Finding Jordan Normal Form And Rank

Question

Let $0\neq A\in M_3(\mathbb{R})$
Such that $R(A)\perp C(A)$
What are the Rank(A) and $A$ jordan normal form?

How should I approach this?

Answer 1

If $C(A)\bot R(A)$, then immediately $A^2=0\ne A$. Therefore, all eigenvalues of the matrix $A$ are zero.

If the matrix $A$ has only one Jordan cell of order $3$, then, by easy calculation, $A^2\ne 0=A^3$, therefore, all Jordan cells in the Jordan normal form are of order at most $2$.

If all cells are of order $1$, then the whole matrix is zero, which is prohibited.

Thus, the Jordan normal form consists of one Jordan cell of order $2$ and one Jordan cell of order $1$; one can say that $A$ is equivalent to $$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix},$$hence $rank(A)=1$.