Find minimal polynomial of matrix with some given property Let $A$ be a $4 \times 4$ non-diagonizable complex matrix that satisfies the relation $A(A-2I_4)^2=0$. I want to find, with proof, all the possible forms that the minimal polynomial $m_A(x)$ of the matrix $A$ could have. For each of the above forms of $m_A(x)$, i want to write all the possible forms that the corresponding Jordan matrix of $A$ could have.
I have thought the following:
$A(A-2I_4)^2=0 \Rightarrow A(A^2-4A+4I_4)=0 \Rightarrow A^3-4A^2+4A=0$
Let $\lambda$ be an eigenvalue of $A$ with eigenvector $x$.
We have that $Ax=\lambda x$.
Multiplying $A^3-4A^2+4A=0$ be $x$ we get that $(A^3-4A^2x+4A)x=0 \Rightarrow A^3x -4A^2x+4Ax=0 \Rightarrow \lambda^3 x-4 \lambda^2 x+4 \lambda x=0 \Rightarrow \lambda x(\lambda^2-4 \lambda+4)=0 \Rightarrow \lambda=0 \text{ or } (\lambda-2)^2=0 \Rightarrow \lambda=0 \text{ or } \lambda=2$.
Thus the only minimal polynomial of $A$ is $m_A(x)=x(x-2)$. Am I right?
Shouldn't one eigenvalue have a higher multiplicity? Or shouldn't we have one more eigenvalue? Have I done something wrong?
 A: Since $A$ satisfies the polynomial $p(X):=X(X-2)^2$, the minimal polynomial $m_A$ must divide $p$. Hence a priori you can have either


*

*$m_A(X)=1$

*$m_A(X)=X$

*$m_A(X)=X-2$

*$m_A(X)=(X-2)^2$

*$m_A(X)=X(X-2)$

*$m_A(X)=X(X-2)^2$


But a matrix is diagonalizable iff each factor appearing in its minimal polynomial (for each eigenvalue) is linear. Thus the only nondiagonalizable ones are cases 4 and 6.
To get the possible Jordan forms, We take into account that:
a. The multiplicity of the eigenvalue in the minimal polynomial determines the size of its largest Jordan block.
b. The sum of the sizes of its corresponding Jordan blocks determines its algebraic multiplicity.
c. The geometric multiplicity is found as the number of corresponding Jordan blocks. 
For case (4): We have just one eigenvalue, $\lambda=2$, so it must have algebraic multiplicity $4$ (since we have matrices of size $4\times 4$). Its largest Jordan block has size $2$, so we can either have two blocks of size $2$ ($2+2=4$) or three blocks, one of size $2$ and two of size $1$ ($2+1+1=4$). In the first case the geometric multiplicity is $2$, in the second one it is $3$ (both lower than $4$, the algebraic multiplicity, so the matrices are indeed nondiagonalizable).
I'll let case (6) for you to complete!
