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Let $ V $ be a vector space of finite dimension on $\mathbb{C} $ and $ \dim(V) \gt 1 $. Show that for every quadratic form $ q : V \to \mathbb{C} $ there exists $ 0 \neq v \in V $ such that $ q(v) = 0 $.

It certainly has something to do with the fact that we are in $\mathbb{C} $ and not $ \mathbb{R} $, but I'm since $ q $ can be represented by any non-singular complex matrix, I don't know how this is even possible.

(edit: V is of finite dimension)

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Is the vector space of finite dimension? – Patrick Da Silva Nov 13 '11 at 22:22
If the vector space is finite dimensional, then this statement can certainly not be true. – Manos Nov 13 '11 at 22:35
Are you sure it's not true? This question appeared in a test as is. Can you prove why it's not true? – dan Nov 13 '11 at 22:42
This can happen even in $\mathbb{R}$ but not to all quadratic forms e.g. $$\pmatrix{1\\0}^T\pmatrix{0&1\\1&0}\pmatrix{1\\0} = 0$$. Note that the matrix is invertible. – user13838 Nov 13 '11 at 22:43
@dan Any positive-definite matrix defines a quadratic form, which in turn defines a norm and there is not chance that a nonzero vector is mapped to zero. – Manos Nov 13 '11 at 22:45
up vote 2 down vote accepted

Consider the case $n=2$. Thus we have $$ q(\begin{bmatrix} c_1 & c_2 \\ \end{bmatrix}) = \begin{bmatrix} c_1 & c_2 \\ \end{bmatrix} \begin{bmatrix} q_{11} & q_{12} \\ q_{21} & q_{22} \\ \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \end{bmatrix} = q_{11} c_1^2 + (q_{12} + q_{21}) c_1 c_2 + q_{22}c_2^2 $$ For every value of $c_2$, we can consider the polynomial in one variable $q_{c_2}(c_1) = q(\begin{bmatrix} c_1 & c_2 \\ \end{bmatrix})$. Since $\mathbb C$ is algebraically closed, this polynomial always has two roots. Thus we have a solution by letting $c_2 \neq 0$.

If $n > 2$, note that $q( \begin{bmatrix} c_1 & c_2 & 0 & \dots & 0 \end{bmatrix} )$ has a non-zero solution since it is a quadratic form in two variables.

If $V $ has infinite dimension, it suffices to take a $2$-dimensional subspace of $V$ and apply the case $n=2$.

Note that this result holds whatever is the matrix that represents $q$, i.e. in the finite dimension case, you could consider singular matrices as well.

Hope that helps,

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Edited to reflect that it's an Hermitian form. The original question also called it quadratic form, so I guess from there the confusion ensued... – dan Nov 13 '11 at 23:05
@dan: actually you have to switch back to "quadratic". In my answer i showed it's not true for Hermitian. But as Patrick shows, it is true for "quadratic". – Manos Nov 13 '11 at 23:07
@dan : You edited what? What is Hermitian? Your comment just makes me more and more confused. – Patrick Da Silva Nov 13 '11 at 23:19
@Patrick: In your comment to my answer that i deleted, maybe if instead of saying that "Hermitian forms are Hermitian forms and Quadratic forms are Quadratic forms" you actually said something more useful, then you would avoid getting confused yourself as well. – Manos Nov 13 '11 at 23:22
You seem to react as if I insulted you... I am confused about dan's comment because he's saying something is an Hermitian form and nothing is Hermitian around here. I did say something useful because I pointed out exactly where your misunderstanding of the question was, so don't take things personally in a community where we're just all trying to help out. Pointing out that someone else is wrong is only insulting when it is done with bad intentions. – Patrick Da Silva Nov 13 '11 at 23:28

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