Let $\mathbf X \equiv [\mathbf x_i]$ be a $m \times n$ array of $n$, $m$-dimensional vectors, and $\mathbf X^+$ be its right Moore-Penrose pseudoinverse ($\mathbf X \mathbf X^+ = I_m$). Prove that $\sum_{i=1}^n \| \mathbf X^+ \mathbf x_i \|^2 = m$.
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1Since $XX^+=I_m$ you must have $m\le n$. However, it is impossible since you will have to have $n$ vectors with dimension $m$ which are independent. – Jlamprong May 08 '14 at 09:50
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Of course, you are right; I struck that requirement. – Emre May 08 '14 at 17:10
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$X^+x_i=X^+Xe_i$ and $X^+X$ is an orthogonal projector $P$ to the co-kernelimage of $X$ resp. the image of $X^T$ (or orthogonal complement of $\ker X$). Thus the sum is the trace of $P^TP=P^2=P$, which is exactly the dimension of the image of $X^T$, i.e., the rank of this matrix.
Lutz Lehmann
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The image of $X^T$ is the coimage of $X$, not the cokernel, but I got it, thanks. – Emre May 17 '14 at 22:57
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First, note that $x_i=Xe_i$ for all $i=1,\dots,n$ where $\{e_1,\dots,e_n\}$ is the basis standard for $\mathbb{R}^n$. Since $XX^+=I_m$ then $m\le n$. This means $X^+=X^T(XX^T)^{-1}$. So, \begin{align} \sum_{i=1}^n \| X^+ x_i \|^2 &=\sum_{i=1}^n \| X^T(XX^T)^{-1}X e_i \|^2\\ &=\sum_{i=1}^n (X^T(XX^T)^{-1}X e_i)^TX^T(XX^T)^{-1}X e_i \\ &=\sum_{i=1}^n e_i^TX^T(XX^T)^{-1}X e_i \\ &=\operatorname{trace}(X^T(XX^T)^{-1}X)\\ &=\operatorname{trace}(XX^T(XX^T)^{-1})\\ &=\operatorname{trace}(I_m)\\ &= m. \end{align}
Jlamprong
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