Find $c$ such that you get real eigenvalues Let 

A=$\begin{bmatrix}
     1& c\\
     1& -1
     \end{bmatrix}$

find all eigenvalues, but do not use characteristic polynomial. Find c for which eigenvalue will be real.
I use $detA=-1-c$ and $tr(A)=0$, since $detA=\lambda_1\lambda_2$ and $tr(A)=\lambda_1+\lambda_2$ I get that $\lambda_1=-\lambda_2$ so $\lambda_2=\pm\sqrt{1+c}$ and $\lambda_1=\mp\sqrt{1+c}$, $c$ must be $\geq-1$ if I want real eigenvalue, I think that this is only way to not use characteristic polynomial , what you think?
 A: By direct method we need to solve
$$\begin{bmatrix}
     1& c\\
     1& -1
     \end{bmatrix}\begin{bmatrix}
     x\\
     y
     \end{bmatrix}=\lambda\begin{bmatrix}
     x\\
     y
     \end{bmatrix} \iff \begin{bmatrix}
     1-\lambda& c\\
     1& -1-\lambda
     \end{bmatrix}\begin{bmatrix}
     x\\
     y
     \end{bmatrix}=\begin{bmatrix}
     0\\
     0
     \end{bmatrix} $$
then use elimination method in order to exclude the trivial solution.
A: We have
$$\begin{bmatrix}
     \lambda x\\
     \lambda y\end{bmatrix} = \lambda\begin{bmatrix}
     x\\
     y\end{bmatrix} = \begin{bmatrix}
     1& c\\
     1& -1
     \end{bmatrix}\begin{bmatrix}
     x\\
     y\end{bmatrix} = \begin{bmatrix}
     x+cy\\
     x-y\end{bmatrix}$$
Therefore $\lambda y = x-y$ so $(\lambda + 1)y = x$.
Also $\lambda x = x+cy$ so $(\lambda - 1)x = cy$ or $$(\lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $\lambda$ is not an eigenvalue. Therefore $\lambda^2 - 1 - c = 0$ so $\lambda = \pm \sqrt{1+c}$.
The eigenvalues are real if and only if $c \ge -1$.
