I have the following matrix:
$P=
\begin{bmatrix}
6 & -3 \\
2 & 1
\end{bmatrix}
$
And its eigenvector is :
$v=
\begin{bmatrix}
4 \\
3
\end{bmatrix}
$
I would like to find its eigenvalue. This is my attempt, however I am doing something wrong:
$
\begin{bmatrix}
6 & -3 \\
2 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} =
\begin{bmatrix}
4 \\
3
\end{bmatrix}
$
Then…
$6x -3y = 4$
$2x + y = 3$
Elimination…
$x = 13/12$
$y = 10/12$
$
\begin{bmatrix}
13/12 \\
10/12
\end{bmatrix} = \lambda
\begin{bmatrix}
4 \\
3
\end{bmatrix}
$
For the final part, I can't figure out a value for $\lambda$ that satisfies both 4 and 3, which means I am doing something wrong. Can someone point me in the right direction?
Best Answer
You need to write $$ \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \end{bmatrix} = \lambda \begin{bmatrix} 4 \\ 3 \end{bmatrix} $$
and then find $\lambda$. (By the way, when I do this in my head, I find that no such lambda exists, i.e., that this is NOT an eigenvector of the given matrix...but maybe my in-the-head arithmetic isn't so reliable.)
Given Amzoti's comment, you need to write $$ \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = a\begin{bmatrix} x \\ y \end{bmatrix} $$ where $a$ is either $4$ or $3$, and find the associated eigenvector. If $v$ is an eigenvector, so is $2v$, so you'll have to pick either $x$ or $y$ and then solve for the other.