Find all possible Jordan forms

Question

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.

Answer 1

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 $A$ ?

Answer 2

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: $$ \begin{bmatrix} 2&&\\&2&\\&&2 \end{bmatrix}\ ,\ \begin{bmatrix} 3&&\\&3&\\&&3 \end{bmatrix}\ ,\ \begin{bmatrix} 2&&\\&3&\\&&3 \end{bmatrix}\ ,\ \begin{bmatrix} 2&&\\&2&\\&&3 \end{bmatrix}\ ,\ \begin{bmatrix} 2&1&\\&2&\\&&2 \end{bmatrix}\ ,\ \begin{bmatrix} 2&1&\\&2&\\&&3 \end{bmatrix}\ . $$

Answer 3

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 $A^3-7A^2+16A-12I_3=0$.