Trouble to find the solution to a linear system where a matrix is not invertible












1












$begingroup$


Hello community I am new here and I have a question which might be pretty basic.



So I am trying to solve an equation. I have 3 matrices



A = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0end{pmatrix}



B = begin{pmatrix}0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



and
K = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0end{pmatrix}



H = begin{pmatrix}A\Bend{pmatrix}



and Z = begin{pmatrix}z1\z2\z3end{pmatrix}



And I have linear system



$begin{pmatrix}H\Kend{pmatrix} v = begin{pmatrix}0\Zend{pmatrix}$



It says that solution for determined system is



$ v = begin{bmatrix}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}H\Kend{pmatrix} end{bmatrix} ^{-1}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}0\Zend{pmatrix} $



Okay when I put this system of matrices in python it shows that matrix inside can't be inverse. I am presumping that



H = begin{pmatrix}A\Bend{pmatrix}
just means that



H is



begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



Two matrices put together. Same I presume for begin{pmatrix}H\Kend{pmatrix}



Where am I wrong?



Ps: sorry im new here hope I wrote everything correctly and understandable. Happy holidays everyone.






matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have taken the liberty to make your title less interchangeable with another one
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:28
















1












$begingroup$


Hello community I am new here and I have a question which might be pretty basic.



So I am trying to solve an equation. I have 3 matrices



A = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0end{pmatrix}



B = begin{pmatrix}0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



and
K = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0end{pmatrix}



H = begin{pmatrix}A\Bend{pmatrix}



and Z = begin{pmatrix}z1\z2\z3end{pmatrix}



And I have linear system



$begin{pmatrix}H\Kend{pmatrix} v = begin{pmatrix}0\Zend{pmatrix}$



It says that solution for determined system is



$ v = begin{bmatrix}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}H\Kend{pmatrix} end{bmatrix} ^{-1}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}0\Zend{pmatrix} $



Okay when I put this system of matrices in python it shows that matrix inside can't be inverse. I am presumping that



H = begin{pmatrix}A\Bend{pmatrix}
just means that



H is



begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



Two matrices put together. Same I presume for begin{pmatrix}H\Kend{pmatrix}



Where am I wrong?



Ps: sorry im new here hope I wrote everything correctly and understandable. Happy holidays everyone.






matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have taken the liberty to make your title less interchangeable with another one
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:28














1












1








1





$begingroup$


Hello community I am new here and I have a question which might be pretty basic.



So I am trying to solve an equation. I have 3 matrices



A = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0end{pmatrix}



B = begin{pmatrix}0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



and
K = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0end{pmatrix}



H = begin{pmatrix}A\Bend{pmatrix}



and Z = begin{pmatrix}z1\z2\z3end{pmatrix}



And I have linear system



$begin{pmatrix}H\Kend{pmatrix} v = begin{pmatrix}0\Zend{pmatrix}$



It says that solution for determined system is



$ v = begin{bmatrix}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}H\Kend{pmatrix} end{bmatrix} ^{-1}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}0\Zend{pmatrix} $



Okay when I put this system of matrices in python it shows that matrix inside can't be inverse. I am presumping that



H = begin{pmatrix}A\Bend{pmatrix}
just means that



H is



begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



Two matrices put together. Same I presume for begin{pmatrix}H\Kend{pmatrix}



Where am I wrong?



Ps: sorry im new here hope I wrote everything correctly and understandable. Happy holidays everyone.






matrices







share|cite|improve this question











$endgroup$




Hello community I am new here and I have a question which might be pretty basic.



