Calculate singular value decomposition












0














I need to find the reduced singular value decomposition of this matrix.



$$begin{pmatrix}
-2 & -1 & 2 \
2 & 1 & -2 \
end{pmatrix}
$$



I formed
$$
A^TA=
begin{pmatrix}
8 & 4 & -8 \
4 & 2 & -4 \
-8 & -4 & 8 \
end{pmatrix}
$$



Found the eigenvalue 18 and eigenvector
$$begin{pmatrix} -1 \ -1/2 \ 1 end{pmatrix}$$



$$
u_1=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}
$$



So
$$
A=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}begin{pmatrix}3sqrt2end{pmatrix}begin{pmatrix} -1 & -1/2 & 1 end{pmatrix}
$$



But this is clearly not correct






linear-algebra matrix-decomposition







share|cite|improve this question
























  • Why stop at one eigenvalue/eigenvector?
    – Michael Burr
    Nov 27 '18 at 3:22










  • The other was zero. My book says to not include them.
    – Brady Dean
    Nov 27 '18 at 3:22
















0














I need to find the reduced singular value decomposition of this matrix.



$$begin{pmatrix}
-2 & -1 & 2 \
2 & 1 & -2 \
end{pmatrix}
$$



I formed
$$
A^TA=
begin{pmatrix}
8 & 4 & -8 \
4 & 2 & -4 \
-8 & -4 & 8 \
end{pmatrix}
$$



Found the eigenvalue 18 and eigenvector
$$begin{pmatrix} -1 \ -1/2 \ 1 end{pmatrix}$$



$$
u_1=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}
$$



So
$$
A=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}begin{pmatrix}3sqrt2end{pmatrix}begin{pmatrix} -1 & -1/2 & 1 end{pmatrix}
$$



But this is clearly not correct






linear-algebra matrix-decomposition







share|cite|improve this question
























  • Why stop at one eigenvalue/eigenvector?
    – Michael Burr
    Nov 27 '18 at 3:22










  • The other was zero. My book says to not include them.
    – Brady Dean
    Nov 27 '18 at 3:22














0












0








0







I need to find the reduced singular value decomposition of this matrix.



$$begin{pmatrix}
-2 & -1 & 2 \
2 & 1 & -2 \
end{pmatrix}
$$



I formed
$$
A^TA=
begin{pmatrix}
8 & 4 & -8 \
4 & 2 & -4 \
-8 & -4 & 8 \
end{pmatrix}
$$



Found the eigenvalue 18 and eigenvector
$$begin{pmatrix} -1 \ -1/2 \ 1 end{pmatrix}$$



$$
u_1=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}
$$



So
$$
A=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}begin{pmatrix}3sqrt2end{pmatrix}begin{pmatrix} -1 & -1/2 & 1 end{pmatrix}
$$



But this is clearly not correct






linear-algebra matrix-decomposition







share|cite|improve this question















I need to find the reduced singular value decomposition of this matrix.



$$begin{pmatrix}
-2 & -1 & 2 \
2 & 1 & -2 \
end{pmatrix}
$$



I formed
$$
A^TA=
begin{pmatrix}
8 & 4 & -8 \
4 & 2 & -4 \
-8 & -4 & 8 \
end{pmatrix}
$$



Found the eigenvalue 18 and eigenvector
$$begin{pmatrix} -1 \ -1/2 \ 1 end{pmatrix}$$



$$
u_1=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}
$$



So
$$
A=begin{pmatrix} frac{3sqrt2}{4} \ frac{-3sqrt2}{4} end{pmatrix}begin{pmatrix}3sqrt2end{pmatrix}begin{pmatrix} -1 & -1/2 & 1 end{pmatrix}
$$



But this is clearly not correct






linear-algebra matrix-decomposition




linear-algebra matrix-decomposition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 3:24

























asked Nov 27 '18 at 3:18









Brady Dean

1353




1353












  • Why stop at one eigenvalue/eigenvector?
    – Michael Burr
    Nov 27 '18 at 3:22










  • The other was zero. My book says to not include them.
    – Brady Dean
    Nov 27 '18 at 3:22


















  • Why stop at one eigenvalue/eigenvector?
    – Michael Burr
    Nov 27 '18 at 3:22










  • The other was zero. My book says to not include them.
    – Brady Dean
    Nov 27 '18 at 3:22
















