Prove this matrix algebraic equation

Question

Let $X$ be an $n \times m$ matrix, $x_i^T$ the $i$th row of $X$. Is it true that $\sum_{i=1}^n x_i^T (X^T X)^{-1} x_i = m$ ? How do I prove it?

Answer 1

We assume $m\leq n$ and $rank(X)=m$. We consider the thin SVD of $X$ cf. https://en.wikipedia.org/wiki/Singular_value_decomposition

$X=U_{nm}\Sigma_{mm} V_{m,m}^*$ where $U^*U=I_m,V^*V=I_m,\Sigma=diag((\sigma_i))$ where $\sigma_i>0$. The $i^{th}$ row of $X$ is $X_i=U_i\Sigma V^*$ where $U_i$ is the $i^{th}$ row of $U$. Note that $X^*X=V\Sigma^2V^*$ and $Y=(X^*X)^{-1}=V\Sigma^{-2}V^*$.

We calculate $\sum_{i=1}^n X_iYX_i^T=\sum_i U_i\Sigma V^*V\Sigma^{-2} V^*V\Sigma U_i^*=\sum_i U_i\Sigma\Sigma^{-2}\Sigma U_i^*=\sum_i U_iU_i^*$.

Remark that $tr(U^*U)=tr(I_m)=m=tr(UU^*)=\sum_i U_iU_i^*$ and we are done.