 13 '17 at 02:52
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    @Eric Yes, that follows from the fact that finite-dimensional subspaces are always closed: look here for example. – mechanodroid Sep 13 '17 at 09:10

1 Answers1

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Consider the space $V=C[0,1]$ of continuous function $[0,1]\to\Bbb R$ with inner product $\langle f,g\rangle:=\int_0^1f(t)g(t)\,\mathrm dt$ and $W$ the subspace of polynomial functions. Then $W^\perp=\{0\}$.

Hagen von Eitzen
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  • So if I get it right, $V=W+W^\perp$ is not true for $W$ being infinite-dimensional subspace right? – Eric Sep 12 '17 at 13:43
  • Is it true for $V$ be infinite-dimensional and $W$ be finite-dimensional? (and is $W^\perp$ at this time must be of finite or infinite dimension?) – Eric Sep 12 '17 at 13:46
  • @Eric Only W must be finite-dimensional. V is arbitrary. – alan23273850 Dec 10 '22 at 13:01