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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edited yesterday

Tianlalu

2,559632

asked yesterday

user463102

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
  

  
  
  
  

add a comment | 

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

edited yesterday

Tianlalu

2,559632

asked yesterday

user463102

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
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

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

edited yesterday

Tianlalu

2,559632

asked yesterday

user463102

14013

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

edited yesterday

Tianlalu

2,559632

asked yesterday

user463102

14013

share|cite|improve this question

edited yesterday

Tianlalu

2,559632

asked yesterday

user463102

14013

share|cite|improve this question

share|cite|improve this question

edited yesterday

Tianlalu

2,559632

edited yesterday

Tianlalu

2,559632

edited yesterday

Tianlalu

2,559632

2,559632

asked yesterday

user463102

14013

asked yesterday

user463102

14013

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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

add a comment | 

1 Answer
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accepted

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

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

answered yesterday

Rima Khouja

965

add a comment | 

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
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accepted

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

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

answered yesterday

Rima Khouja

965

add a comment | 

up vote
1
down vote



accepted

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

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

answered yesterday

Rima Khouja

965

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

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

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

answered yesterday

Rima Khouja

965

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

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

answered yesterday

Rima Khouja

965

share|cite|improve this answer

share|cite|improve this answer

edited 22 hours ago

Moo

5,2683920

edited 22 hours ago

Moo

5,2683920

edited 22 hours ago

Moo

5,2683920

5,2683920

answered yesterday

Rima Khouja

965

answered yesterday

Rima Khouja

965

answered yesterday

Rima Khouja

965

965

add a comment | 

add a comment | 

 

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