Solving overdetermined system by QR decomposition Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What does QR decomposition have to do with least squares method?Check Whether an Overdetermined Linear Equation System is Consistent: General Approach?Is the least squares solution to an overdetermined system a triangle center?Solving a feasible system of linear equations using Linear ProgrammingProving unique solution exists for a system of equationsSolve an overdetermined system of linear equationsSolving overdetermined linear system with $3$ equations in $2$ unknownsOverdetermined System Ax=bSolving a system by using Cholesky Decomposition $(LDL^T)$Solving $Ax = b$ using least squares (minimize $||Ax -b||_2$)How to do Given's rotation for $3×2$ matrix? (QR decomposition)

Why aren't air breathing engines used as small first stages

Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?

English words in a non-english sci-fi novel

Why do people hide their license plates in the EU?

What does this icon in iOS Stardew Valley mean?

What's the purpose of writing one's academic biography in the third person?

How to answer "Have you ever been terminated?"

What is Wonderstone and are there any references to it pre-1982?

Why did the rest of the Eastern Bloc not invade Yugoslavia?

When were vectors invented?

Can I cast Passwall to drop an enemy into a 20-foot pit?

Storing hydrofluoric acid before the invention of plastics

What is the role of the transistor and diode in a soft start circuit?

Echoing a tail command produces unexpected output?

How widely used is the term Treppenwitz? Is it something that most Germans know?

How do I stop a creek from eroding my steep embankment?

ListPlot join points by nearest neighbor rather than order

Identify plant with long narrow paired leaves and reddish stems

Can a non-EU citizen traveling with me come with me through the EU passport line?

Output the ŋarâþ crîþ alphabet song without using (m)any letters

51k Euros annually for a family of 4 in Berlin: Is it enough?

Resolving to minmaj7

When a candle burns, why does the top of wick glow if bottom of flame is hottest?

How to align text above triangle figure



Solving overdetermined system by QR decomposition



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What does QR decomposition have to do with least squares method?Check Whether an Overdetermined Linear Equation System is Consistent: General Approach?Is the least squares solution to an overdetermined system a triangle center?Solving a feasible system of linear equations using Linear ProgrammingProving unique solution exists for a system of equationsSolve an overdetermined system of linear equationsSolving overdetermined linear system with $3$ equations in $2$ unknownsOverdetermined System Ax=bSolving a system by using Cholesky Decomposition $(LDL^T)$Solving $Ax = b$ using least squares (minimize $||Ax -b||_2$)How to do Given's rotation for $3×2$ matrix? (QR decomposition)










3












$begingroup$


I need to solve $Ax=b$ in lots of ways using QR decomposition.



$$A = beginbmatrix
1 & 1 \
-1 & 1 \
1 & 2
endbmatrix, b = beginbmatrix
1 \
0 \
1
endbmatrix$$



This is an overdetermined system. That is, it has more equations than needed for a unique solution.



I need to find $min ||Ax-b||$. How should I solve it using QR?



I know that QR can be used to reduce the problem to
$$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^-1b Vert.$$



but what do I do after this?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    I need to solve $Ax=b$ in lots of ways using QR decomposition.



    $$A = beginbmatrix
    1 & 1 \
    -1 & 1 \
    1 & 2
    endbmatrix, b = beginbmatrix
    1 \
    0 \
    1
    endbmatrix$$



    This is an overdetermined system. That is, it has more equations than needed for a unique solution.



    I need to find $min ||Ax-b||$. How should I solve it using QR?



    I know that QR can be used to reduce the problem to
    $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^-1b Vert.$$



    but what do I do after this?










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      I need to solve $Ax=b$ in lots of ways using QR decomposition.



      $$A = beginbmatrix
      1 & 1 \
      -1 & 1 \
      1 & 2
      endbmatrix, b = beginbmatrix
      1 \
      0 \
      1
      endbmatrix$$



      This is an overdetermined system. That is, it has more equations than needed for a unique solution.



      I need to find $min ||Ax-b||$. How should I solve it using QR?



      I know that QR can be used to reduce the problem to
      $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^-1b Vert.$$



      but what do I do after this?










      share|cite|improve this question









      $endgroup$




      I need to solve $Ax=b$ in lots of ways using QR decomposition.



      $$A = beginbmatrix
      1 & 1 \
      -1 & 1 \
      1 & 2
      endbmatrix, b = beginbmatrix
      1 \
      0 \
      1
      endbmatrix$$



      This is an overdetermined system. That is, it has more equations than needed for a unique solution.



