I have studied before that the rank of a matrix = number of non zero Eigen values. But recently i came across a problem and i dont think it is valid there. I know i am going wrong somewhere.
$$A=
\begin{bmatrix}
0 & 4 & 0 \\
0 & 0 & 4\\
0 & 0 & 0 \\
\end{bmatrix}
\quad
$$
The Rank of this matrix is 2. So there should be 2 non zero eigen values. But I only get 0 as the eigen value(λ) using $$[A-λI]=0$$
Can anybody explain? Thanks
Best Answer
Unfortunately, the answer is no in general, though the claim will hold true for diagonalizable matrices. Not all matrices are diagonalizable, including the matrix that you gave in your example. If your matrix is $n \times n$, then diagonalizability is equivalent to having a set of $n$ linearly independent eigenvectors, and those eigenvectors corresponding to non-zero eigenvalues will form a basis for the range of the matrix; hence rank is obtained (including multiplicities).
However, if you look at $A^T A$, then you can use the eigenvalues of that matrix to obtain the rank, regardless of what $A$ is. This is because $A^T A$ is symmetric, and thus must be diagonalizable, and furthermore one can show that $\mathrm{rank}(A^T A) = \mathrm{rank}(A)$.