Let $S=\{A_{1}, ..., A_{m}\}$ be a set of $n \times n$ symmetric matrices and $m>n$, the rank $r(A_{i})=1$ for each $i$. Suppose that for any $m-1$ matrices $\{A_{i_{1}},...,A_{i_{m-1}}\}$ in $S$, we have the rank of matrix $r(\Sigma_{j}A_{i_{j}}) \leq t$.

My question is:

$r(\Sigma_{i} A_{i}) \leq t$ ?

I think this is true for $n \leq 2$, but I have no idea to prove the general case.

  • 1
    $\begingroup$ If the $A_i$ are all scalar multiples of each other, then you can take $t=1$, while the sum of the ranks will be $m$, contradiction. $\endgroup$ Aug 10, 2013 at 5:53
  • $\begingroup$ @Gerry: we assume $m>n$. $\endgroup$
    – kiseki
    Aug 10, 2013 at 6:09
  • $\begingroup$ You may assume $m$ is as big as you like; my remark still holds. The rank of the sum is $1$; the sum of the ranks is $m$; $m\le1$ is false. $\endgroup$ Aug 10, 2013 at 6:10
  • $\begingroup$ Sorry, I have a typo $\endgroup$
    – kiseki
    Aug 10, 2013 at 6:14
  • $\begingroup$ If $v_i$ spans the image of $A_i$ then your hypothesis implies one cannot find $t+1$ independent $v_i$. To see this, suppose one could find them. Replace every $A_i$ with $SA_iS^t$, with $S$ chosen so that the $t+1$ vectors $Sv_i$ are part of the standard basis. $\endgroup$ Aug 10, 2013 at 8:28

1 Answer 1


If the characteristic of your base field is 0 the answer is yes. If the characteristic is positive, the answer is no in general. First the proof for characteristic 0, since it will also show how to get counterexamples in positive characteristic.

Assume that the sum of all $A_i$ (denote it by $B$) has rank $t+1$. By general theory on symmetric matrices resp. quadratic forms there is an invertible matrix $S$ such that $SBS^t$ is of diagonal form and the first $t+1$ diagonal entries are nonzero whereas the others are $0$. By also transforming the $A_i$ we therefore can assume that already $B$ has this diagonal form.
By assumption on the sum of $m-1$ matrices we have that $B-A_i$ has rank $t$ for all $i$. Since $A_i$ has rank $1$, its columns are multiples of some column vector $w$, and since it is also symmetric, it is even of the form $a_w\cdot w\cdot w^t$ for some number $a_w$. The columns of $B-A_i$ are therefore of the form $d_j \cdot \underline{e}_j-a_w w_j\cdot w$ ($\underline{e}_j$ the $j$-th standard basis vector, $d_j$ the $j$-th diagonal entry of $B$.), and it is not hard to see that the rank is even $t+2$ if $w$ is not in the span of $\underline{e}_1,\dots,\underline{e}_{t+1}$.

Therefore, everything takes place in the $(t+1)\times (t+1)$-upper left submatrix. By ignoring the rest of the matrix, we are therefore reduced to the case $t+1=n$.

Now back to the condition that $B-A_i$ has rank $t=n-1$. Then there is a vector $v\ne 0$ such that $(B-A_i)v=0$, i.e. $$\begin{pmatrix} d_1v_1\\ \vdots \\ d_nv_n\end{pmatrix}=a_w (\sum_{i=1}^n v_iw_i) \begin{pmatrix} w_1\\ \vdots \\ w_n\end{pmatrix} $$ (where we use the $d_j$, $a_w$ and $w$ as above).
Since all the $d_j$ are nonzero, we obtain $v_j=\frac{a_ws}{d_j}w_j$ where $0\ne s:=\sum_{i=1}^n v_iw_i$. This leads to $$s=\sum_{j=1}^n v_jw_j =a_ws \sum_{j=1}^n \frac{w_j^2}{d_j}$$ resp. $$a_w \sum_{j=1}^n \frac{w_j^2}{d_j}=1.\tag{1}$$

After all this preparatory work, we consider the diagonal entries of $B$ which are on one hand $d_j$ and on the other $\sum_{w} a_w w_j^2$. Hence, $$\sum_{w} a_w \frac{w_j^2}{d_j}=1 $$ We finally get: $$n=\sum_{j=1}^n \left(\sum_{w} a_w \frac{w_j^2}{d_j}\right)=\sum_{w} \left( \sum_{j=1}^n a_w \frac{w_j^2}{d_j}\right)=\text{number of }w\text{'s}=m$$

Remark:This even shows more than the question, namely if the rank of the sum of $m$ matrices is $t+1$ and the rank of the sum of arbitrary $m-1$ of those is $t$, then $m=t+1$.

Why does it not work in characteristic $p>0$? Because the last equation is an equation in the field, i.e. we only obtain $n\equiv m \mod{p}$. Indeed, since the condition (1) is sufficient for the $w$ (resp. the matrices $A$) one can construct examples for which $m=n+p$, even for $n=2$.


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