Can $\sum_{i=1}^N \mathbf r_i^T \big (\sum_{j=1}^N \mathbf r_j \mathbf r_j^T \big)^{-1} \mathbf r_i$ be simplified?

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Let $\{\mathbf r_i \in \mathbb R^n \}_{i=1\dots N}$ be some linearly independent column-vectors. I think it must be possible to simplify the expression

$$ \sum_{i=1}^N \mathbf r_i^T \big (\sum_{j=1}^N \mathbf r_j \mathbf r_j^T \big)^{-1} \mathbf r_i, $$

to a constant, however I am unsure.

Note that the bracketed term is a rank-$N$ matrix. You can safely assume here that $N > n$ so that the inverse exists.

Can this expression be reduced to a constant? Please (dis)prove or provide a condition.

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Your quantity (I'll call it $Q$) is a number, but think of it as the trace of the matrix $\sum r_i^T M^{-1} r_i$ where $M=\sum r_i r_i^T$. But $\operatorname{tr} (AB)=\operatorname{tr} (BA),$ so $$Q = \operatorname{tr} \left(M^{-1} \sum r_i r_i^T \right)= \operatorname{tr} (M^{-1} M) = \operatorname{tr} I = n.$$