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?

  • 2
    $\begingroup$ You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"... $\endgroup$
    – Vim
    Sep 8, 2018 at 9:06
  • $\begingroup$ Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae. $\endgroup$
    – qualcuno
    Sep 8, 2018 at 9:06
  • $\begingroup$ I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know $\endgroup$ Sep 8, 2018 at 9:10
  • $\begingroup$ It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ... $\endgroup$
    – Mandelbrot
    Sep 8, 2018 at 9:10

2 Answers 2


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 $c \ge -1$.

  • $\begingroup$ That’s precisely the solution by the definition, that is solve the system $Ax=\lambda x$ with $x\neq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear. $\endgroup$
    – user
    Sep 8, 2018 at 10:43
  • $\begingroup$ Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway? $\endgroup$
    – amd
    Sep 8, 2018 at 21:07
  • $\begingroup$ @amd Yes, but it's more elementary as it avoids determinants. $\endgroup$ Sep 8, 2018 at 21:17

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.

  • $\begingroup$ Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way $\endgroup$ Sep 8, 2018 at 9:11
  • $\begingroup$ @MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials. $\endgroup$
    – user
    Sep 8, 2018 at 9:13
  • $\begingroup$ Yes but that is my teacher he give that on exam $\endgroup$ Sep 8, 2018 at 9:16
  • 3
    $\begingroup$ @MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr). $\endgroup$
    – user
    Sep 8, 2018 at 9:19

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