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Let $A\in \mathcal{R}^{n\times d}$ be a matrix and its singular value decomposition be $A = U_A\Sigma_A V_A^T$. For any orthogonal matrix $U\in \mathcal{R}^{n\times l}$, let $U^\perp \in \mathcal{R}^{n\times (n-l)}$ be an orthogonal matrix whose column are orthonormal basis spanning the subspace of $\mathcal{R}^n$ that is orthogonal to $U$.

Then show that

$Z = \min_{x\in \mathcal{R}^d}\lvert\lvert Ax-b\rvert\rvert_2 = \lvert\lvert U^\perp_A U_A^{\perp^T}b\rvert\rvert_2$.

Note that this is the least square regression problem, and its exact solution is given by

$x^\star = A^+b = V_A\Sigma_A^{-1} U_A^Tb$.

Where $A^+$ is the Moore-Penrose pseudo inverse of $A$.

Reference this, Section 2.2.

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    $\begingroup$ What's your question ? $\endgroup$ – Gabriel Romon Oct 24 '17 at 19:52
  • $\begingroup$ To show that $Ax^\star-b = U_A^\perp U_A^{\perp^T}b$. $\endgroup$ – Sudipta Roy Oct 24 '17 at 20:06
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Hint 1:

$min_x \|f(x)\|_2 \Leftrightarrow min_x \|f(x)\|_2^2$

Hint 2:

$\|x\|_2^2 = x^Tx$

Can you differentiate this and get the point $x$ where the expression is minimized?

%%% Fixed the answer according to the author's comments:

Let $ e = b - Ax^\star$

Then, $ e = b - AA^+b = (I-AA^+)b = (I-U\Sigma \Sigma^+ U^T)b = U(I-\Sigma\Sigma^+)U^Tb$

So, the matrix in the middle (which is like the Eigenvalue decomposition) has diagonal entries to be $0$ corresponding to those $u_i$ which are the basis vectors of rangespace of $A$, but $1$ for the ones which are the basis vectors for the orthocomplement of rangespace of A.

So, $e \in \mathit{R^\perp(A)}$

Also, from the last equality, since only the vectors that contribute to the orthocomplement survive the matrix multiplication, we will have $e = U_\perp U_\perp^T b$.

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  • $\begingroup$ That would yield the normal equation. But how to use the $U_A^\perp$? $\endgroup$ – Sudipta Roy Oct 25 '17 at 7:02
  • $\begingroup$ Let $b^\perp = Ax^\star-b$. Then we know that $A^Tb^\perp=0$. Is is true that $b^\perp = U_A^\perp U_A^{\perp^T}b$? $\endgroup$ – Sudipta Roy Oct 25 '17 at 7:10
  • $\begingroup$ @SudiptaRoy Fixed. $\endgroup$ – learning Oct 25 '17 at 7:57
  • $\begingroup$ did not understand the last line. $\endgroup$ – Sudipta Roy Oct 25 '17 at 8:53
  • $\begingroup$ @SudiptaRoy Did you try matrix multiplication? $\endgroup$ – learning Oct 25 '17 at 18:09

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