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?
Asked
Active
Viewed 57 times
0
-
What ensures the existence of $(X^{\top}X)^{-1}$? What if $n<m$? – Servaes Jun 04 '16 at 12:14
-
Assume that $X$ has full column rank. – ywx Jun 04 '16 at 12:14
1 Answers
0
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.
-
Note that the matrices $U$ and $V$ are square, but I get the idea. In the end we have to calculate $\sum_{i=1}^n u_i^T \Sigma (\Sigma^T \Sigma)^{-1}\Sigma^T u_i$, where $\Sigma (\Sigma^T \Sigma)^{-1}\Sigma^T$ is n-by-n diagonal matrix with $m$ 1's on the diagonal, so the summation reduces to $m$ terms: $\sum_{i=1}^m u_i^T u^i = \sum_{i=1}^m 1 = m$ – ywx Jun 07 '16 at 09:41
-