I am currently studying  answer on the Moore-Penrose pseudoinverse and the Euclidean norm by the user &quot;Etienne dM&quot;. The first point of their answer proceeds as follows:  Let $x$ be $A^+y$.  Let me begin by the second point. For all $z$, we have: \begin{align} \lVert Az-b \rVert_2^2 &amp;= \lVert Ax-b \rVert_2^2 + \lVert A(z-x) \rVert_2^2 + 2 (z-x)^TA^T(Ax-b)\\ &amp; \geq  \lVert Ax-b \rVert_2^2 + 2 (z-x)^TA^T(Ax-b) \end{align} Moreover, because $(AA^+)^T = AA^+$, $$ A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$$ Thus, we prove that for all $z$, $\rVert Az-b \lVert_2^2 \geq\rVert Ax-b \lVert_2^2$, that is to say $A^+b$ is as close as possible to $y$ in term of the Euclidian norm $\lVert Ax-b\rVert_2$.   I realise that $x = A^+ y$, but I don't understand how any of this implies that &quot;$A^+ b$ is as close as possible to $y$ in terms of the Euclidean norm $\lVert Ax-b\rVert_2$&quot;. The way I see it is that we have 4 facts:  $\lVert Az - b \rVert_2^2 = \lVert Ax - b \rVert_2^2 + \lVert A(z - x) \rVert_2^2 + 2(z - x)^T A^T (Ax - b) \ge \lVert Ax - b \rVert_2^2 + 2(z - x)^T A^T (Ax - b)$;  Because $(AA^+)^T = AA^+$, $A^T(Ax - b) = ((AA^+)A)^T b - A^T b = 0$;  Singular value decomposition (SVD): $A^+ = VD^+U^T = V \Sigma^+ U^T$;  $x = A^+ y$.   Supposedly, these 4 facts together prove that &quot;$A^+ b$ is as close as possible to $y$ in terms of the Euclidean norm $\lVert Ax-b\rVert_2$&quot;. However, I do not see how this is so.  answer by &quot;Ben Grossmann&quot; is supposed to show something similar. I can see how this answer is related to what &quot;Etienne dM&quot; did in their proof, but I do not see how it is the same. I suspect that I am lacking some assumed, fundamental knowledge that &quot;Etienne dM&quot; had in constructing this proof, and so I am unable to see how these 4 facts combine to prove what is claimed. I would greatly appreciate it if people would please take the time to carefully explain this to me.  related question is for the limit definition of pseudoinverse.
I am currently studying  answer on the Moore-Penrose pseudoinverse and the Euclidean norm by the user &quot;Etienne dM&quot;. The first point of their answer proceeds as follows:  Let $x$ be $A^+y$.  Let me begin by the second point. For all $z$, we have: \begin{align} \lVert Az-b \rVert_2^2 &amp;= \lVert Ax-b \rVert_2^2 + \lVert A(z-x) \rVert_2^2 + 2 (z-x)^TA^T(Ax-b)\\ &amp; \geq  \lVert Ax-b \rVert_2^2 + 2 (z-x)^TA^T(Ax-b) \end{align} Moreover, because $(AA^+)^T = AA^+$, $$ A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$$ Thus, we prove that for all $z$, $\rVert Az-b \lVert_2^2 \geq\rVert Ax-b \lVert_2^2$, that is to say $A^+b$ is as close as possible to $y$ in term of the Euclidian norm $\lVert Ax-b\rVert_2$.   I realise that $x = A^+ y$, but I don't understand how any of this implies that &quot;$A^+ b$ is as close as possible to $y$ in terms of the Euclidean norm $\lVert Ax-b\rVert_2$&quot;. I would greatly appreciate it if people would please take the time to explain this to me.
How do I access a page's HTTP response headers via JavaScript?  Related to , which was modified to ask about accessing two specific HTTP headers.     Related: Is the W statistic output by wilcox.test() in R the same as the U statistic? Making your Windows 7 a WIFI hotspot for Android? I use a broadband dongle on my Windows 7. I want to expose my laptop as a wifi hotspot so that i can use internet on my samsung galaxy ace.  I have tried Connectify but my chipset is not supported so it simply make the adhoc network. Tried Bzeek. My phone get connected, but no browsing.  Any solution that you guys have?  PS: I am in no mood to root my device. How to get over "device or resource busy"? how to pass a value to a variable that in a other variable I am getting a "bad substitution" error while running this code. 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First Method  $\newcommand{\rd}{\mathrm{d}} \Pr(Z\leq z) = \Pr(X/Y\leq z) = \Pr(X\leq zY) = \int^{1}_{0}\int^{\min(1,zy)}_{0} \rd x \rd y = \int^{1}_{0}\min(1,zy)\ \rd y$ $  = \left\{      \begin{array}{lr}        \int^{1/z}_{0}zy\ \rd y + \int^{1}_{1/z} \rd y&amp; : z &gt; 1\\        \int^{1}_{0}zy\ \rd y &amp; : z \leq 1      \end{array}    \right. $ $  = \left\{      \begin{array}{lr}        1 - \frac{1}{2z} &amp; : z &gt; 1\\        \frac{z}{2} &amp; : z \leq 1      \end{array}    \right. $  This appears correct.  Second Method  $\Pr(X/Y\leq z) = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(zY \geq 1) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(zY &lt; 1) $ by total probability  $ = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(Y \geq 1/z) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(Y &lt; 1/z)$  Taking $z&gt;1$ yields $ (1)(1-\frac{1}{z}) + (\int^{1/z}_{0}\int^{zy}_{0} \rd x \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + (\int^{1/z}_{0}zy\ \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + \frac{1}{2z^{2}}$  This is already different. 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