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?
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$\begingroup$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 $f$.
$\endgroup$ $\begingroup$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
$m_f(X) = (X-1)(X-2)^2$
$\endgroup$ $\begingroup$NOTE: matrix $A$ is diagonalizable iff it contains only linear factors in it's minimal polynomial.
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.
$\endgroup$ 3 $\begingroup$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.
$$ \left( \begin{array}{ccc} 1 & 1 & 0 \\ -1 & -1 & 1 \\ -1 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right) = \left( \begin{array}{rrr} 3 & 1 & 1 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right) $$
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