# Find eigenvalues of a linear map knowing its eigenvectors

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.

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).

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.

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