So I am trying to solve an equation. I have 3 matrices



A = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0end{pmatrix}



B = begin{pmatrix}0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



and
K = begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0end{pmatrix}



H = begin{pmatrix}A\Bend{pmatrix}



and Z = begin{pmatrix}z1\z2\z3end{pmatrix}



And I have linear system



$begin{pmatrix}H\Kend{pmatrix} v = begin{pmatrix}0\Zend{pmatrix}$



It says that solution for determined system is



$ v = begin{bmatrix}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}H\Kend{pmatrix} end{bmatrix} ^{-1}begin{pmatrix}H\Kend{pmatrix}^T begin{pmatrix}0\Zend{pmatrix} $



Okay when I put this system of matrices in python it shows that matrix inside can't be inverse. I am presumping that



H = begin{pmatrix}A\Bend{pmatrix}
just means that



H is



begin{pmatrix}1&0&0&0&0&0&0&0\0&1&1&0&0&0&0&0\0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&1end{pmatrix}



Two matrices put together. Same I presume for begin{pmatrix}H\Kend{pmatrix}



Where am I wrong?



Ps: sorry im new here hope I wrote everything correctly and understandable. Happy holidays everyone.






matrices




matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:26









Jean Marie

30.3k42051




30.3k42051










asked Dec 13 '18 at 8:36









Noob ProgrammerNoob Programmer

103




103












  • $begingroup$
    I have taken the liberty to make your title less interchangeable with another one
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:28


















  • $begingroup$
    I have taken the liberty to make your title less interchangeable with another one
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:28
















$begingroup$
I have taken the liberty to make your title less interchangeable with another one
$endgroup$
– Jean Marie
Dec 13 '18 at 13:28




$begingroup$
I have taken the liberty to make your title less interchangeable with another one
$endgroup$
– Jean Marie
Dec 13 '18 at 13:28










1 Answer
1






active

oldest

votes


















0












$begingroup$

For the sake of simpler notations, let us rename $$L:=begin{pmatrix}H\Kend{pmatrix} text{and} P:=begin{pmatrix}0\Zend{pmatrix} $$



giving :



$$Lv=P tag{1}$$



which is a $7 times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).



Instead of your $$ v = (L^T L)^{-1}L^T P $$



which is a formula valid for overdetermined systems, one must take for an underdetermined system :



$$ v = L^T(L L^T)^{-1} P tag{2}$$



(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").



This will give you a so-called "mean-squares" solution.



Proof of (1) : Plug (2) into (1), you will get :



$$underbrace{L L^T(L L^T)^{-1}}_{text{identity matrix} I} P=P...$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:22












  • $begingroup$
    Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 10:33












  • $begingroup$
    Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:47










  • $begingroup$
    I have added a reference.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:23











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

For the sake of simpler notations, let us rename $$L:=begin{pmatrix}H\Kend{pmatrix} text{and} P:=begin{pmatrix}0\Zend{pmatrix} $$



giving :



$$Lv=P tag{1}$$



which is a $7 times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).



Instead of your $$ v = (L^T L)^{-1}L^T P $$



which is a formula valid for overdetermined systems, one must take for an underdetermined system :



$$ v = L^T(L L^T)^{-1} P tag{2}$$



(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").



This will give you a so-called "mean-squares" solution.



Proof of (1) : Plug (2) into (1), you will get :



$$underbrace{L L^T(L L^T)^{-1}}_{text{identity matrix} I} P=P...$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:22












  • $begingroup$
    Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 10:33












  • $begingroup$
    Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:47










  • $begingroup$
    I have added a reference.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:23
















0












$begingroup$

For the sake of simpler notations, let us rename $$L:=begin{pmatrix}H\Kend{pmatrix} text{and} P:=begin{pmatrix}0\Zend{pmatrix} $$



giving :



$$Lv=P tag{1}$$



which is a $7 times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).



Instead of your $$ v = (L^T L)^{-1}L^T P $$



which is a formula valid for overdetermined systems, one must take for an underdetermined system :



$$ v = L^T(L L^T)^{-1} P tag{2}$$



(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").



This will give you a so-called "mean-squares" solution.