Why stop at one eigenvalue/eigenvector?
– Michael Burr
Nov 27 '18 at 3:22




Why stop at one eigenvalue/eigenvector?
– Michael Burr
Nov 27 '18 at 3:22












The other was zero. My book says to not include them.
– Brady Dean
Nov 27 '18 at 3:22




The other was zero. My book says to not include them.
– Brady Dean
Nov 27 '18 at 3:22










1 Answer
1






active

oldest

votes


















1














Guide:



This is a rank $2$ matrix, your computation of $A^TA$ is wrong.



octave:1> A = [-2 -1 1; 2 1 -2]

A =



  -2  -1   1

   2   1  -2



octave:2> A'*A

ans =



   8   4  -6

   4   2  -3

  -6  -3   5


Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.



Edit:



begin{align}
begin{bmatrix} -2 & -1 & 2 \2 & 1 & -2end{bmatrix} &= begin{bmatrix} 1 \ -1end{bmatrix}begin{bmatrix} -2 & -1 & 2 end{bmatrix} \
&=begin{bmatrix} frac{1}{sqrt2} \ -frac1{sqrt{2}}end{bmatrix}3sqrt{2}begin{bmatrix} -frac23 & -frac13 & frac23 end{bmatrix}
end{align}



Singular vectors are of unit length.






share|cite|improve this answer























  • Sorry just realized I typed the wrong number in A. I updated the question
    – Brady Dean
    Nov 27 '18 at 3:25










  • updated the answer.
    – Siong Thye Goh
    Nov 27 '18 at 3:35










  • How did you get [-2/3 -1/3 2/3]?
    – Brady Dean
    Nov 27 '18 at 3:41






  • 1




    Just normalize the vector.
    – Siong Thye Goh
    Nov 27 '18 at 3:42






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 '18 at 3:47











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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%2f3015283%2fcalculate-singular-value-decomposition%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Guide:



This is a rank $2$ matrix, your computation of $A^TA$ is wrong.



octave:1> A = [-2 -1 1; 2 1 -2]

A =



  -2  -1   1

   2   1  -2



octave:2> A'*A

ans =



   8   4  -6

   4   2  -3

  -6  -3   5


Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.



Edit:



begin{align}
begin{bmatrix} -2 & -1 & 2 \2 & 1 & -2end{bmatrix} &= begin{bmatrix} 1 \ -1end{bmatrix}begin{bmatrix} -2 & -1 & 2 end{bmatrix} \
&=begin{bmatrix} frac{1}{sqrt2} \ -frac1{sqrt{2}}end{bmatrix}3sqrt{2}begin{bmatrix} -frac23 & -frac13 & frac23 end{bmatrix}
end{align}



Singular vectors are of unit length.






share|cite|improve this answer























  • Sorry just realized I typed the wrong number in A. I updated the question
    – Brady Dean
    Nov 27 '18 at 3:25










  • updated the answer.
    – Siong Thye Goh
    Nov 27 '18 at 3:35










  • How did you get [-2/3 -1/3 2/3]?
    – Brady Dean
    Nov 27 '18 at 3:41






  • 1




    Just normalize the vector.
    – Siong Thye Goh
    Nov 27 '18 at 3:42






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 '18 at 3:47
















1














Guide:



This is a rank $2$ matrix, your computation of $A^TA$ is wrong.



octave:1> A = [-2 -1 1; 2 1 -2]

A =



  -2  -1   1

   2   1  -2



octave:2> A'*A

ans =



   8   4  -6

   4   2  -3

  -6  -3   5


Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.



Edit:



begin{align}
begin{bmatrix} -2 & -1 & 2 \2 & 1 & -2end{bmatrix} &= begin{bmatrix} 1 \ -1end{bmatrix}begin{bmatrix} -2 & -1 & 2 end{bmatrix} \
&=begin{bmatrix} frac{1}{sqrt2} \ -frac1{sqrt{2}}end{bmatrix}3sqrt{2}begin{bmatrix} -frac23 & -frac13 & frac23 end{bmatrix}
end{align}



Singular vectors are of unit length.






share|cite|improve this answer























  • Sorry just realized I typed the wrong number in A. I updated the question
    – Brady Dean
    Nov 27 '18 at 3:25










  • updated the answer.
    – Siong Thye Goh
    Nov 27 '18 at 3:35










  • How did you get [-2/3 -1/3 2/3]?
    – Brady Dean
    Nov 27 '18 at 3:41






  • 1




    Just normalize the vector.
    – Siong Thye Goh
    Nov 27 '18 at 3:42






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 '18 at 3:47














1












1








1






Guide:



This is a rank $2$ matrix, your computation of $A^TA$ is wrong.



octave:1> A = [-2 -1 1; 2 1 -2]

A =



  -2  -1   1

   2   1  -2



octave:2> A'*A

ans =



   8   4  -6

   4   2  -3

  -6  -3   5


Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.



