singular value decomposition of simple $2times2$ matrix.











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Consider the matrix $A=begin{bmatrix}1& varepsilon\ 0 & 0 end{bmatrix}$ where $varepsilon >0$.



Determine a singular value decomposition of $A$.



As far as I know the SVD can be calculated as follows:



$A=USigma V^T$
in which $U$ are the eigenvectors of $AA^T$ and $V$ are the eigenvectors of $A^{T}A$.



The singular values are $0$ and $sqrt{varepsilon^2+1}quad$
So $Sigma=begin{bmatrix}sqrt{varepsilon^2+1 }&0\0& 0 end{bmatrix}$



$AA^T=begin{bmatrix}1+varepsilon^2 & 0\0&0end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



$A^TA=begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2 end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



In order to calculate the matix $U$, we take $(AA^T-lambda I) underline{u}_i = 0$.



For $lambda = 1+varepsilon^2$ this gives: $begin{bmatrix}0&0\0&-1-varepsilon^2 end{bmatrix} underline{u}_1 = 0 quad$ so $underline{u}_1=begin{bmatrix}1\0 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1+varepsilon^2&0\0&0end{bmatrix} underline{u}_2=0 quad$so $underline{u}_2=begin{bmatrix}0\1 end{bmatrix}$



So $U = begin{bmatrix}1&0\0&1end{bmatrix}$



Next we calculate $V$: using $(A^TA-lambda I)underline{v}_i=0.$



For $lambda=1+varepsilon^2$ this gives: $begin{bmatrix}-varepsilon^2&varepsilon\varepsilon&-1end{bmatrix}underline{v}_i = 0 quad$ so $underline{v}_1=begin{bmatrix}frac{1}{varepsilon}\1 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2end{bmatrix} underline{v}_2=0 quad$so $underline{v}_2=begin{bmatrix}1\ -frac{1}{varepsilon} end{bmatrix}$



So $V=begin{bmatrix}1&frac{1}{varepsilon}\-frac{1}{varepsilon}&1 end{bmatrix}$.



But if I now use $A=USigma V^T$ I do not get the correct $A$.



I don't know what I'm doing wrong anymore.






linear-algebra matrices matrix-calculus







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  • 4




    Your $V$ is not orthogonal.
    – Yves Daoust
    yesterday






  • 1




    Your columns of $V$ are in the wrong order I think
    – Richard Martin
    yesterday






  • 1




    And then you have to rescale $V$ so it becomes orthogonal
    – Richard Martin
    yesterday















up vote
1
down vote

favorite
1












Consider the matrix $A=begin{bmatrix}1& varepsilon\ 0 & 0 end{bmatrix}$ where $varepsilon >0$.



Determine a singular value decomposition of $A$.



As far as I know the SVD can be calculated as follows:



$A=USigma V^T$
in which $U$ are the eigenvectors of $AA^T$ and $V$ are the eigenvectors of $A^{T}A$.



The singular values are $0$ and $sqrt{varepsilon^2+1}quad$
So $Sigma=begin{bmatrix}sqrt{varepsilon^2+1 }&0\0& 0 end{bmatrix}$



$AA^T=begin{bmatrix}1+varepsilon^2 & 0\0&0end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



$A^TA=begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2 end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



In order to calculate the matix $U$, we take $(AA^T-lambda I) underline{u}_i = 0$.



For $lambda = 1+varepsilon^2$ this gives: $begin{bmatrix}0&0\0&-1-varepsilon^2 end{bmatrix} underline{u}_1 = 0 quad$ so $underline{u}_1=begin{bmatrix}1\0 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1+varepsilon^2&0\0&0end{bmatrix} underline{u}_2=0 quad$so $underline{u}_2=begin{bmatrix}0\1 end{bmatrix}$



So $U = begin{bmatrix}1&0\0&1end{bmatrix}$



Next we calculate $V$: using $(A^TA-lambda I)underline{v}_i=0.$



For $lambda=1+varepsilon^2$ this gives: $begin{bmatrix}-varepsilon^2&varepsilon\varepsilon&-1end{bmatrix}underline{v}_i = 0 quad$ so $underline{v}_1=begin{bmatrix}frac{1}{varepsilon}\1 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2end{bmatrix} underline{v}_2=0 quad$so $underline{v}_2=begin{bmatrix}1\ -frac{1}{varepsilon} end{bmatrix}$



So $V=begin{bmatrix}1&frac{1}{varepsilon}\-frac{1}{varepsilon}&1 end{bmatrix}$.



But if I now use $A=USigma V^T$ I do not get the correct $A$.



I don't know what I'm doing wrong anymore.






linear-algebra matrices matrix-calculus







share|cite|improve this question




















  • 4




    Your $V$ is not orthogonal.
    – Yves Daoust
    yesterday






  • 1




    Your columns of $V$ are in the wrong order I think
    – Richard Martin
    yesterday






  • 1




    And then you have to rescale $V$ so it becomes orthogonal
    – Richard Martin
    yesterday













up vote
1
down vote

favorite
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up vote
1
down vote

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1





Consider the matrix $A=begin{bmatrix}1& varepsilon\ 0 & 0 end{bmatrix}$ where $varepsilon >0$.



Determine a singular value decomposition of $A$.



As far as I know the SVD can be calculated as follows:



$A=USigma V^T$
in which $U$ are the eigenvectors of $AA^T$ and $V$ are the eigenvectors of $A^{T}A$.



The singular values are $0$ and $sqrt{varepsilon^2+1}quad$
So $Sigma=begin{bmatrix}sqrt{varepsilon^2+1 }&0\0& 0 end{bmatrix}$



$AA^T=begin{bmatrix}1+varepsilon^2 & 0\0&0end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



$A^TA=begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2 end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



In order to calculate the matix $U$, we take $(AA^T-lambda I) underline{u}_i = 0$.



