1
$\begingroup$

A differentail equations exercise asks for the eigenvectors of $$\begin{bmatrix} 3&2&4\\ 2&0&2\\ 4&2&3 \end{bmatrix}.$$

The answer in the back of the book gives $$\begin{bmatrix} 2\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}, \begin{bmatrix} 1\\ -4\\ 1 \end{bmatrix};$$ the first two have eigenvalues $-1$, the second has eigenvalue $8$. Mathematica confirms this:

A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}}; v2 = {{2, 1, 2}, {1, 0, -1}, {1, -4, 1}} Table[A.v2[[i]], {i, 1, 3}] {{2, 1, 2}, {1, 0, -1}, {1, -4, 1}} {{16, 8, 16}, {-1, 0, 1}, {-1, 4, -1}}

However, Mathematica gives the eigenvectors as $$\begin{bmatrix} 2\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -2\\ 0 \end{bmatrix};$$ and then confirms them as eigenvectors:

A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}}; v = Eigenvectors[A] Table[A.v[[i]], {i, 1, 3}] {{2, 1, 2}, {-1, 0, 1}, {-1, 2, 0}} {{16, 8, 16}, {1, 0, -1}, {1, -2, 0}}

Why is this happening?

$\endgroup$
6
  • $\begingroup$ If there is any eigenvector at all, there will always be infinitely many. $\endgroup$ Commented Nov 10, 2014 at 2:36
  • $\begingroup$ I guess I was just being stupid. Clearly $A(3v) = 3Av = 3\lambda v = \lambda3v$. $\endgroup$ Commented Nov 10, 2014 at 2:43
  • 2
    $\begingroup$ A linear combination of eigenvectors with exactly the same eigenvalues is another eigenvector $\endgroup$ Commented Nov 10, 2014 at 2:43
  • $\begingroup$ Thank you for the help. Should I delete the question or leave it for the other poor souls that make the same mistake? $\endgroup$ Commented Nov 10, 2014 at 2:44
  • 1
    $\begingroup$ @JacobBond if you feel that you understand the answer well enough now, feel free to post an answer yourself $\endgroup$ Commented Nov 10, 2014 at 2:45

1 Answer 1

Reset to default
1
$\begingroup$

Simple answer:

Any scalar multiple $\mu$ of an eigenvector $v$ with eigenvalue $\lambda$ is an eigenvector: $$A(\mu v) = \mu Av = \mu \lambda v = \lambda(\mu v).$$

More complete answer:

As Will Jagy points out in the comments, any linear combination of eigenvectors with the same eigenvalue $\lambda$ with be an eigenvector for $\lambda$:

$$A\sum_{i = 1}^{n}\mu_{i}v_{i} = \sum_{i = 1}^{n}\mu_{i} Av_{i} = \sum_{i = 1}^{n}\mu_{i}\lambda v_{i} = \lambda\sum_{I = 1}^{n}\mu_{i}v_{i}.$$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.