# Finding eigenvalues and eigenvector from a linear transformation

Problem:

Let T: $$ℝ^3$$$$ℝ^3$$ be a linear transformation for which

$$T\left( \begin{bmatrix} 1 \\ 0 \\ -2 \\ \end{bmatrix} \right)=\begin{bmatrix} -3 \\ 0 \\ 6 \\ \end{bmatrix}$$

and

$$T\left( \begin{bmatrix} 2 \\ 1 \\ -1 \\ \end{bmatrix} \right)=\begin{bmatrix} -2 \\ -1 \\ 1 \\ \end{bmatrix}$$

(a) From the above information, find two eigenvalues and their corresponding eigenvectors.

(b) If det(T) = 6, what is the third eigenvalue of T?

(c) Is it possible that T is a stochastic matrix? Why or why not?

So I am pretty sure that I would be able to figure out the answers for part b and c however I am stuck on part a. I can't seem to think of the link between the eigenvalues/vectors and am just completely stumped and I'm can't go anywhere. I'm pretty sure for part b I could use the determinant and trace to find the eigenvalue, and for part C it would depend on the three eigenvectors/values. If it has 1 for a eigenvalue and if the eigenvalues are complex numbers they must come in complex conjugates otherwise if 1 is an eigenvalue it is possible that it is a stochastic matrix.

• You say $T:\mathbb R^2\to\mathbb R^2$, but the vectors are $3$-tuples. Are you sure this is right?
– user403337
Commented Dec 1, 2018 at 22:47
• No it was supposed to be 3, I fixed it. Commented Dec 1, 2018 at 22:51

You know that$$T\left(\begin{bmatrix} 1 \\ 0 \\ -2 \\ \end{bmatrix}\right)=(-3)\times\begin{bmatrix}1 \\0 \\-2 \\ \end{bmatrix}$$and that$$T\left( \begin{bmatrix} 2 \\ 1 \\ -1 \\ \end{bmatrix} \right)=(-1)\times\begin{bmatrix}2 \\1 \\-1 \\\end{bmatrix}.$$Doesn't this give you a clue about the eigenvalues of $$T$$?