Let $A$ be a $5 \times 5$ matrix in $\mathbb R.$ If $A^3=0$ and $\dim(\ker(A))+\dim(\ker(A^{2})=8$. Find the Jordan canonical form of $A.$

Question

I think $A^3=0$, the minimal polynomial should divide $t^3$. The possible Jordan Canonical forms are $(J_3(0), J_1(0), J_1(0)),$

$(J_1(0), J_1(0), J_1(0),J_1(0),J_1(0))$

$(J_2(0), J_1(0), J_1(0),J_1(0))$

$(J_2(0), J_2(0), J_1(0))$

$(J_3(0), J_2(0))$

The second condition state that (sum of block of order $1$ and block of order $2$ is eight).

Can anyone help me to think about second option?

Thanks

Answer 1

Hint: If Jordan blocks of $A$ are $(J_{r_1}(0),\, J_{r_2}(0),\, \dots)$, then $$\dim\ker(A^k) =\sum_i\min(r_i, k)$$ so you can verify all the combinations.

Answer 2

For convenience set

• $k_1 := \dim(\ker(A))$ and $k_2=\dim(\ker(A^{2}))$

So, you have $$0 < k_1 \leq k_2 \leq 5 \mbox{ and } k_1 + k_2 = 8$$

The only possible solutions are $k_1=3, k_2=5$ or $k_1=k_2 = 4$.

$k_1=4$ must be ruled out (note that $\dim (\operatorname{im}(A)) = 1$ in this case) because this corresponds to $A \sim (J_2(0),J_1(0),J_1(0),J_1(0))$ and then $k_2=5$, which is a contradiction.

For $k_1=3, k_2=5$ you find $A\sim (J_2(0),J_2(0),J_1(0))$, which satisfies the given condition.