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I was looking at this answer and when it states:

Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$.

And then it proceed to state:

$\beta\cup\gamma$ generates $V$.

I understand that $v=v_1+v_2$ will produce some of the elements in $V$ but that doesn't necessarily mean that it can produce all of the elements in $V$.

Is there a proof for this?

ianc1339
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1 Answers1

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Yes, of course. It follows from Hilbert's projection theorem (https://en.wikipedia.org/wiki/Hilbert_projection_theorem).

amsmath
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  • Oh, I haven't learnt that yet. I will have a read about it then. Thanks! – ianc1339 Feb 28 '21 at 08:12
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    In finite dimensions it's much easier. Choose a basis ${v_1,\ldots,v_r}$ of $V$, complete it to a basis ${v_1,\ldots,v_n}$ of $\mathbb R^n$ and perform Gram-Schmidt to get an orthonormal basis ${e_1,\ldots,e_n}$ of $\mathbb R^n$. Then ${e_1,\ldots,e_r}$ will span $V$ and the rest spans $V^\perp$, hence $\mathbb R^n = V\oplus V^\perp$. – amsmath Feb 28 '21 at 08:17