Finding the rank of an non-invertible matrix

Question

I have a $3\times3$ matrix with three different eigenvalues $0,1, 2$.

The question is: what is the rank of this matrix? If the matrix was invertible, I could say that the rank was equal to $n=3$. But as zero is an eigenvalue of this matrix, this matrix does not satisfy the Invertible Matrix Theorem.

How should I determine the rank? Thanks in advance.

Answer 1

1. All eigenvalues are different, then the matrix is diagonalizable.
2. The corresponding diagonal matrix has the eigenvalues on the diagonal, i.e. $$ S^{-1}AS=D=\left[\matrix{2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0}\right]. $$
3. The matrices $A$ and $D$ have the same rank.

Answer 2

In a simplistic sense, for a square matrix-

Rank = Number of non-zero eigen-values

Answer 3

By the rank-nullity theorem, the rank is always $n$ (the size of the matrix) minus the nullity. The nullity is always the geometric multiplicity (number of associated linearly independent eigenvectors) of the eigenvalue $0$.

In your case, the nullity has to be $1$, which means that the rank is $2$.