In my statistics classes it was stated that if an $n\times n$ 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?

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    $\begingroup$ 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 Mar 27 at 11:43
  • $\begingroup$ Maybe this can be useful math.stackexchange.com/questions/2340541/… $\endgroup$ – Widawensen Mar 27 at 12:07
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    $\begingroup$ 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 $n-k$. $\endgroup$ – Joppy Mar 27 at 12:12
  • $\begingroup$ @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 Mar 28 at 19:41
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    $\begingroup$ 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 Mar 28 at 21:14

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