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Let $V$ a vector subspace of $\mathbb{R}^4$. The bilinear form $f(x,y) = x_1y_1 + x_2y_2 + x_3y_3 - x_4y_4$ where $x,y \in \mathbb{R}$. Let $V^\bot = \{ y \in \mathbb{R}^4 : f(x, y) = 0 \ \ \ \forall x \in V \}$.

Prove that $ \text{dim}(V) + \text{dim}(V^\bot) = 4$.


This problem currently had multiple subproblems which I've already solved successfully (e.g. prove that $V^\bot$ is a subspace, find $V$ that $ V \cap V^\bot \ne \{ 0 \} $ ).

I've tried giving a basis of $V \cap V^\bot$ and extending it to a basis of $V$ and a basis of $V^\bot$ but I couldn't go further.

I'll appreciate any hint. Thanks.

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up vote 10 down vote accepted

This doesn't have much to do with the particular field $\mathbf R$, number $4$, or form $f$—we get a similar formula for any non-degenerate bilinear form on a finite dimensional vector space. We know that $f$ is non-degenerate because the matrix of $f$ with respect to the standard basis has determinant $-1$.

Thus $f$ induces an isomorphism $\mathbf R^4 \to (\mathbf R^4)^*$ sending $x$ to the functional $y \mapsto f(x, y)$. Now compose with the restriction map $(\mathbf R^4)^* \to V^*$, which is surjective. What is the kernel of this composition?

[The map $(\mathbf R^4)^* \to V^*$ is dual to the inclusion $V \to \mathbf R^4$. It is surjective because we can find a decomposition $\mathbf R^4 = V \oplus W$ and use the universal property of direct sums to extend any functional on $V$.]

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I've thought on this before, I will give a try again. The dual space have the same dimension as $R^4$. Then I can use the dimension formula as the kernel of the composition is the $V^\perp$ subspace. So, the key is to find the restriction map (or matrix representation) and show that is surjective? – Rolando Dec 21 '11 at 5:05
Another way to think about this: you have an exact sequence $0 \to V \to \mathbf R^4 \to \mathbf R^4/V \to 0$. This yields an exact sequence $0 \to (\mathbf R^4/V)^* \to (\mathbf R^4)^* \to V^* \to 0$. Now, the dual space of $\mathbf R^4/V$ is precisely the subspace of $\mathbf R^4$ consisting of functionals that vanish on $V$. Under our isomorphism $V \to V^*$, these correspond to vectors in the annihilator! – Dylan Moreland Dec 21 '11 at 21:06

Write the bilinear form as $f(x, y) = \langle Tx, y \rangle$ where $T : \mathbb R^4 \to \mathbb R^4$ is the invertible linear transformation given by $T(x_1, x_2, x_3, x_4) = (x_1, x_2, x_3, -x_4)$. Then $V^\perp$ is just the orthogonal complement of $T(V)$; so $\dim \ V^{\perp} + \dim \ T(V) = 4$. The given claim then follows by noting that $\dim \ T(V) = \dim \ V$ by invertibility of $T$. $\qquad \square$

Just like Dylan's elegant answer, the above proof works for all non-degenerate bilinear forms $f(x,y)$. This is because a bilinear form $f(x,y)$ is non-degenerate if and only if it can be written as $\langle Tx, y \rangle$ for an invertible $T : \mathbb R^4 \to \mathbb R^4$.

As a final note, I think that the above proof is closely related to Dylan's answer. Indeed, where I use the linear transformation $T$, he considers the isomorphism $y \mapsto \langle Tx, y \rangle$. Similarly, I use the relation between the dimensions of a subspace and it's orthogonal complement, while he invokes the rank-nullity theorem.

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Good one. If I understand right, you make of f a inner product and the orthogonality complement theorem is applicable. – Rolando Dec 21 '11 at 4:59
@RolandE Yes, your understanding seems correct to me. – Srivatsan Dec 21 '11 at 5:59

Just FYI, Dylan's argument is a standard one in the theory of bilinear forms, and can be found in Chapter 1 of most treatises on that subject. For instance, see Proposition 7 of these notes.

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Thanks for the reference—I've used your notes on quadratic forms before but didn't remember this proposition. I bet the Conrads have expositions as well. The only book I own that touches on this sort of thing is Lang, and sadly I don't think that he ever spells this out. – Dylan Moreland Dec 22 '11 at 0:34
Thanks for the link. I'm choosing Dylan's answer mainly because I had a lecture on dual spaces which was based in Steven Roman's Advance Linear Algebra book, although we don't get deep neither learnt the 'nondegenerate' definition. – Rolando Dec 22 '11 at 4:04

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