Matrices of the form $A^k = A$

Question

Let $A$ be a $n\times n$ matrix. Choose the correct option.

a) if $A^2 =0$ then $A$ is diagonalisable.

b) if $A^2 =I$ then $A$ is diagonalisable.

c) if $A^2 - A =0$ then $A$ is diagonalisable.

Now... for a) I tried by transforming them using minimal polynomial (cayley Hamilton) so it should satisfy $t^2 =0$ hence the eigen values are $0$ (twice) hence its not convertible and not diagonalisable... similar arguments for b) and c).

But I am not sure whether I am right... please guide me..

Answer 1

• For a) : if $A^2=0$ then $A$ is nilpotent and hence isn't diagonalizable, unless it's the null matrix
• For b) : $A$ is the root of a polynomial($X^2-1$) with distinct roots and hence is diagonalizable
• For c) : $A$ is the root of a polynomial ($X^2-X$) with distinct roots and hence is diagonalizable

Answer 2

For (c), it is an idempotent matrix, and idempotent matrices are always diagonalizable.

Answer 3

Hint: If $p(A) = 0$, then the minimal polynomial of $A$ divides $p(t)$. $A$ is diagonalizable if and only if its minimal polynomial is the product of distinct linear factors (so, for example: if $A$ has minimal polynomial $t^2$ it is not diagonalizable, but if it has minimal polynomial $t$ then it is).

Answer 4

One simple proof for $A^2=0$

Let $A^2=0$. if $A$ is diagonalisable Then $A=PDP^{-1}$ for some invertible $P$ $$ A^2=0\implies PD^2P^{-1}=0\implies D^2=0 \implies D=0\implies A=0 $$