Suppose $A$ is a 2 x 2 matrix, prove that if $A^{3} = 0$, then $A^{2} = 0$

Question

I was given this equation as a hint:

$$A^{2} - tr(A)A + det(A)I = 0$$

where $tr(A)$ is the sum of the diagonal entries of $A$.

My attempt:

$$A^{2} - tr(A)A = -det(A)I$$ $$A(A - tr(A)) = -det(A)I$$ $$A[-\frac{1}{det(A)}A(A - tr(A))] = I$$ $A$ is invertible, thus $det(A)$ is not equal to $0$. Multiplying $A^{2}$ to both sides of equation 2:

$$A^{3}(A - tr(A)) = A^{2}(-det(A)I)$$ Since $A^{3} = 0$ and since $-det(A)I$ is not equal to $0$, then $A^{2}$ must be equal. QED.

Is this proof correct?

EDIT: There seems to be a similar question, but my question is more specific in that I have to use the equation given in the hint.

if fact, your $\text{det}$ can be 0, so it is not completely valid to divide by $\text{det}$ — Andrei Kulunchakov, Feb 26 '17 at 13:45
@AndreiKulunchakov Hmm, is it possible to get rid of $det(A)$ in the equation given by the hint? — Kyoma, Feb 26 '17 at 13:52
"There seems to be a similar question, but my question is more specific in that I have to use the equation given in the hint." Indeed, and a solution on the linked page does exactly that. — Did, Feb 26 '17 at 13:54

score 3 · Accepted Answer · answered Feb 26 '17 at 13:52

If $A^3=0$, $\det{A^3} = (\det{A})^3 = 0$, so the equation becomes $$ A^2 -A\operatorname{tr}{A} = 0. \tag{1} $$ Multiplying by $A$, $$ 0 = A^3-A^2\operatorname{tr}{A} = 0-A^2\operatorname{tr}{A}, $$ so either $\operatorname{tr}{A}=0$ or $A^2=0$. But if $\operatorname{tr}{A}=0$, then (1) becomes $A^2=0$ anyway.

Suppose $A$ is a 2 x 2 matrix, prove that if $A^{3} = 0$, then $A^{2} = 0$

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