If square matrix A satisfying $A^2-4A+4I=0$ does it follow that A is diagonizable?

Question

I am given the following statement and asked to determine whether it is true or false:

If $A$ is an $n\times n$-matrix, and $A^2-4A+4I=0$, then $A$ is diagonizable.

Any help is appreciated, thank you.

Answer 1

"Hint": Consider the matrix $$A:=\begin{pmatrix}2&1\\0&2\end{pmatrix}.$$

Answer 2

The minimum polynomial is $$ m(\lambda)=\lambda^2-4\lambda+4=(\lambda-2)^2 $$ So the only eigenvalue of $A$ is $2$. Since the $m(\lambda)$ is not of linear factor, $A$ is not diagonizable. A Jordan block has a maximum size of $2$.

An example of $A$ is $$ A=\pmatrix{2 & 1 & 0 \\0 & 2&0 \\ 0 &0 &2} $$

Answer 3

The equation $A^2 -4A +4I = 0$ can be written as $(A -2I)^2 =0$ and hence $A-2I$ must be a nilpotent matrix of order $2$. Therefore the only eigenvalue of $A-2I$ is $0$ and so the only eigenvalue of $A$ is $2$. Now it is easy to see that $A$ and $A-2I$ have the same eigenvectors and therefore the dimension of their respective eigenspaces must be the same. It follows that $A$ is diagonalizable if and only if $A-2I$ is diagonalizable. Since the only diagonalizable nilpotent matrix is the zero matrix we conclude that $A$ is not diagonalizable unless it equals $2I$.

Answer 4

Even simpler answer is to calculate the minimal polynomial, which (by dropping exponents of all factors of the characteristic polynomial) : $\lambda-2$. So what we need to check is if ${\bf A}-2 {\bf I} = 0$ or not.