      I need to find $min ||Ax-b||$. How should I solve it using QR?



      I know that QR can be used to reduce the problem to
      $$Vert Ax - b Vert = Vert QRx - b Vert = Vert Rx - Q^-1b Vert.$$



      but what do I do after this?







      linear-algebra numerical-methods numerical-linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 12 at 15:37









      Guerlando OCsGuerlando OCs

      16321856




      16321856




















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbbR^m times n,R in mathbbR^n times n$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            Apr 12 at 15:45










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            Apr 12 at 15:46



















          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = beginbmatrix y_1 \ y_2 \ 0endbmatrix $$
          , so if $$ Q^-1b = beginbmatrix z_1 \ z_2 \ z_3endbmatrix$$
          then $|| Rx - Q^-1b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = beginbmatrix R_1 \ 0endbmatrix$ and intoduce $P=beginbmatrix1 & 0 & 0 \ 0 & 1& 0endbmatrix$, then you have
          $$ R_1x = PQ^-1b$$
          $$ x = (R_1)^-1PQ^-1b$$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            Apr 12 at 20:34










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
            $endgroup$
            – Adam Latosiński
            Apr 13 at 11:14












          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185239%2fsolving-overdetermined-system-by-qr-decomposition%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbbR^m times n,R in mathbbR^n times n$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            Apr 12 at 15:45










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            Apr 12 at 15:46
















          7












          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbbR^m times n,R in mathbbR^n times n$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            Apr 12 at 15:45










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            Apr 12 at 15:46














          7












          7








          7





          $begingroup$

          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbbR^m times n,R in mathbbR^n times n$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.






          share|cite|improve this answer











          $endgroup$



          The most straightforward way I know is to pass through the normal equations:



          $$A^T A x = A^T b$$



          and substitute in the $QR$ decomposition of $A$ (with the convention $Q in mathbbR^m times n,R in mathbbR^n times n$). Thus you get



          $$R^T Q^T Q R x = R^T Q^T b.$$



          But $Q^T Q=I_n$. (Note that in this convention $Q$ isn't an orthogonal matrix, so $Q Q^T neq I_m$, but this doesn't matter here.) Thus:



          $$R^T R x = R^T Q^T b.$$



          If $A$ has linearly independent columns (as is usually the case with overdetermined systems), then $R^T$ is injective, so by multiplying both sides by the left inverse of $R^T$ you get



          $$Rx=Q^T b.$$



          This system is now easy to solve numerically.



          For numerical purposes it's important that the removal of $Q^T Q$ and $R^T$ from the problem is done analytically, and in particular $A^T A$ is never constructed numerically.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 12 at 19:46

























          answered Apr 12 at 15:44









          IanIan

          69.2k25393




          69.2k25393







          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            Apr 12 at 15:45










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            Apr 12 at 15:46













          • 1




            $begingroup$
            Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
            $endgroup$
            – Martin Argerami
            Apr 12 at 15:45










          • $begingroup$
            @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
            $endgroup$
            – Ian
            Apr 12 at 15:46








          1




          1




          $begingroup$
          Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
          $endgroup$
          – Martin Argerami
          Apr 12 at 15:45




          $begingroup$
          Is there any reason to make this so convoluted? From $Ax=b$ you have $QRx=b$, multiply by $Q^T$ on the left.
          $endgroup$
          – Martin Argerami
          Apr 12 at 15:45












          $begingroup$
          @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
          $endgroup$
          – Ian
          Apr 12 at 15:46





          $begingroup$
          @MartinArgerami Because actually the least squares solution usually does not satisfy $Ax=b$. This simple perspective only shows you that this approach gives you a solution when a solution exists. Now you could argue directly that multiplying both sides by $Q^T$ furnishes an equation whose solution is the least squares solution. (Such an argument would resemble the usual geometric argument for deriving the normal equations.) This would make a good alternative answer to mine.
          $endgroup$
          – Ian
          Apr 12 at 15:46












          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = beginbmatrix y_1 \ y_2 \ 0endbmatrix $$
          , so if $$ Q^-1b = beginbmatrix z_1 \ z_2 \ z_3endbmatrix$$
          then $|| Rx - Q^-1b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = beginbmatrix R_1 \ 0endbmatrix$ and intoduce $P=beginbmatrix1 & 0 & 0 \ 0 & 1& 0endbmatrix$, then you have
          $$ R_1x = PQ^-1b$$
          $$ x = (R_1)^-1PQ^-1b$$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            Apr 12 at 20:34










