What is the minimal polynomial of A? 
Let $f$ be an endomorphism of $\mathbb{R}^3$ with its matrix $A$ in the canonical basis $\mathcal{B}$ as
  $$ A = \begin{pmatrix} 3 & 1 & 1 \\ -1 & 1 & - 1 \\ 0 & 0 & 1 \end{pmatrix}.$$
  What is the minimal polynomial of $f$?

The characteristic polynomial of $f$ is: 
$$P_{f}(X) = (1 - X)(X - 2)^2.$$
The minimal polynomial $m_{f}$ is the polynomial with the least degree that  divides $P_{f}$, has the eigenvalues of $f$ as roots and $m_{f}(A) = 0$. 
In this case, we have $m_{f}(X)= (1 - X)(2 - X)$ but $m_f(f) \neq 0 $.
How can I find the minimal polynomial of $A$ and what is the fastest method to determine it?
 A: Since : 


*

*$m_f$ divides $P_f$;

*they share the same roots;

*$m_f$ is monic,


one can conclude that, in this case, there are only two possibilites : 


*

*either $m_f(X) = (1-X)(X-2)$

*or $m_f(X) = (1-X)(X-2)^2$


Now which one is it ? Well, plug in $f$ in the first one. This doesn't evaluate to $0$, so the only possibility left is $m_f = P_f$, which we know evaluate to $0$ when we plug in $f$. 
A: The matrix $A$ has eigenvalues 1,2,2. as minimal polynomial of $A$ has the eigenvalues of $A$ as roots and divides characteristic polynomial so only possibilities for minimal polynomial are :
$m_f(X)$ = $(X-1)(X-2)$ or $(X-1)(X-2)^2$
For eigenvalue 2  
Algebraic multiplicity = 2
Geometric Multiplicity = 3 [no. of columns in A] -  $\rho(A-2I)$ [Rank of $A-2I$] = 3 - 2 = 1
this gives A.M. is not equal to G.M so $A$ is not diagonalizable and only option left is
$m_f(X) =  (X-1)(X-2)^2$

NOTE: matrix $A$ is diagonalizable iff it contains only linear factors in it's minimal polynomial.

A: Your minimal polynomial and characteristic polynomial have same roots. This what you can use then you have to check whether the matrix vanishes at $(1-X)(2-X)$ or not. In general otherwise, you have to use Jordan-Block theorem and see what is the dimension of each eigenspace. Most of the cases it is calculative.
A: The minimal polynomial and characteristic polynomial agree, which is equivalent to each eigenvalue occurring in exactly one Jordan block.
$$
R =
\left(
\begin{array}{rrr}
1 & 1  & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 0
\end{array}
\right)
$$
$$
R^{-1} =
\left(
\begin{array}{rrr}
0 & 0  & -1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right)
$$
and
$$  R^{-1} A R = J. $$
$$
J =
\left(
\begin{array}{ccc}
1 & 0  & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
$$
The direction that is actually useful is $R J R^{-1} = A.$ Useful for finding $e^A$ or $A^{100}$ or any $f(A)$ with $f$ single-variable analytic.
$$
\left(
\begin{array}{ccc}
1 & 1  & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0  & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 0  & -1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right) =
\left(
\begin{array}{rrr}
3 & 1  & 1 \\
-1 & 1 & -1 \\
0 & 0 & 1
\end{array}
\right)
$$
