Question

What are the eigenvectors of \[\left[\begin{matrix}7 & -\sqrt{3} \\ -\sqrt{3} & 5\end{matrix}\right] \]

Answer 1

\[\left[\begin{matrix}7- \lambda & - \sqrt{3} \\ - \sqrt{3} & 5- \lambda \end{matrix}\right]\]

Solve for Eigenvalues \( \lambda \) first, then you can discuss the eigenvectors.

Answer 2

The eigenvalues are 8 and 4

Answer 3

What would I do now?

Answer 4

Exactly, OpenStudy is kicking me pretty often today so bear with me.

You can substitute back the Eigenvalues. \( \lambda_1=4 \)

You get the following system of equations:

\[3x - \sqrt{3}y=0 \\- \sqrt{3}x + y=0 \]

Answer 5

If you multiply the lower equation by \( - \sqrt{3} \) you will see that they are identical. So you can only solve one of these systems.

Answer 6

Yeah, having the same problem with os. Alright, so I found the eigenvectors as:
\[\left(\begin{matrix}-\sqrt{3} \\1\end{matrix}\right) \ and \ \left(\begin{matrix}1/\sqrt{3} \\ 1\end{matrix}\right)\]
are these correct?

Answer 7

If you look at the first equation you have

\[ 3x-\sqrt{3}y=0 \] This means that:

\[\vec{n}=\left(\begin{matrix}3 \\ -\sqrt{3}\end{matrix}\right)\] is a normal vector of this equation, so if you want the eigenvector you just switch the columns and change one sign, so you should get:

\[\left(\begin{matrix}\sqrt{3} \\ 3\end{matrix}\right)\] as an Eigenvector.

Answer 8

Wouldn't the equations be:
-x-sqrt(3)=0

3x-sqrt(3)y=0

Answer 9

Wait, seems you have that. Both equations work.

Answer 10

Well if it works the way you did it stick to it, the general outcome is that one equation becomes redundant, because it's just a multiple of the other equation, so you can forget about it and solve for Eigenvectors using one equation.

Answer 11

Alright, thanks

Answer 12

welcome