Confused about Eigenvectors Consider the Matrix $\pmatrix{E&t\\t&-E}$
The eigenvalues are $\lambda_1=\sqrt{E^2+t^2}$ , $\lambda_2=-\sqrt{E^2+t^2}$.
Consider the first (+ve) eigenvalue.
To find the eigenvectors, we write:
$$\pmatrix{E-\lambda_1 &t\\t&-E-\lambda_1}  \pmatrix{a\\b} =\pmatrix{0\\0}$$
The problem is that there are two different ways to proceed:
$$(1)\qquad(E-\lambda_1)a + tb  = 0  \Longrightarrow b=\frac{-E+\lambda_1}{t}$$
$$(2)\qquad ta + (-E-\lambda_1)b = 0  \Longrightarrow a=\frac{E+\lambda_1}{t}$$
I know the general solutions is any multiple integer of the eigenvector, but what we have above is NOT! that's something different, what is above is different eigenvectors for the same eigenvalue (not simple integer multiples...)
So what's wrong? Which one of 1) or 2) is right?
 A: It is no problem that you have different eigenvector to an eigenvalue even you may be have eigenvector space with $3$ dimension for a eigenvalue!at this problem also it is not problem!
Only it must satisfy  this relation :
 dimension ( eigenvector space of $\lambda_1$ )+dimension (eigenvector space of $\lambda_2$ )$\le 2$ (because our matrix is $2 \times 2$)
attention if $=2$ then matrix is diagonalizable.
A: An eigenvalue doesn't give an eigenvector, but an eigenspace, formed of infinite eigenvectors.
This is because of the linearity of those applications, $v$ is an eigenvector if $f(v)=\lambda v$, but then any multiple of $v$, $\mu v$, will also be an eigenvector: $f(\mu v)=\mu f(v)=\lambda\mu v$.
So in the case you were writing ($2$ dimensional spaces), the possibility is that the dimension of the eigenspace is $1$ or $2$. Basically, the eigenspace is given by the equation of that same operation you have written $(A-\lambda)v=0$. So it might be the case that the dimension of the eigenspace is not $1$ but $2$ and therefore the same eigenvalue $\lambda$ will give you different independent eigenvectors. This will be one of the keys to understand diagonalization of linear functions.
A: Assuming that the value of $a$ in (1) is $1$ and the value of $b$ in (2) is $1$ (which you don't say explicitly), the vectors you got are indeed multiples (note that they do not have to necessarily be integer multiples as you say) of each other. Hint: what happens if you multiply your $b$ from (1) with your $a$ from (2)?
