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$$
\newcommand{\miao}[1]{\begin{pmatrix}#1\end{pmatrix}}
\newcommand\tr{{\mathrm{tr}}}
$$

$$
A=\miao{-3&-4&4\\2&-1&-12\\1&6&13}
$$

$$
\tr(A)=-3-1+13=9
$$

$$
\det\miao{-3&-4\\2&-1}=13
$$

$$
\det\miao{-3&4\\1&13}=-43
$$

$$
\det\miao{-1&-12\\6&13}=59
$$

The sum of principal minor of size 2 = 27

Now calculate the determinant of $A$, we see

$$
\det\miao{-3&-4&4\\2&-1&-12\\1&6&13}=\det\miao{-3&-4&4\\2&-1&-12\\3&5&1}=\det\miao{0&1&5\\2&-1&-12\\3&5&1}
$$

$$
=\det\miao{0&1&5\\2&0&-7\\3&5&1}=50-21-2=27
$$

The characteristic polynomial

$$
\lambda^3-9\lambda^2+27\lambda-27=(\lambda-3)^3
$$

This means, eigenvalue is $3,3,3$. It is **not diagonalizable**



$$
A-3I=\miao{-6&-4&4\\2&-4&-12\\1&6&10}
$$

I saw the result. Didn't you?

$$\dim(\ker(A-3I))=1$$
$$\dim(\ker(A-3I)^2)=2$$
$$\dim(\ker(A-3I)^3)=3$$

So for this $A$, it must have only one Jordan block of eigenvalue 3.

Now let us find the Zeno's eigenvector. We have to take $\epsilon^2\neq\epsilon^3=0$

$$
A-3I-\epsilon I =
$$

$$
\miao{-6-\epsilon&-4&4\\2&-4-\epsilon&-12\\1&6&10-\epsilon}
$$

To solve the following linear equation, 
$$
\miao{-6-\epsilon&-4&4\\2&-4-\epsilon&-12\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

first row + 3 times the second row. Second row minus 2 times the thrid row

$$
\miao{-\epsilon&-16-3\epsilon&-32\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$


first row + $\epsilon$ times the last row
$$
\miao{0&-16+3\epsilon&-32+10\epsilon-\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

First row minus the second row

$$
\miao{0&4\epsilon&8\epsilon-\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

multiply the first row by $4$

$$
\miao{0&16\epsilon&32\epsilon-4\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

first row + second row $\times 16$

$$
\miao{0&-\epsilon^2&-2\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

first row $\times 16$

$$
\miao{0&-16\epsilon^2&-32\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

first row -$\epsilon^2\times$ second row


$$
\miao{0&0&0\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0}
$$

I can solve this equation now.  

Observation: second column and thrid column is not colinear, which means to find a non-trivial linear dependence relation among columns, I must use the first column.

I can observe a solution

$$
\miao{0&0&0\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{*\\-32+2\epsilon\\16+\epsilon}=\miao{0\\{0}\\{\color{red}0}}
$$

OK 

$$
*+6(-32+2\epsilon)+(10-\epsilon)(16+\epsilon)=0
$$
$$
*=-6(-32+2\epsilon)-(10-\epsilon)(16+\epsilon)
$$

$$
*=192-12\epsilon-160+6\epsilon+\epsilon^2=32-6\epsilon+\epsilon^2
$$

Because the Zeno-eigenvector for eigenvalue $3+\epsilon$ is

$$
\vec v=\miao{32-6\epsilon+\epsilon^2\\-32+2\epsilon\\16+\epsilon}
$$

My eigenvalue is $3+\epsilon$

The matrix $M$ such that $M^2+2M=A$, I can just assume that this M also has $\vec v$ as an eigenvector. Suppose the corresponding eigenvalue for $M$ is $\lambda$.

$$
\lambda^2+2\lambda=3
$$

$$
\lambda^2+2\lambda-3=(\lambda-1)(\lambda+3)
$$

Let's think about $\lambda=1$

We actually want (here $x$ is some nilpotent symbol)

$$
(1+x)^2+2(1+x)=3+\epsilon
$$

$$
(4x+x^2)=\epsilon
$$

For this $x$ the vector 

$$
\vec v=\miao{32-6\epsilon+\epsilon^2\\-32+2\epsilon\\16+\epsilon}
$$

is a vector of eigenvalue $1+x$

$$
\vec v=\miao{32-6(4x+x^2)+(4x+x^2)^2\\-32+2(4x+x^2)\\16+(4x+x^2)}
$$

So now we have eigenvalue of $M$ is $1$ and $\vec v$ is the zeno's eigenvector, so we can find matrix $M$ out.

**Is that clear?**


$$
\vec v=\miao{32-6(4x+x^2)+16x^2\\-32+2(4x+x^2)\\16+(4x+x^2)}
$$

$$
\vec v=\miao{32-24x+10x^2\\-32+8x+2x^2\\16+4x+x^2}
$$


Canonical basis 

$$
\left\{
\miao{32\\-32\\16},\miao{-24\\8\\4},\miao{10\\2\\1}
\right\}
$$

Let $$
P=\miao{32&-24&10\\-32&8&2\\16&4&1}
$$

$$
MP=P\miao{1&1&0\\0&1&1\\0&0&1}
$$

$$
M=\miao{32&-24&10\\-32&8&2\\16&4&1}\miao{1&1&0\\0&1&1\\0&0&1}\miao{32&-24&10\\-32&8&2\\16&4&1}^{-1}
$$

Is that enough? or you want me to calculate it out?

</br></br></br></br></br></br></br></br></br></br></br></br>

## Method of Lagurange polynomial,

You only need to find a polynomial to map $3+\epsilon$ to $1+x$ where $(4x+x^2)=\epsilon$

Then apply this polynomial to the matrix $A$ then you can get the matrix $M$

We need to find a polynomial to map $3+4x+x^2$ to $1+x$. I construct the Taylor polynomial.

$$
f(t)=1+a(t-3)+b(t-3)^2
$$
$$
f(3+4x+x^2)=1+a(4x+x^2)+16bx^2=1+4ax+(a+16b)x^2$$

Solve the equation
$$
4a=1\qquad a+16b=0
$$

$$
a=\frac14\qquad b=-\frac1{64}
$$

Then the polynomial is 

$$
f(t)=1+\frac14(t-3)-\frac1{64}(t-3)^2
$$

So we only need to take
$$
M = 1+\frac14(A-3)-\frac1{64}(A-3)^2
$$
</br></br></br></br></br></br>