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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 $n \gg k$?

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  • $\begingroup$ 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? $\endgroup$ – darij grinberg Aug 31 '10 at 16:53
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    $\begingroup$ The answer to this is no. Take for instance the elementary matrices $E_{1j}$ for $1 \le j \le n$. $\endgroup$ – Keivan Karai Aug 31 '10 at 17:13
  • $\begingroup$ 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... $\endgroup$ – Will Jagy Aug 31 '10 at 17:22
  • $\begingroup$ Keivan, the matrices would be symmetric. $\endgroup$ – Will Jagy Aug 31 '10 at 17:23
  • $\begingroup$ 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. $\endgroup$ – Keivan Karai Aug 31 '10 at 17:30
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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.

The equivalent formulation in terms of symmetric matrices is that any linear combination of \begin{equation*} \begin{pmatrix} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ 0 & 0 & 0 \end{pmatrix} \quad\mbox{and}\quad \begin{pmatrix} 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ 1 & 0 & 0 \end{pmatrix}\end{equation*} is singular, but if we put the matrices side by side, then \begin{equation*} \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 1 \\\ 1 & 0 & 0 & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} \end{equation*} has full rank.

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