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Let $E$ be a vector space endowed with a non-degenerate symmetric bilinear form. Show $\dim F+\dim F^{\perp}=\dim E=\dim\left(F+F^{\perp}\right)+\dim\left(F\cap F^{\perp}\right)$

Lang uses this formula in proposition 1.2, page 573 of his book "Algebra" (Graduate).

I did not find a straight forward argument to show it, the only argument I found was really long, and not very enlightening. But I assume if he stated it without proof

Here is what I want to show:

Let $E$ be a finite-dimensional vector space over a field K.
Let $g$ be a symmetric/alternating/hermitian form (it suffices to me if you prove it for symmetric and hopefuly I can go from there).
Assume g is non-degenerate.
Let F be a subspace of E.
I want to show the following: $ \dim F+\dim F^{\perp}=\dim E=\dim\left(F+F^{\perp}\right)+\dim\left(F\cap F^{\perp}\right) $
It may be that not all hypothesis are necessary, he just uses this formula in a proof but doesn't derive it.

Thanx to anyone who helps! cheers

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    $\begingroup$ In $\LaTeX$, you may type \dim for $\dim$. $\endgroup$ – user1551 Nov 27 '13 at 19:04
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For arbitrary subspaces $U,V$ of an arbitrary vector space $W$, we have the dimension formula

$$\dim U + \dim V = \dim (U+V) + \dim (U\cap V).$$

(Take a basis $B_1$ of $U\cap V$, and extend it by systems $B_2$ to a basis of $U$, and $B_3$ to a basis of $V$. Then $B_1 \cup B_2 \cup B_3$ is a basis of $U+V$.)

That yields the

$$\dim F + \dim F^\perp = \dim (F + F^\perp) + \dim (F\cap F^\perp)$$

part. It remains to see that $\dim F + \dim F^\perp = \dim E$. Let's denote the non-degenerate bilinear form by $\beta$. $\beta$ induces a linear map $\Phi \colon E \to E^\ast$ ($E^\ast$ is the dual space of $E$, the space of all linear maps $E\to K$) via

$$\Phi(x)(y) = \beta(x,y).$$

The non-degeneracy of $\beta$ is equivalent to the injectivity of $\Phi$. Since the spaces are finite-dimensional, we have $\dim E^\ast = \dim E$, and thus $\Phi$ is an isomorphism. For a subspace $F \subset E$, the image of $F^\perp$ under $\Phi$ is the annihilator of $F$,

$$\Phi(F^\perp) = F^0 = \{ \lambda \in E^\ast : F\subset \ker\lambda\}.$$

If you already know that $\dim F + \dim F^0 = \dim E$ for all subspaces of finite-dimensional spaces, that's it. Otherwise, to see that, choose a basis $B_0 = \{v_1,\dotsc, v_f\}$ of $F$, extend it to a basis $B = B_0 \cup B_1 = \{v_1,\dotsc,v_f,v_{f+1},\dotsc,v_e\}$ of $E$, and consider the dual basis $B^\ast = \{\lambda_1,\dotsc,\lambda_e\}$ of $E^\ast$, defined by

$$\lambda_i(v_j) = \delta_{ij} = \begin{cases}1 &, i = j\\ 0 &, i \neq j. \end{cases}$$

It is then easy to see that $F^0 = \operatorname{span} \{\lambda_{f+1},\dotsc,\lambda_e\}$, whence $\dim F^0 = e - f = \dim E - \dim F$.

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  • $\begingroup$ Thank you for this proof. Does the form really need to be symmetric? It does not seem to get used. $\endgroup$ – JDZ Nov 12 '16 at 23:30
  • $\begingroup$ @JDZ The form does indeed not need to be symmetric. Only the non-degeneracy is needed (for the $\dim F + \dim F^\perp = \dim E$ part). $\endgroup$ – Daniel Fischer Nov 13 '16 at 9:36
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Have you read about complements of a vector space? Finite-dimensional vector spaces have the good property that every subspace is of finite dimension and has a complement. If you have a subspace $F \subseteq E$, then you get the complement $F^\perp$ why considering all elements of $E$ that are ortogonal with respect to your bilinear form, ie. all $x \in E$ with $$\langle x,y \rangle = 0 \text{ for all } y \in F$$ Then you get a decomposition of $E$ with respect to your subspace $F$ and your bilinear form $\langle,\rangle$ into $$ E = F \oplus F^\perp$$ So the dimension forumlar should be clear. If the bilinear form degenerates, the sum coudl be not direct so you have to take the dimension of "the rest" $F \cap F^\perp$ into account.

You can read the Wikipedia article on orthogonal complements to get informed about this subject as well.

Hope, that helps you!

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  • $\begingroup$ I believe the sum could be not direct even if the form is non-degenerate, that g is not-degenerate only means that its kernel is 0. But not that the kernel of a restriction into any one subspace is 0. So it should be possible to have a non-degenerate form but still have a subspace which intersects its own orthogonal complement. Maybe I should have been clearer, g is non-degenerate as a bilinear form on E. $\endgroup$ – fiftyeight Nov 27 '13 at 12:47
  • $\begingroup$ No, you're actually wrong about that. If $g$ is non-degenerate and $F$ finite-dimensional, the orthogonal complement $F^\perp$ is a direct complement: $E = F \oplus F^\perp$ as you can see by choosing an orthonormal basis of $F$ and adding linearly independent orthonmal vectors to a full basis of $E$. Then the span of those added vectors is actually equal to $F^\perp$. $\endgroup$ – BIS HD Nov 27 '13 at 13:40
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    $\begingroup$ Over finite fields we can have a symmetric non-degenerate bilinear form and a subspace $C$ such that $C\cap C^\perp$ is not zero, in fact we could have $C\subseteq C^\perp$. The dimension formula still holds, in fact it's true for any pair of subspaces $E$ and $F$ (where $E=F^\perp$ above). $\endgroup$ – Chris Godsil Nov 27 '13 at 13:46
  • $\begingroup$ Thanks for clearing that! :-) I was (wrongly) assuming $K$ to be a real/complex vector space. $\endgroup$ – BIS HD Nov 27 '13 at 13:47
  • $\begingroup$ Even for real spaces. Consider $\langle (x,y), (u,v)\rangle = xu - yv$, and the subspace $C = \mathbb{R}\cdot (1,1)$. Then $C^\perp = C$. $\endgroup$ – Daniel Fischer Nov 27 '13 at 19:11

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