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Number of zero eigenvalues equal to the number of dimensions “lost”?

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In my statistics classes it was stated that if an $ntimes n$ matrix $A$ has $k$ zero eigenvalues, we have $rank(A) = n-k$. Is there any straightforward proof of this? Are there any limitations on the matrix needing to be symmetric?

From the relations $det(A) = Pi_i lambda_i$ and $ rank(A) < n leftrightarrow det(A) = 0 $, I understand that one zero eigenvalue must imply that the tank of $A$ is deficient. But in case of several zeros, why must the number of them equal the number of ranks lost?

linear-algebra eigenvalues-eigenvectors determinant matrix-rank

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asked yesterday

JhonnyJhonny

127110

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  The point is that the kernel and the eigenspace with eigenvalue $0$ are the same space. Note, however, that $0$ could be a defective eigenvalue, so when you say "has $k$ zero eigenvalues" you really mean "has $0$ as an eigenvalue with geometric multiplicity $k$".
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  – Ian
  yesterday
  

  
  
  
  


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  Maybe this can be useful math.stackexchange.com/questions/2340541/…
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  – Widawensen
  yesterday
  
  

  
  
  
  


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  As stated it might be wrong, consider for example the matrix $beginpmatrix 0 & 1 \ 0 & 0 endpmatrix$ which has a zero eigenvalue with multiplicity $2$, but rank 1. What is true is that if $k$ is the dimension of the zero eigenspace, then the rank of the matrix is $n-k$.
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  – Joppy
  yesterday
  

  
  
  
  


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  @Joppy interesting, thank you! I don't think we have studied this in greater depth (restricted to symmetric matrices). In what cases would the dimension of an eigenspace not be equal to the algebraic multiplicity?
  $endgroup$
  – Jhonny
  7 hours ago
  

  
  
  
  


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  Symmetric matrices are always diagonalisable, so it will never happen for them. If you would like to look up some examples, you should search for the “Jordan form” of a matrix, which is a kind of replacement for diagonalisation.
  $endgroup$
  – Joppy
  5 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

In my statistics classes it was stated that if an $ntimes n$ matrix $A$ has $k$ zero eigenvalues, we have $rank(A) = n-k$. Is there any straightforward proof of this? Are there any limitations on the matrix needing to be symmetric?

From the relations $det(A) = Pi_i lambda_i$ and $ rank(A) < n leftrightarrow det(A) = 0 $, I understand that one zero eigenvalue must imply that the tank of $A$ is deficient. But in case of several zeros, why must the number of them equal the number of ranks lost?

linear-algebra eigenvalues-eigenvectors determinant matrix-rank

share|cite|improve this question

asked yesterday

JhonnyJhonny

127110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The point is that the kernel and the eigenspace with eigenvalue $0$ are the same space. Note, however, that $0$ could be a defective eigenvalue, so when you say "has $k$ zero eigenvalues" you really mean "has $0$ as an eigenvalue with geometric multiplicity $k$".
  $endgroup$
  – Ian
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe this can be useful math.stackexchange.com/questions/2340541/…
  $endgroup$
  – Widawensen
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As stated it might be wrong, consider for example the matrix $beginpmatrix 0 & 1 \ 0 & 0 endpmatrix$ which has a zero eigenvalue with multiplicity $2$, but rank 1. What is true is that if $k$ is the dimension of the zero eigenspace, then the rank of the matrix is $n-k$.
  $endgroup$
  – Joppy
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joppy interesting, thank you! I don't think we have studied this in greater depth (restricted to symmetric matrices). In what cases would the dimension of an eigenspace not be equal to the algebraic multiplicity?
  $endgroup$
  – Jhonny
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Symmetric matrices are always diagonalisable, so it will never happen for them. If you would like to look up some examples, you should search for the “Jordan form” of a matrix, which is a kind of replacement for diagonalisation.
  $endgroup$
  – Joppy
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In my statistics classes it was stated that if an $ntimes n$ matrix $A$ has $k$ zero eigenvalues, we have $rank(A) = n-k$. Is there any straightforward proof of this? Are there any limitations on the matrix needing to be symmetric?

From the relations $det(A) = Pi_i lambda_i$ and $ rank(A) < n leftrightarrow det(A) = 0 $, I understand that one zero eigenvalue must imply that the tank of $A$ is deficient. But in case of several zeros, why must the number of them equal the number of ranks lost?

linear-algebra eigenvalues-eigenvectors determinant matrix-rank

share|cite|improve this question

asked yesterday

JhonnyJhonny

127110

$endgroup$

share|cite|improve this question

asked yesterday

JhonnyJhonny

127110

share|cite|improve this question

asked yesterday

JhonnyJhonny

127110

asked yesterday

JhonnyJhonny

127110

asked yesterday

JhonnyJhonny

127110