Can this equation give $A=A^\ast$

Question

I have this equation: $$ (A-2I)(A^2-9I)=0 $$$$ \rightarrow A^3-2A^2-9A+18I=0 /()^* $$$$ \rightarrow (A^*)^3-2(A^*)^2-9A^*+18I=0 $$$$ \rightarrow A^3-2A^2-9A+18I=(A^*)^3-2(A^*)^2-9A^*+18I $$ And got to this: $$ (A^*)^3-2(A^*)^2-9A^*=A^3-2A^2-9A $$

Could this mean that A is normal ($AA^*=A^*A$) or the other way around? and how?

Thanks.

Answer 1

The answer is: no, $A$ does not have to be normal.

A counterexample: take $$ A=\begin{bmatrix}2 & 0 & 1\\0 & 3 & 0\\0 & 0 & -3\end{bmatrix}. $$ It satisfies the matrix equation, but $AA^T\ne A^TA$.

The equation you have $$ (A-2I)(A-3I)(A+3I)=0 $$ simply means that the polynomial $q(x)=(x-2)(x-3)(x+3)$ is annihilating. The conclusion you can get from the equation is that the matrix $A$ has eigenvalues among $2$, $3$ and $-3$ and is diagonalizable. However, it does not have to be unitary diagonalizable (=normal).

Answer 2

You can apply to the equation $A^3-2A^2-9A=0$ the adjoint operator to have the other identity so your question is to prove that if $A $is a root of $A^3-2A^2-9A$ than A is normal.

$A^3-2A^2-9A=A((A-I)^2-10I)=$

$A(A-(1+\sqrt{10})I)(A+(-1+\sqrt{10})I)=0$

So the roots are

$A_1=0$

$A_2= (1+\sqrt{10})I$

$A_3= (1-\sqrt{10})I$

That are all normal