1
$\begingroup$

I have a matrix

$$ A = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} $$

I can find eigenvectors in Maple with Eigenvectors(A) from which I get the eigenvalues

$$ \lambda_1 = 1 \qquad \lambda_2 = -1 $$

and the eigenvectors

$$ v_1 = (-1,1) \qquad v_2 = (1,1) $$

which is all fine.

But if I want to find the eigenvectors more 'manually' I will first define the characteristic matrix $K_A(\lambda) = A-\lambda I$ and use v[1] = LinearSolve(K[A](lambda[1]),<0,0>,free='t') and v[2] = LinearSolve(K[A](lambda[2]),<0,0>,free='t') in Maple. Now I get the eigenvectors

$$ v_1 = (-t_2,t_2) \qquad v_2 = (t_2,t_2) $$

They are almost correct; however, I have to use subs(t[2]=1, %) in order to get the correct result.

Why does it use $t_2$ instead of $1$? Why is $t_2$ and not just $t$? How can I get the eigenvectors correct? I want to define an automatic procedure which gives me the eigenvectors, so I don't want to substitute $t_2$ manually afterwards.

$\endgroup$
1
$\begingroup$

Maple is looking for ALL of the solutions when it runs LinearSolve. Eigenvectors only finds a basis for the eigenspace.

$\endgroup$
  • $\begingroup$ Can I do something to also find a basis for the eigenspace using LinearSolve? $\endgroup$ – Jamgreen Sep 11 '15 at 6:47
1
$\begingroup$

The second Maple result is correct, and the first is wrong... OK, maybe not exactly wrong, but not completely correct. For any eigenvalue of any matrix, there will always be infinitely many eigenvectors. Remember the definition: $v$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$ means $$Av=\lambda v\ ,\quad v\ne0\ .$$ If this is true and $t$ is a scalar, then $$A(tv)=\lambda(tv)\ ,$$ so $tv$ is also an eigenvector of $A$ corresponding to $\lambda$, as long as $t\ne0$.

$\endgroup$
  • $\begingroup$ The two functions are solving different problems, and returning the correct solutions to those problems. $\endgroup$ – Christopher Carl Heckman Sep 11 '15 at 6:36
  • $\begingroup$ @CarlHeckman OK, maybe I should have said that the OP's use of the first one is wrong, if the intention is to find a complete solution of the eigenvector problem. $\endgroup$ – David Sep 11 '15 at 6:38
0
$\begingroup$

You can find a basis for each of the eigenspaces associated with the eigenvalues, "by hand", as follows:

restart:

with(LinearAlgebra):

A:=Matrix([[0,-1],[-1,0]]):

evals:=Eigenvalues(A);

                                      [ 1]
                             evals := [  ]
                                      [-1]


NullSpace(CharacteristicMatrix(A,evals[1]));

                                 [-1]
                                {[  ]}
                                 [ 1]

NullSpace(CharacteristicMatrix(A,evals[2]));

                                  [1]
                                 {[ ]}
                                  [1]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.