Why is the eigenvector for λ=3 equal to j-hat?

[kq6up]

Homework Statement

Find the eigenvectors of: $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 5 & 0 & \sqrt{3} \\ 0 & 3 & 0 \\ \sqrt{3} & 0 & 3 \end{array}\right)$

Homework Equations

$(\mathbf{A}-\lambda\mathbf{I})\cdot\mathbf{x}=0$

The Attempt at a Solution

I get the correct eigenvectors for $\lambda=2,6$, but I don't understand why the eigenvector is $\hat{j}$ when $\lambda=3$.

When $\lambda=3$, the matrix becomes $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 2 & 0 & \sqrt{3} \\ 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 \end{array}\right)$. The first row yields a function $2x-\sqrt{3}z=0$. The points that satisfy this equation do not lay along $\hat{j}$. What am I missing?

Thanks,
Chris

 

[vela]

The third row implies $x=0$. Do you see that?

 

[Mark44]

The first row yields a function 2x− 3√ z=0

The first row yields the equation 2x + 3 √ z=0.

As vela notes, the third row implies that x = 0. Neither equation involves y, so there are no constraints on y.

 

[kq6up]

Ah, that makes sense.

Thanks,
Chris