11
$\begingroup$

The professor teaching a class I am taking wants me to find the eigenvalues and the eigenvectors for the following matrix below.

$$\begin{bmatrix}-5 & 5\\4 & 3\end{bmatrix}$$

I have succeeded in getting the eigenvalues, which are $\lambda= \{ 5,-7 \}$. When finding the eigenvector for $\lambda= 5$, I get $\begin{bmatrix}1/2\\1 \end{bmatrix}$. However, the correct answer is $\begin{bmatrix}1\\2 \end{bmatrix}$ .

I have tried doing this question using multiple online matrix calculators. One of which gives me $\begin{bmatrix}1/2\\1 \end{bmatrix}$, and the other gives me $\begin{bmatrix}1\\2 \end{bmatrix}$.

The online calculator that gave me $\begin{bmatrix}1\\2 \end{bmatrix}$ explains, that y=2, hence $\begin{bmatrix}1/2·2\\1·2 \end{bmatrix} = \begin{bmatrix}1\\2 \end{bmatrix}$.

What I do not understand is, why is y must equal to 2?Is it because there cannot be a fraction in an eigenvector?

$\endgroup$
  • 22
    $\begingroup$ Both are eigenvectors. Eigenvectors are not unique in general. $\endgroup$ – Arctic Char Nov 30 at 9:32
  • 10
    $\begingroup$ A constant times an eigenvector is also an eigenvector. $\endgroup$ – Ameet Sharma Nov 30 at 10:06
  • 4
    $\begingroup$ Conclusion: Any scalar that is multiplied with an eigenvector is still an eigenvector (In the same eigenspace). Even when the eigenvector is multiplied with different scalars, the λ would be the same, Av = λV. *I cannot choose which comment is the best since all of it helped me in understanding the given problem. Thank you all. $\endgroup$ – Kamarul Adha Nov 30 at 10:36
  • 4
    $\begingroup$ For some reason most problems I've encountered (in college) have asked me to find eigenvalues and the corresponding eigenvectors, when in reality what we care about are the eigenvalues and the corresponding eigenspaces. $\endgroup$ – BallpointBen Nov 30 at 22:43
  • 5
    $\begingroup$ @KamarulAdha: Second conclusion: If anyone tells you that $v$ is an eigenvector of a matrix but $2v$ is not, that one does not know mathematics. $\endgroup$ – user21820 Dec 1 at 15:28
31
$\begingroup$

By definition, an eigenvalue $\lambda$ and one corresponding eigenvector $v$ must satisfy the following equation:

$$Av = \lambda v.$$

Now, consider the vector

$$w = \alpha v,$$

where $\alpha$ is a non null number.

Then, notice that:

$$Aw = A(\alpha v) = \alpha (Av) = \alpha (\lambda v) = \lambda (\alpha v) =\lambda w.$$

Therefore:

$$Aw = \lambda w,$$

and hence $w$ is another eigenvector associated to the eigenvalue $\lambda$.

In general, it is not true that there is only one eigenvector associated to the eigenvalue $\lambda$. Instead, there is a linear subspace, also known as the eigenspace associated to $\lambda$. Informally, there are infinitely many eigenvectors to $\lambda$, which belong to a certain eigenspace. Given one eigenvector (say $v$), then all the multiples of $v$ except for $0$ (i.e. $w = \alpha v$ with $\alpha \neq 0$) are also eigenvectors.

$\endgroup$
8
$\begingroup$

There are matrices with eigenvectors that have irrational components, so there is no rule that your eigenvector must be free of fractions or even radical expressions. As an example:

$$\begin{bmatrix}1 & 2\\1 & 1\end{bmatrix}$$

Which has eigenvalues of $1 \pm \sqrt{2}$ and eigenvectors of $\begin{bmatrix}\pm\sqrt{2} \\ 1\end{bmatrix}$

(Additionally, because of the unsolvability of the quintic, there are even eigenvectors that cannot be expressed in elementary functions.)

Because of the form of the equations that you solve to get the eigenvectors, you have infinite solutions to the eigenvectors. Additionally, an eigenvector is only really valuable as a direction. So if any eigenvector can be said to be the "correct" or "most special" one, it's the one that has a norm of 1, or a norm of the associated eigenvalue. However, these can be actually a bit ugly to express.

For our example they are: norm 1, $\begin{bmatrix}\pm\sqrt{\frac{2}{3}} \\ \sqrt{\frac1{3}}\end{bmatrix}$ norm $\lambda$: $\frac1{3}\begin{bmatrix}\mp 2\sqrt{3} \pm\sqrt{6} \\ \sqrt{3} \pm \sqrt{6} \end{bmatrix}$.

That being said, it is, of course, nicer to work with whole numbers when the opportunity arises.

$\endgroup$
4
$\begingroup$

$$\begin{bmatrix}-5 & 5\\4 & 3\end{bmatrix}\begin{bmatrix}\frac12\\1 \end{bmatrix}=\begin{bmatrix}\frac52\\5\end{bmatrix}=5\begin{bmatrix}\frac12\\1 \end{bmatrix}$$

so that $\begin{bmatrix}\frac12\\1 \end{bmatrix}$ is undisputably an Eigenvector associated to the Eigenvalue $5$.

$\endgroup$
3
$\begingroup$

Calculations with whole numbers are easier than fractions, so people often take eigenvectors with whole numbers for convenience.
For example, $(1/2,1)$ and $(1,2)$ both normalize to the same vector, but the calculations give you $$\left(\frac{1/2}{\sqrt{5/4}},\frac1{\sqrt{5/4}}\right)\text{ and }\left(\frac1{\sqrt5},\frac2{\sqrt5}\right)$$

$\endgroup$

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.