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.
Best Answer
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 $T$?