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
            $endgroup$
            – Adam Latosiński
            Apr 13 at 11:14
















          2












          $begingroup$

          Note that $Rx$ has the form
          $$Rx = beginbmatrix y_1 \ y_2 \ 0endbmatrix $$
          , so if $$ Q^-1b = beginbmatrix z_1 \ z_2 \ z_3endbmatrix$$
          then $|| Rx - Q^-1b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = beginbmatrix R_1 \ 0endbmatrix$ and intoduce $P=beginbmatrix1 & 0 & 0 \ 0 & 1& 0endbmatrix$, then you have
          $$ R_1x = PQ^-1b$$
          $$ x = (R_1)^-1PQ^-1b$$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            Apr 12 at 20:34










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
            $endgroup$
            – Adam Latosiński
            Apr 13 at 11:14














          2












          2








          2





          $begingroup$

          Note that $Rx$ has the form
          $$Rx = beginbmatrix y_1 \ y_2 \ 0endbmatrix $$
          , so if $$ Q^-1b = beginbmatrix z_1 \ z_2 \ z_3endbmatrix$$
          then $|| Rx - Q^-1b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = beginbmatrix R_1 \ 0endbmatrix$ and intoduce $P=beginbmatrix1 & 0 & 0 \ 0 & 1& 0endbmatrix$, then you have
          $$ R_1x = PQ^-1b$$
          $$ x = (R_1)^-1PQ^-1b$$






          share|cite|improve this answer









          $endgroup$



          Note that $Rx$ has the form
          $$Rx = beginbmatrix y_1 \ y_2 \ 0endbmatrix $$
          , so if $$ Q^-1b = beginbmatrix z_1 \ z_2 \ z_3endbmatrix$$
          then $|| Rx - Q^-1b||$ will be minimal for $y_1 = z_1$, $y_2=z_2$. This set of equation is no longer overdetermined.



          Using matrix notation, if tou write $R = beginbmatrix R_1 \ 0endbmatrix$ and intoduce $P=beginbmatrix1 & 0 & 0 \ 0 & 1& 0endbmatrix$, then you have
          $$ R_1x = PQ^-1b$$
          $$ x = (R_1)^-1PQ^-1b$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 12 at 15:56









          Adam LatosińskiAdam Latosiński

          7408




          7408







          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            Apr 12 at 20:34










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
            $endgroup$
            – Adam Latosiński
            Apr 13 at 11:14













          • 1




            $begingroup$
            The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
            $endgroup$
            – Ian
            Apr 12 at 20:34










          • $begingroup$
            @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
            $endgroup$
            – Adam Latosiński
            Apr 13 at 11:14








          1




          1




          $begingroup$
          The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
          $endgroup$
          – Ian
          Apr 12 at 20:34




          $begingroup$
          The key trick in this answer is that by the orthogonality, $| Ax - b | = | Rx - Q^T b |$.
          $endgroup$
          – Ian
          Apr 12 at 20:34












          $begingroup$
          @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
          $endgroup$
          – Adam Latosiński
          Apr 13 at 11:14





          $begingroup$
          @Ian, That's something that OP has alredy obtained on his own (since $Q$ is orthogonal, $Q^-1=Q^T$).
          $endgroup$
          – Adam Latosiński
          Apr 13 at 11:14


















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185239%2fsolving-overdetermined-system-by-qr-decomposition%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Tamil (spriik) Luke uk diar | Nawigatjuun

          Align equal signs while including text over equalitiesAMS align: left aligned text/math plus multicolumn alignmentMultiple alignmentsAligning equations in multiple placesNumbering and aligning an equation with multiple columnsHow to align one equation with another multline equationUsing \ in environments inside the begintabularxNumber equations and preserving alignment of equal signsHow can I align equations to the left and to the right?Double equation alignment problem within align enviromentAligned within align: Why are they right-aligned?

          Training a classifier when some of the features are unknownWhy does Gradient Boosting regression predict negative values when there are no negative y-values in my training set?How to improve an existing (trained) classifier?What is effect when I set up some self defined predisctor variables?Why Matlab neural network classification returns decimal values on prediction dataset?Fitting and transforming text data in training, testing, and validation setsHow to quantify the performance of the classifier (multi-class SVM) using the test data?How do I control for some patients providing multiple samples in my training data?Training and Test setTraining a convolutional neural network for image denoising in MatlabShouldn't an autoencoder with #(neurons in hidden layer) = #(neurons in input layer) be “perfect”?