Find all possible Jordan forms Let $A \in M_3(\mathbb{C})$, and:
$\frac1{12}A=[A^2-7A+16I_3]^{-1}$
Find all possible Jordan Forms of A
(no need to show different block orders)
I was thinking of Algebraic manipulations such as:
$12A^{-1}=A^2-7A+16I_3 \space/*A$
$12I_3=A(A^2-7A+16I_3)\space/*A$
$A(A^2-7A+16I_3)-12I_3=0 \space/$
$A^3-7A^2+16A-12I_3=0$
$(A-3)(A-2)^2$
Meaning Eigenvalues are 2 and 3.
Notice that Geometric Multiplicity of $\lambda_2$ can be 1 or 2.
So biggest Joran block can be of the sizes 1 or 2
finally:
$$
    \begin{pmatrix}
    2 & 1 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
    \end{pmatrix}
$$
or
$$
    \begin{pmatrix}
    2 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
    \end{pmatrix}
$$
I'd be happy if you could confirm my solution.
 A: Ok, i will type it... As in the later form of the question, we know that $A$ satisfies $p(A)=0$ for 
$$
p(x)=(x-2)^2(x-3)\ .
$$
So the minimal polynomial of $A$ is a divisor of the above $p$, it is thus one among:
$$
(x-2)\ ,\ 
(x-3)\ ,\
(x-2)(x-3)\ ,\
(x-2)^2\ ,\
(x-2)^2(x-3)\ .
$$
Correspondingly we have the possible Jordan forms:
$$
\begin{bmatrix}
2&&\\&2&\\&&2
\end{bmatrix}\ ,\
\begin{bmatrix}
3&&\\&3&\\&&3
\end{bmatrix}\ ,\
\begin{bmatrix}
2&&\\&3&\\&&3
\end{bmatrix}\ ,\
\begin{bmatrix}
2&&\\&2&\\&&3
\end{bmatrix}\ ,\
\begin{bmatrix}
2&1&\\&2&\\&&2
\end{bmatrix}\ ,\
\begin{bmatrix}
2&1&\\&2&\\&&3
\end{bmatrix}\ .
$$
A: Hint: from $\frac1{12}A=[A^2-7A+16I_3]^{-1}$ we get $A^3-7A^2+16A-12I_3=0$, hence the char. polynomial of $A$ is given by
$p(x)=x^3-7x^2+16x-12$.
Then show that $p(x)=(x-2)^2(x-3)$.
Can you proceed with these informations concerning the eigenvalues of $A$ ?
A: You obtained $A^3-7A^2+16A-12I_3=0$ therefore the minimal polynomial $m_A$ divides the polynomial $x^3-7x^2+16x-12=(x-2)^2(x-3)$.
The options for the minimal polynomial and respective Jordan forms are:


*

*$m_A(x) = x-2$
$$\pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$

*$m_A(x) = (x-2)^2$
$$\pmatrix{2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$

*$m_A(x) = x-3$
$$\pmatrix{3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3}$$

*$m_A(x) = (x-2)(x-3)$
$$\pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3} \quad\text{ or} \quad\pmatrix{2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3}$$

*$m_A(x) = (x-2)^2(x-3)$
$$\pmatrix{2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3}$$
Note that in all these cases the matrix $A^2 -7A+16I_3$ is invertible because the polynomial $x^2-7x+16$ has complex zeros, so $\frac1{12}A=[A^2-7A+16I_3]^{-1}$ is indeed equivalent to $A^3-7A^2+16A-12I_3=0$.
