I have the following square matrix
$$ A = \begin{bmatrix} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 &-1 \end{bmatrix} $$
I found the eigenvalues:
-
$2$ with algebraic and geometric multiplicity $1$ and eigenvector $(1,2,7/3)$.
-
$-1$ with algebraic multiplicity $2$ and geometric multiplicity $1$; one eigenvector is $(0,0,1)$.
Thus, matrix $A$ is not diagonizable. My questions are:
-
How can I find the Jordan normal form?
-
How I can find the dimension of the eigenspace of eigenvalue $-1$?
-
In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?
Best Answer
Most Jordan Normal Form questions, in integers, intended to be done by hand, can be settled with the minimal polynomial. The characteristic polynomial is $\lambda^3 - 3 \lambda - 2 = (\lambda -2)(\lambda + 1)^2.$ the minimal polynomial is the same, which you can confirm by checking that $A^2 - A - 2 I \neq 0.$ Each linear factor of the characteristic polynomial must appear in the minimal polynomial, which exponent at least one, so the quadratic shown is the only possible alternative as minimal.
Next, $$ A+I = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 0 \\ 1 & 3 & 0 \end{array} \right) $$ with genuine eigenvector $t(0,0,1)^T$ with convenient multiplier $t$ if desired.
$$ (A+I)^2 = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 18 & 0 & 0 \\ 21 & 0 & 0 \end{array} \right) $$
The description I like is that we now take $w$ with $(A+I)w \neq 0$ and $(A+I)^2 w = 0.$ I choose $$ w = \left( \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right) $$ This $w$ will be the right hand column of $P$ in $P^{-1}A P = J.$ The middle column is $$ v = (A+I)w, $$ so that $v \neq 0$ but $(A+I)v = (A+I)^2 w = 0$ and $v$ is a genuine eigenvector. You already had the $2$ eigenvector, I take a multiple to give integers. i like integers.
$$ P = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) $$ with $$ P^{-1} = \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) $$
leading to $$ \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) $$
It is the reverse direction $PJP^{-1} = A$ that allows us to evaluate functions of $A$ such as $e^{At},$
$$ \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right) $$