How can I find the dimension of an eigenspace?

Question

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:

1. How can I find the Jordan normal form?

2. How I can find the dimension of the eigenspace of eigenvalue $-1$?

3. In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?

Answer 1

The SageMath commands to compute anything about this matrix are easy to discover.

Define the matrix:

sage: a = matrix(ZZ, 3, [2, 0, 0, 6, -1, 0, 1, 3, -1])

and then type a.jor<TAB> and then a.eig<TAB>, where <TAB> means hit the TAB key. This will show you the methods that can be applied to a that start with jor and with eig.

Then, once you found the method a.jordan_form, read its documentation by typing a.jordan_form? followed by TAB or ENTER.

You will find that you can call a.jordan_form() to get the Jordan form, or a.jordan_form(transformation=True) to also get the transformation matrix.

sage: j, p = a.jordan_form(transformation=True)
sage: j
[ 2| 0  0]
[--+-----]
[ 0|-1  1]
[ 0| 0 -1]
sage: p
[  1   0   0]
[  2   0   1]
[7/3   3   0]

Here is an exploration of the eigenvalues, eigenspaces, eigenmatrix, eigenvectors.

sage: a.eigenvalues()
[2, -1, -1]
sage: a.eigenspaces_right()
[
(2, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[  1   2 7/3]),
(-1, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[0 0 1])
]
sage: a.eigenmatrix_right()
(
[ 2  0  0]  [  1   0   0]
[ 0 -1  0]  [  2   0   0]
[ 0  0 -1], [7/3   1   0]
)
sage: j, p
(
[ 2| 0  0]               
[--+-----]  [  1   0   0]
[ 0|-1  1]  [  2   0   1]
[ 0| 0 -1], [7/3   3   0]
)
sage: a.eigenvectors_right()
[(2, [
  (1, 2, 7/3)
  ], 1), (-1, [
  (0, 0, 1)
  ], 2)]

Answer 2

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 $e^{At},$

$$ \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right) $$