# why the zero “eigenvector”

Find the eigenvectors of $G$:

G=$\pmatrix{5/4 & \sqrt3/4 & 0 \\ \sqrt3/4 & 3/4 & 0 \\ 0 & 0 & 2}$

I compute the characteristic polynomial and find that the eigenvalues are $\lambda=1,2$ , with $2$ having multiplicity $2$. I then calculate the eigenvector of $\lambda=1$ to get and eigenvector $(1, \sqrt3,0)$.

Then looking at $\lambda=2$ we get the equation:

$\pmatrix{-3/4 & \sqrt3/4 & 0 \\ \sqrt3/4 & -1/4 & 0 \\ 0 & 0 & 0} \pmatrix{ v_1\\v_2 \\v_3}=2\pmatrix{ v_1\\v_2 \\v_3}$.

Which gives the equations:

$\frac{-3}{4}v_1 + \frac{\sqrt3}{4}v_2=2v_1$

$\frac{\sqrt3}{4}v_1 - \frac{1}{4}v_2=2v_2$

$0v_3=2v_3$

But this would suggest an zero eigenvector which is not possible by definition of an eigenvector. Have I gone wrong or does the multiplicity of 2 have something to do with the problem.

• You try to solve the equation $(G - 2I)v = 2v$, but to find the eigenvectors, you ought to solve $(G-2I)v = 0$. – Daniel Fischer Mar 8 '14 at 15:22
• Or solve $Gv=2v$. – Michael Hoppe Mar 8 '14 at 15:26

You must solve $(G-\lambda I) = 0$. The equation you have written is $(G-\lambda I) = \lambda I$

If you write the correct equations, you will get:

$$\frac{-3v_1}{4} + \frac{\sqrt{3}v_2}{4} = 0\\ \frac{\sqrt{3}v_1}{4} - \frac{v_2}{4} = 0\\ 0 = 0$$

When you go for (G−2I)v=2v,to find v as eigen vector you will no longer be finding eigen vector corresponding to that e-value:
since this implies Gv-2v=2v or Gv=4v