Is the space of symmetric matrices generated by rank-$1$ matrices?

Question

Let $\mbox{Sym}_n (\mathbb K)$ the $\mathbb K$-linear space of symmetric matrices over a field $\mathbb K$.

1. Is $\mbox{Sym}_n(\mathbb K)$ generated by symmetric rank 1 matrices ?

2. If yes, can an explicit basis of rank-$1$ matrices be described ?

If $\mathbb K = \mathbb R$, then the answer of question 1 seems to be "yes" because of the singular value decomposition. But I have no idea on how to answer question 2 by giving $\binom{n+1}{2}$ linearly independent rank-$1$ symmetric matrices.

I am also wondering if question 1 would depend on the field $\mathbb K$.

Many thanks!

Answer 1

1. yes

2. Denote $e_k = (0,\dots,0,1,0,\dots ,0)^T$ the usual $k$-th unit vector.

The claim is that the matrices $$\begin{split} v_{ij} &= (e_i + e_j)(e_i + e_j)^T \quad i=1\dots n, j=i+1\dots n,\\ v_{ii} &= e_ie_i^T \quad i=1\dots n \end{split}$$ are a basis of the space of symmetric matrices. First, note that the number of matrices of this type is precisely the dimension of this space. It remains to show linear independence.

The matrices $v_{ii}$ are diagonal with exactly one zero on the diagonal and all other entries zero. The matrices $v_{ij}$, $i\ne j$ contain four non-zero entries equal to one at positions $(i,j),(j,i),(i,i),(j,j)$.

(New proof of linear independence) Let numbers $a_{i,j}$, $i=1\dots n$, $j=i\dots n$ be given such that $$ \sum_{i=1}^n\sum_{j=1}^{n} a_{ij} v_{ij}=0. $$ Let $1\le k < l \le n$ be two indices. Then the $(k,l)$-entry of $\sum_{i=1}^n\sum_{j=1}^n a_{ij} v_{ij}$ is equal to $a_{kl}$. This shows $a_{kl}=0$ for all off-diagonal entries $k\ne l$. The equation reduces to $$ \sum_{i=1}^n a_{ii} v_{ii}=0. $$ The $(k,k)$-entry of this linear combination of $v_{ii}$'s is equal to $a_{kk}$, so $a_{kk}=0$ for all $k$.

---

(Old proof, which was accepted) I will prove per induction with respect to $m$ that the set of vectors $$ \{ v_{ij} : i = 1\dots m,\ j=i\dots m\} $$ is linearly independent. The claim is true for $m=1$. Now suppose this claim was proven for some $m$ with $n>m\ge0$.

Consider the case $m+1$. Let numbers $a_{i,j}$, $i=1\dots m+1$, $j=i\dots m+1$ be given such that $$ \sum_{i=1}^{m+1}\sum_{j=i}^{m+1} a_{ij} v_{ij}=0. $$ This is a equation in vector space of matrices. Now let us look at the row $m+1$ of the above linear combination. The entry in row $m+1$, column $k$ is equal to $a_{k,m+1}$. Hence the numbers $a_{k,m+1}$ are zero, $k=1\dots m+1$.

By the induction hypothesis it follows that all the other $a_{ij}$ are zero as well. Hence, the vectors are linearly independent, and thus form a basis.