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I am quite confused about this. I know that zero eigenvalue means that null space has non zero dimension. And that the rank of matrix is not the whole space. But is the number of distinct eigenvalues ( thus independent eigenvectos ) is the rank of matrix?

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    $\begingroup$ The identity matrix has $1$ as its only eigenvalue. It has many eigenvectors. $\endgroup$ – André Nicolas Jul 5 '15 at 6:10
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Well, if $A$ is an $n \times n$ matrix, the rank of $A$ plus the nullity of $A$ is equal to $n$; that's the rank-nullity theorem. The nullity is the dimension of the kernel of the matrix, which is all vectors $v$ of the form: $$Av = 0 = 0v.$$ The kernel of $A$ is precisely the eigenspace corresponding to eigenvalue $0$. So, to sum up, the rank is $n$ minus the dimension of the eigenspace corresponding to $0$. If $0$ is not an eigenvalue, then the kernel is trivial, and so the matrix has full rank $n$. The rank depends on no other eigenvalues.

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  • $\begingroup$ Thanks for the answer. If I know that a matrix has 1 eigenvalue which is zero, does it mean that dimension of kernal is 1? And rank of matrix is (rank of whole space - 1)? $\endgroup$ – Shifu Jul 5 '15 at 6:20
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    $\begingroup$ Well, consider the $0$ matrix. It has one eigenvalue: $0$, and the dimension of its eigenspace is $n$, since it sends everything to $0$. If you have $n$ distinct eigenvalues, one of them zero, then the eigenspace for $0$ must have dimension $1$, hence the rank is $n - 1$. $\endgroup$ – Theo Bendit Jul 5 '15 at 6:26
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    $\begingroup$ This is what I have understood. Please correct me if i am wrong. 1) If a matrix has 1 eigenvalue as zero, the dimension of its kernel may be 1 or more (depends upon the number of other eigenvalues). 2) If it has n distinct eigenvalues its rank is atleast n. 3) The number of independent eigenvectors is equal to the rank of matrix. $\endgroup$ – Shifu Jul 5 '15 at 6:33
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    $\begingroup$ 2) is not right. If it has $n$ non-zero eigenvalues, then the nullity is $0$ and the rank is $n$. If one of the $n$ is $0$, then it has rank $n-1$ as I said. 1) is almost right. Yes, if $1$ of the eigenvalues is $0$, then the kernel has dimension at least $1$, maybe more. However, it doesn't just depend on the number of other eigenvalues. It is possible to have only $0$ as an eigenvalue, but still only have a nullity of $1$. 3) is again, not quite right. The rank is equal to the number of independent generalised eigenvectors. Look up Jordan Bases. $\endgroup$ – Theo Bendit Jul 5 '15 at 7:29
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    $\begingroup$ Eigenvectors/values are only applicable to square matrices. But $A^T A$ and $AA^T$ are square matrices with equal rank to $A$. $\endgroup$ – Theo Bendit May 20 '17 at 14:50

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