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For $i=1, \ldots, N$, let $x^i := [x^i_1, \ldots, x_n^i]^\top$ be a $n \times 1$ matrix in which every entry is non-negative and their sum is equal to $1$. This means $x^i$ belongs to the probability simplex of $\mathbb R^n$. I'm interested in when the sum $$ S :=\sum_{i=1}^N x^i (x^i)^\top $$ is invertible. If $n=2$, I get $$ \det S = \sum_{i=1}^N (x^i_1)^2 \sum_{i=1}^N (x^i_2)^2 - \left (\sum_{i=1}^N x^i_1 x^i_2 \right )^2. $$

By Cauchy-Schwarz inequality, $$ \det S = 0 \iff \frac{x_1^i}{x^i_2} = \frac{x_1^j}{x^j_2} \quad \forall i,j = 1,\ldots, N. $$

On the other hand, each $x^i_1+x^i_2 = 1$ for all $i$. Then the equality happens if and only if $x^i=x^j$ for all $i,j = 1,\ldots, N$.

Can we generalize above result to the case $n>2$?

Update: Let $X := [x^1, \ldots, x^N]$. Then $S = XX^\top$. Then $S$ is invertible if and only if $\{x^1, \ldots, x^N\}$ is linearly independent.

Analyst
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1 Answers1

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Let $X := [x^1, \ldots, x^N]$. Then $S = XX^\top$ which is the Gram matrix of $X$. Then $S$ is invertible if and only if $\operatorname{rank} S = n$ if and only if $\operatorname{rank} X = n$.

Analyst
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    What about $X=[e_1,e_1,e_2,e_2,\ldots,e_n,e_n]$, where $e_1,\ldots,e_n$ is the canonical basis? Then $S$ is just 2 times the identity matrix which is invertible. Your condition is for the case $N=n$. I suspect $\operatorname{span}(X)=\Bbb R^n$ is enough. – Surb Apr 04 '22 at 08:48
  • @Surb You're right! I made a mistake. – Analyst Apr 04 '22 at 09:06
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    @Analyst Then correct your answer. – KBS Apr 04 '22 at 10:15
  • Also, the comment/hint of @nsigma under the OP is the key for a complete answer. – Surb Apr 04 '22 at 12:28