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How could I go about proving that $$\dim \ U = \dim \ V + \dim \ V^{\bot}$$

Where $U$ is a finite dim vector space. If I know that $V$ is a subspace of $U$, and let $V^{\bot}$ be the set of all vectors $s \in U$ and that the inner product $\langle v,s \rangle = 0$ for all $v \in V$.

Can I use the fact $U = V \oplus V^{\bot}$?

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Do you mean $U=V\oplus V^\perp$? – Shitikanth Mar 17 '13 at 23:56
Possible duplicate of Non-degenerate symmetric bilinear form; dimension formulae. – Gianluca Jul 12 at 18:34

Yes, if you can prove that $U = V \oplus V^⊥$ then the dimension result follows almost straight away. Recall that $\dim (U+V)+ \dim(U\cap V) = \dim(U) + \dim (V)$ and the definition of direct sum.

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Yes, use $U = V \oplus V^\perp$. Show that if you have a basis for $A$ and a basis for $B$ then together they give a basis for $A \oplus B$. This gives $\dim A \oplus B = \dim A + \dim B$.

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