Proof of (1) : Plug (2) into (1), you will get :



$$underbrace{L L^T(L L^T)^{-1}}_{text{identity matrix} I} P=P...$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:22












  • $begingroup$
    Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 10:33












  • $begingroup$
    Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:47










  • $begingroup$
    I have added a reference.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:23














0












0








0





$begingroup$

For the sake of simpler notations, let us rename $$L:=begin{pmatrix}H\Kend{pmatrix} text{and} P:=begin{pmatrix}0\Zend{pmatrix} $$



giving :



$$Lv=P tag{1}$$



which is a $7 times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).



Instead of your $$ v = (L^T L)^{-1}L^T P $$



which is a formula valid for overdetermined systems, one must take for an underdetermined system :



$$ v = L^T(L L^T)^{-1} P tag{2}$$



(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").



This will give you a so-called "mean-squares" solution.



Proof of (1) : Plug (2) into (1), you will get :



$$underbrace{L L^T(L L^T)^{-1}}_{text{identity matrix} I} P=P...$$






share|cite|improve this answer











$endgroup$



For the sake of simpler notations, let us rename $$L:=begin{pmatrix}H\Kend{pmatrix} text{and} P:=begin{pmatrix}0\Zend{pmatrix} $$



giving :



$$Lv=P tag{1}$$



which is a $7 times 8$ system (thus an underdetermined system : less constraints than the number of unknowns).



Instead of your $$ v = (L^T L)^{-1}L^T P $$



which is a formula valid for overdetermined systems, one must take for an underdetermined system :



$$ v = L^T(L L^T)^{-1} P tag{2}$$



(see https://see.stanford.edu/materials/lsoeldsee263/08-min-norm.pdf for the difference of formulas between under- and over-determined systems ; in this document, a matrix like $L$ with more columns than lines is called "fat").



This will give you a so-called "mean-squares" solution.



Proof of (1) : Plug (2) into (1), you will get :



$$underbrace{L L^T(L L^T)^{-1}}_{text{identity matrix} I} P=P...$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 13:22

























answered Dec 13 '18 at 8:59









Jean MarieJean Marie

30.3k42051




30.3k42051












  • $begingroup$
    Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:22












  • $begingroup$
    Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 10:33












  • $begingroup$
    Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:47










  • $begingroup$
    I have added a reference.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:23


















  • $begingroup$
    Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:22












  • $begingroup$
    Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 10:33












  • $begingroup$
    Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
    $endgroup$
    – Noob Programmer
    Dec 13 '18 at 10:47










  • $begingroup$
    I have added a reference.
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 13:23
















$begingroup$
Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
$endgroup$
– Noob Programmer
Dec 13 '18 at 10:22






$begingroup$
Thank you so much for the time and effort. I tried your way and indeed i got a solution, but i think its the wrong one. since i still get that v4 = 0 whiere it should be v4 = v5 = 1/3(vz1+vz3 -vz2). This is just an example of the error, there are more of them. And I have another question. How do I know if its pseudo-inverse, where it doesn't say in the exercise?
$endgroup$
– Noob Programmer
Dec 13 '18 at 10:22














$begingroup$
Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
$endgroup$
– Jean Marie
Dec 13 '18 at 10:33






$begingroup$
Did you notice that I have completely re-written my answer. I had made a mistake (I had sticked to your formula which is not the right one). In particular, I do not use any more the pseudo inverse. Besides, I am going to add a proof for the formula I gave.
$endgroup$
– Jean Marie
Dec 13 '18 at 10:33














$begingroup$
Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
$endgroup$
– Noob Programmer
Dec 13 '18 at 10:47




$begingroup$
Sorry didn't notice, didn't refresh my page. Thank you for your effort. It solved my problem :)
$endgroup$
– Noob Programmer
Dec 13 '18 at 10:47












$begingroup$
I have added a reference.
$endgroup$
– Jean Marie
Dec 13 '18 at 13:23




$begingroup$
I have added a reference.
$endgroup$
– Jean Marie
Dec 13 '18 at 13:23


















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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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