What is the minimal polynomial of A?

Question

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?

Answer 1

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 $f$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

$$ \left( \begin{array}{ccc} 1 & 1 & 0 \\ -1 & -1 & 1 \\ -1 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right) = \left( \begin{array}{rrr} 3 & 1 & 1 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right) $$