Pythagorean theorem in linear regression using matrix notation












1












$begingroup$


I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
enter image description here



First of all I am confused of what "cost" is and then how we get from



$$arg min_x ||A(x-x') - bb'||_2^2 $$ to



$$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.



Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?



Thank you!






linear-algebra regression







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
    enter image description here



    First of all I am confused of what "cost" is and then how we get from



    $$arg min_x ||A(x-x') - bb'||_2^2 $$ to



    $$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.



    Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?



    Thank you!






    linear-algebra regression







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
      enter image description here



      First of all I am confused of what "cost" is and then how we get from



      $$arg min_x ||A(x-x') - bb'||_2^2 $$ to



      $$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.



      Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?



      Thank you!






      linear-algebra regression







      share|cite|improve this question









      $endgroup$




      I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
      enter image description here



      First of all I am confused of what "cost" is and then how we get from



      $$arg min_x ||A(x-x') - bb'||_2^2 $$ to



      $$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.



      Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?



      Thank you!






      linear-algebra regression




      linear-algebra regression






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 19:54









      YohanRothYohanRoth

      6261714




      6261714






















          1 Answer
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          1












          $begingroup$

          The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.



          The "Pythagorean theorem" refers to the fact that



          $$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.



          In general (Hilbert spaces over $Bbb{C}$), we have



          begin{align}
          ||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
          &= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
          &= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
          &= ||A(x-x')||_2^2 + ||b'||_2^2
          end{align}



          To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am not very clear on this math notation. What is x' or b'? What is Re? )
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:10










          • $begingroup$
            I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 7 '18 at 21:15










          • $begingroup$
            one more question . Why b = Ax' + b' must be orthogonal to col space of A?
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:28












          • $begingroup$
            @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 8 '18 at 9:11













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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.



          The "Pythagorean theorem" refers to the fact that



          $$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.



          In general (Hilbert spaces over $Bbb{C}$), we have



          begin{align}
          ||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
          &= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
          &= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
          &= ||A(x-x')||_2^2 + ||b'||_2^2
          end{align}



          To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am not very clear on this math notation. What is x' or b'? What is Re? )
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:10










          • $begingroup$
            I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 7 '18 at 21:15










          • $begingroup$
            one more question . Why b = Ax' + b' must be orthogonal to col space of A?
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:28












          • $begingroup$
            @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 8 '18 at 9:11


















          1












          $begingroup$

          The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.



          The "Pythagorean theorem" refers to the fact that



          $$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.



          In general (Hilbert spaces over $Bbb{C}$), we have



          begin{align}
          ||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
          &= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
          &= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
          &= ||A(x-x')||_2^2 + ||b'||_2^2
          end{align}



          To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am not very clear on this math notation. What is x' or b'? What is Re? )
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:10










          • $begingroup$
            I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 7 '18 at 21:15










          • $begingroup$
            one more question . Why b = Ax' + b' must be orthogonal to col space of A?
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:28












          • $begingroup$
            @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 8 '18 at 9:11
















          1












          1








          1





          $begingroup$

          The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.



          The "Pythagorean theorem" refers to the fact that



          $$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.



          In general (Hilbert spaces over $Bbb{C}$), we have



          begin{align}
          ||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
          &= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
          &= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
          &= ||A(x-x')||_2^2 + ||b'||_2^2
          end{align}



          To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.






          share|cite|improve this answer











          $endgroup$



          The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.



          The "Pythagorean theorem" refers to the fact that



          $$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.



          In general (Hilbert spaces over $Bbb{C}$), we have



          begin{align}
          ||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
          &= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
          &= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
          &= ||A(x-x')||_2^2 + ||b'||_2^2
          end{align}



          To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 9:09

























          answered Dec 7 '18 at 20:21









          GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

          12.8k72445




          12.8k72445












          • $begingroup$
            I am not very clear on this math notation. What is x' or b'? What is Re? )
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:10










          • $begingroup$
            I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 7 '18 at 21:15










          • $begingroup$
            one more question . Why b = Ax' + b' must be orthogonal to col space of A?
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:28












          • $begingroup$
            @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 8 '18 at 9:11




















          • $begingroup$
            I am not very clear on this math notation. What is x' or b'? What is Re? )
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:10










          • $begingroup$
            I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 7 '18 at 21:15










          • $begingroup$
            one more question . Why b = Ax' + b' must be orthogonal to col space of A?
            $endgroup$
            – YohanRoth
            Dec 7 '18 at 21:28












          • $begingroup$
            @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
            $endgroup$
            – GNUSupporter 8964民主女神 地下教會
            Dec 8 '18 at 9:11


















          $begingroup$
          I am not very clear on this math notation. What is x' or b'? What is Re? )
          $endgroup$
          – YohanRoth
          Dec 7 '18 at 21:10




          $begingroup$
          I am not very clear on this math notation. What is x' or b'? What is Re? )
          $endgroup$
          – YohanRoth
          Dec 7 '18 at 21:10












          $begingroup$
          I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 7 '18 at 21:15




          $begingroup$
          I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 7 '18 at 21:15












          $begingroup$
          one more question . Why b = Ax' + b' must be orthogonal to col space of A?
          $endgroup$
          – YohanRoth
          Dec 7 '18 at 21:28






          $begingroup$
          one more question . Why b = Ax' + b' must be orthogonal to col space of A?
          $endgroup$
          – YohanRoth
          Dec 7 '18 at 21:28














          $begingroup$
          @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 8 '18 at 9:11






          $begingroup$
          @YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
          $endgroup$
          – GNUSupporter 8964民主女神 地下教會
          Dec 8 '18 at 9:11




















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