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 $n \gg k$?



Rank of a linear combination of quadratic forms
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.


