Finding Eigenvectors/values given matrix defined by bra-ket notation

[slam7211]

Homework Statement

sorry about the lack of LaTex but I don't know how to do bra-ket notation in tex
vectors |1> and |2> are a complete set of normalized basis vectors.
the hamiltonian is defined as |1><1|-|2><2|+|1><2|+|2><1| find the eigenvalues and eigenvectors in ters of |1> and |2>

Homework Equations

for a normal matrix eigenvalues are

Det(A-Iλ)=0 solve for λ
Eigenvectors are then
A*v=λv

The Attempt at a Solution

I tried just making up 2 generic 1X2 vectors and plugging it into mathematica, but its ugly and I am assuming its not what they want, is there some stupid trick I am missing here?
If the first sign in the hamiltonian was + instead of minus I know the first set of terms would equal 1 but its minus so I am just lost here

 

[HallsofIvy]

I may be misunderstanding your hamiltonian but it looks to me like that corresponds to matrix
$$\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$

If that is correct, then the characteristic equation is
$$\left|\begin{array}{cc} 1-\lambda & 1 \\ 1 & -1-\lambda\end{array}\right|= (1-\lambda)(-1- \lambda)- 1= 0$$
$$= \lambda^2- 2= 0$$

so the eigenvalues are $\sqrt{2}$ and $-\sqrt{2}$.

The corresponding eigenvectors are given by
$$\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}x & y\end{bmatrix}= \sqrt{2}\begin{bmatrix} x \\ y \end{bmatrix}$$
so that $x+ y= \sqrt{2}x$ and $x- y= \sqrt{2}y$

and
$$\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}x & y\end{bmatrix}= -\sqrt{2}\begin{bmatrix} x \\ y \end{bmatrix}$$
so that $x+ y= -\sqrt{2}x$ and $x- y= -sqrt{2}y$

 

[slam7211]

How did you know the hamiltonian corresponded to that matrix?