2
$\begingroup$

Let $\;f: R^{3} \rightarrow R^{3}\;$ be an endomorphism whose eigenvectors are $(0,1,-2),(1,0,4),(1,0,-2)$. Knowing that $f(0,1,0) = (2,1,2)$ find the eigenvalues of $f$.

What is the methodology to solve this problem?

Thank you.

$\endgroup$

2 Answers 2

5
$\begingroup$

The key property here is linearity.

Notice that the eigenvectors form a basis for $\mathbb{R^3}$; this means that you can reconstruct the $(0,1,0)$ vector, whose output you know, out of linear combinations of the eigenvectors. This will look like something of the form:

$$f(0,1,0) = f(\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3) = (2,1,2)$$

where the $\alpha$'s are constants. Then you use the properties of linear operators to distribute the appropriate action of the operator on each of the eigenvectors, resulting in something that looks like:

$$\alpha_1 \lambda_1 e_1 + \alpha_2 \lambda_2 e_2 + \alpha_3 \lambda_3 e_3 = (2,1,2)$$

Given the values for the $\alpha$'s that you solved for in the previous step, you then find the eigenvalues ($\lambda$'s).

$\endgroup$
2
$\begingroup$

A more pedestrian approach is to use the matrix for $f$. The condition that $f(0,1,0)=(2,1,2)$ means that the middle column of the matrix must be $(2,1,2)$.

Also $f(1,0,4)=(.,0,.)$ and $f(1,0,-2)=(.,0,.)$ so the middle row of the matrix must be 0,1,0. So we can write the matrix as $$\begin{pmatrix}a&2&b\\ 0&1&0\\ c&2&d\end{pmatrix}$$ Hence $f(0,1,-2)=(2-2b,1,2-2d)$. But $(0,1,-2)$ is an eigenvector, so $b=1$, the eigenvalue is 1 and $d=2$.

Also $f(1,0,4)=(a+4,0,c+8)$, so $c+8=4(a+4)$. And $f(1,0,-2)=(a-2,0,c-4)$, so $c-4=-2a+4$. Hence $a=0$ and $c=8$, so the matrix is $$\begin{pmatrix}0&2&1\\ 0&1&0\\ 8&2&2\end{pmatrix}$$ and it is easy to check that $f(0,1,0)=(2,1,2),f(0,1,-2)=(0,1,-2)$, $f(1,0,4)=4(1,0,4)$ and $f(1,0,-2)=-2(1,0,-2)$.

So the eigenvalues are 1,4 and -2.

$\endgroup$
2
  • $\begingroup$ Hi, almagest. I don't understand why the middle column of the matrix must be 0. $\endgroup$
    – user9867
    Commented Jan 14, 2020 at 18:53
  • $\begingroup$ Sorry, don't know how that crept in. Fixed. $\endgroup$
    – almagest
    Commented Jan 14, 2020 at 18:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .