# $A$ is diagonalizable and $A^3 = A^2$

If $A$ is diagonalizable and $A^3 = A^2$. Is it necessary true that $A^2 = A$?

• If A is diagonalizable, does it mean the A is non-singular? If yes, than $A^2 = A$, if not - then It's not necessary true that $A^2 = A$... Ideas? – Ilan Aizelman WS Jul 2 '14 at 13:35
• Hint: First solve your problem for diagonal matrices. – Moishe Kohan Jul 2 '14 at 13:38
• "If A is diagonalizable, does it mean the A is non-singular?" No, e.g., the zero matrix is diagonalizable and singular. Incidentally, without the diagonalizability hypothesis, $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ would be a counterexample. – Jonas Meyer Jul 2 '14 at 13:42

Since we can diagonalize $A$ it is sufficient to prove this for a diagonal $D=diag(d_1,...,d_n)$. This would mean that the diagonal entries have the property $(d_i)^3 = (d_i)^2$. If $d_i = 0$ then $(d_i)^2 = d_i$. If $d_i\neq 0$ then we may divide by $d_i$ to find $(d_i)^2 = d_i$.
The justification for switching to a diagonal matrix $D$ follows since we may write $A=Q^{-1}DQ$ which means $$Q^{-1} D^3 Q = A^3 = A^2 = Q^{-1} D^2 Q$$
The polynomial $P=X^3-X^2=X^2(X-1)$ satisfies $P(A)=0$.
Now, if $A$ is diagonalizable, the roots of the minimal polynomial $\mu_A$ of $A$ are of multiplicity one. Since $P(A)=0$, then $P$ is a multiple of the minimal polynomial of $A$, hence $\mu_A$ must divide $X(X-1)$ hence we must have $A(A-I)=0$, i.e., $A^2=A$.