# Rank of a linear combination of quadratic forms

Suppose we have a set of quadratic forms $Q_i (x_1, \dots, x_n)$ for $1 \leq i \leq k$ in $n$ variables, defined over $\mathbb{R}$. We suppose these are 'collectively nondegenerate' in the sense that there does not exist a change of variables which takes us into a set of quadratic forms with less than $n$ variables.

I am looking at linear combinations of these forms: $$Q_{\boldsymbol{\lambda}}(\textbf{x})=\sum_i \lambda_i Q_i(x_1, \dots, x_n)$$ for $\boldsymbol{\lambda} = (\lambda_1, \dots , \lambda_k) \in \mathbb{R}^k$. My question is whether we are guaranteed a set of $\lambda$s which gives us a quadratic form of full rank i.e. $n$? Edit:: this has been shown to be untrue, so...

Is there anything we can do to guarantee a 'high' rank, say bigger than 5? For example by taking $n \gg k$?

• In matrixspeak: Given $k$ square matrices whose kernels have trivial intersection, can we find a nonsingular matrices which can be written as a linear combination of our $k$ matrices? – darij grinberg Aug 31 '10 at 16:53
• The answer to this is no. Take for instance the elementary matrices $E_{1j}$ for $1 \le j \le n$. – Keivan Karai Aug 31 '10 at 17:13
• Could you please show what happens with only 2 variables? Your command boldsymbol does not work for me, I am still using jsMath, but you might switch to \bf or leave it out... – Will Jagy Aug 31 '10 at 17:22
• Keivan, the matrices would be symmetric. – Will Jagy Aug 31 '10 at 17:23
• Will: I agree and I don't know how to do it for symmetric matrices. My comment was in refernce to darji's comment, basically saying that the extra assumption is needed. – Keivan Karai Aug 31 '10 at 17:30

The answer to the first part (about finding a linear combination which has full rank) is no. A counterexample with $n=3$ and $k=2$ is given by the quadratic forms $xy$ and $xz$. A general linear combination of these two is of the form $\lambda_1 xy + \lambda_2 xz = x(\lambda_1 y + \lambda_2 z)$, which obviously has rank 2.