Edit:



begin{align}
begin{bmatrix} -2 & -1 & 2 \2 & 1 & -2end{bmatrix} &= begin{bmatrix} 1 \ -1end{bmatrix}begin{bmatrix} -2 & -1 & 2 end{bmatrix} \
&=begin{bmatrix} frac{1}{sqrt2} \ -frac1{sqrt{2}}end{bmatrix}3sqrt{2}begin{bmatrix} -frac23 & -frac13 & frac23 end{bmatrix}
end{align}



Singular vectors are of unit length.






share|cite|improve this answer














Guide:



This is a rank $2$ matrix, your computation of $A^TA$ is wrong.



octave:1> A = [-2 -1 1; 2 1 -2]

A =



  -2  -1   1

   2   1  -2



octave:2> A'*A

ans =



   8   4  -6

   4   2  -3

  -6  -3   5


Also, rather than dealing with $A^TA$, perhaps working with $AA^T$ is simpler.



Edit:



begin{align}
begin{bmatrix} -2 & -1 & 2 \2 & 1 & -2end{bmatrix} &= begin{bmatrix} 1 \ -1end{bmatrix}begin{bmatrix} -2 & -1 & 2 end{bmatrix} \
&=begin{bmatrix} frac{1}{sqrt2} \ -frac1{sqrt{2}}end{bmatrix}3sqrt{2}begin{bmatrix} -frac23 & -frac13 & frac23 end{bmatrix}
end{align}



Singular vectors are of unit length.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 3:34

























answered Nov 27 '18 at 3:21









Siong Thye Goh

99.5k1464117




99.5k1464117












  • Sorry just realized I typed the wrong number in A. I updated the question
    – Brady Dean
    Nov 27 '18 at 3:25










  • updated the answer.
    – Siong Thye Goh
    Nov 27 '18 at 3:35










  • How did you get [-2/3 -1/3 2/3]?
    – Brady Dean
    Nov 27 '18 at 3:41






  • 1




    Just normalize the vector.
    – Siong Thye Goh
    Nov 27 '18 at 3:42






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 '18 at 3:47


















  • Sorry just realized I typed the wrong number in A. I updated the question
    – Brady Dean
    Nov 27 '18 at 3:25










  • updated the answer.
    – Siong Thye Goh
    Nov 27 '18 at 3:35










  • How did you get [-2/3 -1/3 2/3]?
    – Brady Dean
    Nov 27 '18 at 3:41






  • 1




    Just normalize the vector.
    – Siong Thye Goh
    Nov 27 '18 at 3:42






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 '18 at 3:47
















Sorry just realized I typed the wrong number in A. I updated the question
– Brady Dean
Nov 27 '18 at 3:25




Sorry just realized I typed the wrong number in A. I updated the question
– Brady Dean
Nov 27 '18 at 3:25












updated the answer.
– Siong Thye Goh
Nov 27 '18 at 3:35




updated the answer.
– Siong Thye Goh
Nov 27 '18 at 3:35












How did you get [-2/3 -1/3 2/3]?
– Brady Dean
Nov 27 '18 at 3:41




How did you get [-2/3 -1/3 2/3]?
– Brady Dean
Nov 27 '18 at 3:41




1




1




Just normalize the vector.
– Siong Thye Goh
Nov 27 '18 at 3:42




Just normalize the vector.
– Siong Thye Goh
Nov 27 '18 at 3:42




1




1




Gram-Schmidt, of which part of the procedure involve normalization.
– Siong Thye Goh
Nov 27 '18 at 3:47




Gram-Schmidt, of which part of the procedure involve normalization.
– Siong Thye Goh
Nov 27 '18 at 3:47


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3015283%2fcalculate-singular-value-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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more