Find $c$ such that you get real eigenvalues

Question

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?

Answer 1

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 $c \ge -1$.

Answer 2

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.