# What is the relation between rank of a matrix, its eigenvalues and eigenvectors

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?

• The identity matrix has $1$ as its only eigenvalue. It has many eigenvectors. Jul 5, 2015 at 6:10
• if my matrix is of size $[n, n]$, and rank $r$, wouldn't there be $r$ eigenvalues and we can get $r$ independent eigenvectors? Oct 19, 2021 at 21:25
• Oct 19, 2021 at 21:26

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.
• 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 $n - 1$. Jul 5, 2015 at 6:26
• 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. Jul 5, 2015 at 7:29
• Eigenvectors/values are only applicable to square matrices. But $A^T A$ and $AA^T$ are square matrices with equal rank to $A$. May 20, 2017 at 14:50