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 at 3:22










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










  • The other was zero. My book says to not include them.
    – Brady Dean
    Nov 27 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 at 3:24

























asked Nov 27 at 3:18









Brady Dean

1353




1353












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










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


















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










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
















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




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












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




The other was zero. My book says to not include them.
– Brady Dean
Nov 27 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 at 3:25










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










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






  • 1




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






  • 1




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











Your Answer





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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 at 3:25










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










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






  • 1




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






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 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 at 3:25










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










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






  • 1




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






  • 1




    Gram-Schmidt, of which part of the procedure involve normalization.
    – Siong Thye Goh
    Nov 27 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 at 3:34

























answered Nov 27 at 3:21









Siong Thye Goh

99k1464117




99k1464117












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










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










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






  • 1




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






  • 1




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


















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










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










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






  • 1




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






  • 1




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
















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




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












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




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












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




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




1




1




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




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




1




1




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




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


















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