Linear combination of symmetric matrices

Let $A, B, C, D$ be four linearly independent symmetric 3 x 3-matrices over $\mathbb K$. Show that some linear combination of these matrices is a matrix of rank 1.

I know it is supposed to be a question in Algebraic Geometry, so it must be interesting :)

Solution 1:

First, you need to assume that $\mathbb{K}$ is algebraically closed. For instance, over $\mathbb{R}$, take

$$ A=\left(\begin{array}{ccc}1& 0& 0 \\0&-1&0\\0&0&0 \end{array}\right),B=\left(\begin{array}{ccc}0& 0& 0 \\0&-1&0\\0&0&1 \end{array}\right),C=\left(\begin{array}{ccc}0& 0& 1 \\0&0&0\\1&0&0 \end{array}\right),D=\left(\begin{array}{ccc}0& 0& 0 \\0&0&1\\0&1&0 \end{array}\right) $$ The matrices are linearly independent, and each has rank 2. But no linear combination gives a rank 1 matrix. (Over $\mathbb{C}$ you can take $c_1=1,c_2=-1,c_3=i,c_4=0$.)

So suppose $\mathbb{K}$ is algebraically closed. You may as well show that some combination of the matrices factors as the product $\alpha\alpha'$ for some non-zero vector $\alpha\in \mathbb{K}^3$. Thus you want to solve for $$ \alpha\alpha'=\sum_{i}c_iA_i $$ where the left hand side is $$ \alpha\alpha'=\left(\begin{array}{ccc}\alpha_1^2& \alpha_1\alpha_2& \alpha_1\alpha_3\\\alpha_1\alpha_2&\alpha_2^2&\alpha_2\alpha_3\\\alpha_1\alpha_3&\alpha_2\alpha_3&\alpha_3^2 \end{array}\right). $$ By symmetry, you only need to solve for the upper half. This gives you 7 indeterminates $(\alpha_i,c_i)$'s and 6 equations - call them $f_j$'s. Now each equation drops the dimension of $\mathbb{K}[\alpha_i,c_i]$ by at most 1. More precisely, set $X=\mathrm{Spec}(K[\alpha_i,c_i]/(f_j))$. Then $X$ has dimension at least 1 as a variety. Now if $\mathbb{K}$ is algebraically closed, the points on $X$ are geometric, i.e., they are elements in $\mathbb{K}^3$. Now (by the assumption of linear independence), the set of points on $X$ at which $\alpha\alpha'$ has rank 0 is a subvariety of dimension 0 (namely, it is the origin). So you can pick any point outside of it (which exists by dimension reasons) and this gives you a pair $(\alpha_i,c_i)$ such that $\alpha\alpha'$ is rank 1.