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Let $V$ a finite dimensional linear space and $F$ a subspace of $V$. I want to prove that the direct sum of $F$ and its orthogonal complement $F^\perp$ is the whole space $V$: $$F\oplus F^\perp=V$$ I know that the proof is pretty easy if we use an orthogonal basis, but how we prove it without using the orthogonal basis? The problem is how to prove that $F+F^\perp=V$?

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    $\begingroup$ Is the claim true for infinite-dimensional $V$? How would you distinguish finite from infinte-dimensional without referencing dimension and hence bases? $\endgroup$ – Hagen von Eitzen Jan 4 '18 at 18:53
  • $\begingroup$ Here's a try, assuming we're working over the reals: given $v$ in $V$, let $w$ be the element of $F$ closest to $v$, and show that $v-w$ is in $F^{\perp}$. $\endgroup$ – Gerry Myerson Jan 4 '18 at 18:58
  • $\begingroup$ Do you get to know about dual spaces? Even without an inner product, for finite-dimensions, we can get the right dimension-counting via dimension theorem applied to the image of the algebraic transpose of some linear map $T$ whose kernel is $F$. $\endgroup$ – Christopher A. Wong Jan 4 '18 at 19:03
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Let $(r_1,\ldots,r_p)$ a basis of $F$ and denote $A=(r_1,...,r_p)^T\in \mathcal M_{p,n}(\Bbb R)$ the matrix where $r_i$ is its $i$-th row and $n=\dim V$. We have $$x\in F^\perp \iff \langle r_i,x\rangle =0 \forall i\iff Ax=0\iff x\in\ker A$$ So we get $F^\perp=\ker A$, hence due to rank-nullity theorem $$\dim F^\perp= \dim \ker A=n-rk(A)=n-rk(A^T)=n-p$$ Finally, with the fact $F\cap F^\perp =\{0\}$ we can conclude the desired results.

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