For $lambda = 1+varepsilon^2$ this gives: $begin{bmatrix}0&0\0&-1-varepsilon^2 end{bmatrix} underline{u}_1 = 0 quad$ so $underline{u}_1=begin{bmatrix}1\0 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1+varepsilon^2&0\0&0end{bmatrix} underline{u}_2=0 quad$so $underline{u}_2=begin{bmatrix}0\1 end{bmatrix}$



So $U = begin{bmatrix}1&0\0&1end{bmatrix}$



Next we calculate $V$: using $(A^TA-lambda I)underline{v}_i=0.$



For $lambda=1+varepsilon^2$ this gives: $begin{bmatrix}-varepsilon^2&varepsilon\varepsilon&-1end{bmatrix}underline{v}_i = 0 quad$ so $underline{v}_1=begin{bmatrix}frac{1}{varepsilon}\1 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2end{bmatrix} underline{v}_2=0 quad$so $underline{v}_2=begin{bmatrix}1\ -frac{1}{varepsilon} end{bmatrix}$



So $V=begin{bmatrix}1&frac{1}{varepsilon}\-frac{1}{varepsilon}&1 end{bmatrix}$.



But if I now use $A=USigma V^T$ I do not get the correct $A$.



I don't know what I'm doing wrong anymore.






linear-algebra matrices matrix-calculus







share|cite|improve this question















Consider the matrix $A=begin{bmatrix}1& varepsilon\ 0 & 0 end{bmatrix}$ where $varepsilon >0$.



Determine a singular value decomposition of $A$.



As far as I know the SVD can be calculated as follows:



$A=USigma V^T$
in which $U$ are the eigenvectors of $AA^T$ and $V$ are the eigenvectors of $A^{T}A$.



The singular values are $0$ and $sqrt{varepsilon^2+1}quad$
So $Sigma=begin{bmatrix}sqrt{varepsilon^2+1 }&0\0& 0 end{bmatrix}$



$AA^T=begin{bmatrix}1+varepsilon^2 & 0\0&0end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



$A^TA=begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2 end{bmatrix}$ with eigenvalues $0$ and $1+varepsilon^2$.



In order to calculate the matix $U$, we take $(AA^T-lambda I) underline{u}_i = 0$.



For $lambda = 1+varepsilon^2$ this gives: $begin{bmatrix}0&0\0&-1-varepsilon^2 end{bmatrix} underline{u}_1 = 0 quad$ so $underline{u}_1=begin{bmatrix}1\0 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1+varepsilon^2&0\0&0end{bmatrix} underline{u}_2=0 quad$so $underline{u}_2=begin{bmatrix}0\1 end{bmatrix}$



So $U = begin{bmatrix}1&0\0&1end{bmatrix}$



Next we calculate $V$: using $(A^TA-lambda I)underline{v}_i=0.$



For $lambda=1+varepsilon^2$ this gives: $begin{bmatrix}-varepsilon^2&varepsilon\varepsilon&-1end{bmatrix}underline{v}_i = 0 quad$ so $underline{v}_1=begin{bmatrix}frac{1}{varepsilon}\1 end{bmatrix}$



For $lambda=0$ this gives:
$begin{bmatrix}1&varepsilon\varepsilon&varepsilon^2end{bmatrix} underline{v}_2=0 quad$so $underline{v}_2=begin{bmatrix}1\ -frac{1}{varepsilon} end{bmatrix}$



So $V=begin{bmatrix}1&frac{1}{varepsilon}\-frac{1}{varepsilon}&1 end{bmatrix}$.



But if I now use $A=USigma V^T$ I do not get the correct $A$.



I don't know what I'm doing wrong anymore.






linear-algebra matrices matrix-calculus




linear-algebra matrices matrix-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Tianlalu

2,559632




2,559632










asked yesterday









user463102

14013




14013








  • 4




    Your $V$ is not orthogonal.
    – Yves Daoust
    yesterday






  • 1




    Your columns of $V$ are in the wrong order I think
    – Richard Martin
    yesterday






  • 1




    And then you have to rescale $V$ so it becomes orthogonal
    – Richard Martin
    yesterday














  • 4




    Your $V$ is not orthogonal.
    – Yves Daoust
    yesterday






  • 1




    Your columns of $V$ are in the wrong order I think
    – Richard Martin
    yesterday






  • 1




    And then you have to rescale $V$ so it becomes orthogonal
    – Richard Martin
    yesterday








4




4




Your $V$ is not orthogonal.
– Yves Daoust
yesterday




Your $V$ is not orthogonal.
– Yves Daoust
yesterday




1




1




Your columns of $V$ are in the wrong order I think
– Richard Martin
yesterday




Your columns of $V$ are in the wrong order I think
– Richard Martin
yesterday




1




1




And then you have to rescale $V$ so it becomes orthogonal
– Richard Martin
yesterday




And then you have to rescale $V$ so it becomes orthogonal
– Richard Martin
yesterday










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$V$ contains two vectors the first is $v_1=[-epsilon~~ 1]$ the second is $v_2=[1 ~~epsilon]$






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      $V$ contains two vectors the first is $v_1=[-epsilon~~ 1]$ the second is $v_2=[1 ~~epsilon]$






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        $V$ contains two vectors the first is $v_1=[-epsilon~~ 1]$ the second is $v_2=[1 ~~epsilon]$






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        $V$ contains two vectors the first is $v_1=[-epsilon~~ 1]$ the second is $v_2=[1 ~~epsilon]$







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        edited 22 hours ago









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