Confirm by multiplication that x is an eigenvector of A, and find the corresponding eigenvalue.
Given:
\begin{align}
A = \begin{pmatrix} 1&2\\3&2\\\end{pmatrix}, &&
x = \begin{pmatrix} 1\\-1\\\end{pmatrix}
\end{align}
I know: $Ax = \lambda x$
My work:
I know $\lambda I – A$
\begin{pmatrix} \lambda – 1&-2\\-3&\lambda – 2\\\end{pmatrix}
From there I know the characteristic polynomial is $\lambda^2 – 3\lambda – 4 = 0$ through ad-bc (since this is a 2 x 2 matrix)
I can successively trying out each factor of c (which is 4) : positive and negative of 1,2,4.
4 turns out to be the only one that works. So $\lambda – 4 = 0$. So the $\lambda$ = 4.
I also know I can divide the characteristic polynomial by $\lambda – 4$ and get $\lambda + 1$. Setting $\lambda + 1 = 0$. $\lambda$ is $-1$.
Answer: So I got two eigenvalues which is $-1$ and $4$.
Dilemma I am having with eigenvector:
The problem is I am not sure if the given eigenvector applies for both the left and right side of the equation Ax = $\lambda$x. Or is it just the left side?
Work I have done using the given eigenvector x:
I know Ax = $\lambda$x
\begin{align}
\begin{pmatrix} 1&2\\3&2\\\end{pmatrix} \cdot \begin{pmatrix} 1\\-1\\\end{pmatrix} = \begin{pmatrix} 1*(1) + &2 (-1)\\3*(1)&2(-1)\\\end{pmatrix} = \begin{pmatrix}
-1\\1\\\end{pmatrix} = Ax.
\end{align}
Problem I am facing:
What do I do after this step? Do I use the given value of the eigenvector $x$ on the right side of the equation $Ax = \lambda x$ along with the eigenvalue I find to see if the equation satisfies itself? How do I know if the given eigenvector is actually correct?
Best Answer
The directions are confirm by multiplication. All you need do is compute $Ax$ for the given $A$ and $x$ then compare that result to the given $x$.