Claim:
For any matrix, the number of non-zero eigenvalues – including algebraic multiplicity – seems to always be equal to the rank of the matrix. Here is a non-counterexample:
\begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{bmatrix}
The above matrix has an eigenvalue of 3 with an algebraic multiplicity of 3. The rank is also three. I have read about this topic: What is the relation between rank of a matrix, its eigenvalues and eigenvectors
However, I see a few people use this counterexample:
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
They claim: (1) the rank is 1, and (2) the eigenvalue is 0 with an algebraic multiplicity of 2. I dispute the latter.
Why not just switch row 1 and row 2?
Is there something illegal about that? Then the rank would still be 1 and the eigenvalues would be 1 and 0. Thus, the claim that the number of non-zero eigenvalues is the rank of a matrix holds true.
Is there a better counterexample matrix that disproves my original claim?
Best Answer
Because that changes the eigenvalues, their multiplicities (geometric and algebraic), and the eigenspaces. Elementary row operations do not preserve eigenanythings. Note that $1$ is now an eigenvalue, and $0$ is now has multiplicity $1$.
Fun fact: what does work is, for every elementary row operation you perform, you perform immediately afterwards the corresponding inverse elementary column operation! For example, if you multiply the $i$th row by $\alpha \neq 0$, then divide the $i$th column by $\alpha$. This will actually preserve eigenvalues (but not eigenvectors).
So, in your example, after swapping the rows, if you swap the columns, you get $$\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$ which is another counterexample with rank $1$ and eigenvalue $0$ with multiplicity $2$.