Given a nilpotent matrix $A$ of index 3, prove that its transpose $A^T$ is also nilpotent of index 3

Question

Given the matrix $A$ which is nilpotent of index 3 (which implies $A^3=0$ and $A^2\neq0$), and using matrix and nil-potency properties I have to prove that its transpose is also nilpotent of index 3 such that: $$(A^T)^3 = 0 \land (A^T)^2\neq0$$I tried to solve it using the properties i mentioned above $A^3=0 $ and $A^2\neq0$: $$(A^T)^3 = A^T*A^T*A^T = (A^3)^T = 0$$ $$(A^2)^T = A^T*A^T=(A^2)^T \neq0$$but I am just not confident enough in this solution. Am I missing something or is my approach totally wrong? Furthermore, I feel like the solution is really simple so any help/hint would be beneficial.

Answer 1

it is a correct way to solve it,

(A^3)^T = 0

using the property of matrix transpose (A^T)^n = (A^n)^T, we can write:

(A^T)^3 = 0

hence transpose of A^3 is a zero matrix which means that transpose of A is nilpotent of index 3

to prove that (A^T)^2 ≠ 0:

Again, using the property of matrix transpose (A^T)^n = (A^n)^T, we can write the nilpotency property A^2 ≠ 0 as:

(A^T)^2 ≠ 0

This shows that the transpose of A raised to the power of 2 is not the zero matrix, which means that the transpose of A is also not nilpotent of index 2.