Question

Find the eigenvalues and eigenvectors of the 2x2 Matrix:
|-3 1|
| 4 2|

Answer 1

\[\left[\begin{matrix}-3 & -1 \\ 4 & 2\end{matrix}\right]\]
The eigenvalues are 1 and -2

Answer 2

r = 1\[\left(\begin{matrix}1 \\ -4\end{matrix}\right)\]r = -2\[\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]

Answer 3

Ok, that is correct. But how do I calculate the eigenvectors?

Answer 4

\[\left[\begin{matrix}-3-r & -1 \\ 4 & 2-r\end{matrix}\right]\]Plug in your values for r. So for r = 1 you get\[\left[\begin{matrix}-4 & -1 \\ 4 & 1\end{matrix}\right]\]Convert that to a linear eq to get \[4x + 1y\]Solving you get that \[-4x = y\] Leaving you with\[\left[\begin{matrix}1 \\ -4\end{matrix}\right]\]

Answer 5

Ok, silly question here but how to I get from\[\left[\begin{matrix}-4 & -1 \\ 4 & 1\end{matrix}\right]\]
To the linear equation

Answer 6

without getting to complicated, that matrix is equivalent to \[-4x -1y\]\[4x +1y\] It's just the coefficients for each variable

Answer 7

I see in this case the two equations are equivalent, is this always the case?

Answer 8

when using eigenvalues to find eigenvectors, yes

Answer 9

Thanks

Answer 10

\[Ax = \lambda x\]\[Ax - \lambda x = 0\]\[(A - \lambda I)x = 0\]

So we have
\[(A - \lambda I)x = 0\]
Solve for x (this is also equivalent to a system of linear equations).

Answer 11

Its better to know where it comes from otherwise it seems like magic.

Answer 12

For [abcd], det(A-lamI) = [a-lamb,b,c,d-lam]

= lam^2 - (a+d)lam +(ad-bc), roots are eigenvalues.

Eigenvectors

a-lamx +by =0 and

cx + (d -lam)y =0 which reduces to a single equation.