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

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

ViktorStein
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Jhonny
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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 $k$". – Ian Mar 27 '19 at 11:43
  • Maybe this can be useful https://math.stackexchange.com/questions/2340541/relation-between-rank-and-number-of-distinct-eigen-values/2340547#2340547 – Widawensen Mar 27 '19 at 12:07
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    As stated it might be wrong, consider for example the matrix $\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$ 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 $n-k$. – Joppy Mar 27 '19 at 12:12
  • @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? – Jhonny Mar 28 '19 at 19:41
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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. – Joppy Mar 28 '19 at 21:14
  • Indeed it only holds for the geometric multiplicity – FShrike Dec 28 '22 at 23:15

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