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What are the implications when matrix's lowest eigenvalue is equal to 0?

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3

$begingroup$

I have a task to solve where only eigenvalues are given and I need to calculate a matrix condition number. The formula for it requires division by the lowest eigenvalue (which is zero). In such case the condition number cannot be calculated.

Can equation system be solved if the lowest matrix eigenvalue is 0?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

$endgroup$

add a comment | 

3

$begingroup$

I have a task to solve where only eigenvalues are given and I need to calculate a matrix condition number. The formula for it requires division by the lowest eigenvalue (which is zero). In such case the condition number cannot be calculated.

Can equation system be solved if the lowest matrix eigenvalue is 0?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

$endgroup$

add a comment | 

$begingroup$

I have a task to solve where only eigenvalues are given and I need to calculate a matrix condition number. The formula for it requires division by the lowest eigenvalue (which is zero). In such case the condition number cannot be calculated.

Can equation system be solved if the lowest matrix eigenvalue is 0?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

$endgroup$

share|cite|improve this question

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

share|cite|improve this question

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

edited 7 hours ago

belford

1033

edited 7 hours ago

belford

1033

asked 8 hours ago

MikeMike

1236

asked 8 hours ago

MikeMike

1236

2 Answers
2

active

oldest

votes

2

$begingroup$

Recall that a matrix is invertible if and only if its determinant is non-zero, and the determinant is the product of the eigenvalues. So, having an eigenvalue of zero means that the matrix is not invertible.

Since the condition number measures how close to singular your matrix is, with larger values implying closer to singular, it makes sense that you cannot calculate it!

It is worth mentioning that for a non-singular, square matrix, the condition number induced by the $2$-norm is $kappa_2(A)=sigma_1/sigma_n$, where $sigma_1$ and $sigma_n$ denote the largest and smallest singular values of $A$, not the eigenvalues. These are not always the same as the eigenvalues of $A$ (although, if $A$ is singular, then $A^TA$ does not have full rank, and so we will have a singular value of $0$).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

cmkcmk

2,245319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoted because...?
  $endgroup$
  – cmk
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

As the previous answer suggests, the matrix is not invertible. From a different perspective we can argue that if an eigenvalue is zero, then at least two columns (or rows) are dependent. Thus, since the colums (or rows) are not independent, the matrix is singular. Therefore, it is not invertible.

share|cite|improve this answer

answered 8 hours ago

DunkelDunkel

401211

$endgroup$

add a comment | 

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2

$begingroup$

Recall that a matrix is invertible if and only if its determinant is non-zero, and the determinant is the product of the eigenvalues. So, having an eigenvalue of zero means that the matrix is not invertible.

Since the condition number measures how close to singular your matrix is, with larger values implying closer to singular, it makes sense that you cannot calculate it!

It is worth mentioning that for a non-singular, square matrix, the condition number induced by the $2$-norm is $kappa_2(A)=sigma_1/sigma_n$, where $sigma_1$ and $sigma_n$ denote the largest and smallest singular values of $A$, not the eigenvalues. These are not always the same as the eigenvalues of $A$ (although, if $A$ is singular, then $A^TA$ does not have full rank, and so we will have a singular value of $0$).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

cmkcmk

2,245319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoted because...?
  $endgroup$
  – cmk
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Recall that a matrix is invertible if and only if its determinant is non-zero, and the determinant is the product of the eigenvalues. So, having an eigenvalue of zero means that the matrix is not invertible.

Since the condition number measures how close to singular your matrix is, with larger values implying closer to singular, it makes sense that you cannot calculate it!

It is worth mentioning that for a non-singular, square matrix, the condition number induced by the $2$-norm is $kappa_2(A)=sigma_1/sigma_n$, where $sigma_1$ and $sigma_n$ denote the largest and smallest singular values of $A$, not the eigenvalues. These are not always the same as the eigenvalues of $A$ (although, if $A$ is singular, then $A^TA$ does not have full rank, and so we will have a singular value of $0$).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

cmkcmk

2,245319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Downvoted because...?
  $endgroup$
  – cmk
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Recall that a matrix is invertible if and only if its determinant is non-zero, and the determinant is the product of the eigenvalues. So, having an eigenvalue of zero means that the matrix is not invertible.

Since the condition number measures how close to singular your matrix is, with larger values implying closer to singular, it makes sense that you cannot calculate it!

It is worth mentioning that for a non-singular, square matrix, the condition number induced by the $2$-norm is $kappa_2(A)=sigma_1/sigma_n$, where $sigma_1$ and $sigma_n$ denote the largest and smallest singular values of $A$, not the eigenvalues. These are not always the same as the eigenvalues of $A$ (although, if $A$ is singular, then $A^TA$ does not have full rank, and so we will have a singular value of $0$).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

cmkcmk

2,245319

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

cmkcmk

2,245319

answered 8 hours ago

cmkcmk

2,245319

answered 8 hours ago

cmkcmk

2,245319

3

$begingroup$

As the previous answer suggests, the matrix is not invertible. From a different perspective we can argue that if an eigenvalue is zero, then at least two columns (or rows) are dependent. Thus, since the colums (or rows) are not independent, the matrix is singular. Therefore, it is not invertible.

share|cite|improve this answer

answered 8 hours ago

DunkelDunkel

401211

$endgroup$

add a comment | 

3

$begingroup$

As the previous answer suggests, the matrix is not invertible. From a different perspective we can argue that if an eigenvalue is zero, then at least two columns (or rows) are dependent. Thus, since the colums (or rows) are not independent, the matrix is singular. Therefore, it is not invertible.

share|cite|improve this answer

answered 8 hours ago

DunkelDunkel

401211

$endgroup$

add a comment | 

$begingroup$

As the previous answer suggests, the matrix is not invertible. From a different perspective we can argue that if an eigenvalue is zero, then at least two columns (or rows) are dependent. Thus, since the colums (or rows) are not independent, the matrix is singular. Therefore, it is not invertible.

share|cite|improve this answer

answered 8 hours ago

DunkelDunkel

401211

$endgroup$

share|cite|improve this answer

answered 8 hours ago

DunkelDunkel

401211

answered 8 hours ago

DunkelDunkel

401211

answered 8 hours ago

DunkelDunkel

401211