For what values of t does the eigenvalues $\lambda_1$,$\lambda_2$ has the value $\lambda_1$+$\lambda_2$ = 1.

Question

Do not know how to solve it, could anyone help me please?

Answer 1

\begin{align}\det(M-\lambda I)&=(\cos t-\lambda)^2+\sin^2 t=\cos^2 t-2\lambda \cos t+\lambda^2+\sin^2t\\&=\lambda^2-(2\cos t)\lambda+1\end{align}Hence, by the sum of roots formula: $$\lambda_1+\lambda_2=2\cos t$$ and therefore $$\lambda_1+\lambda_2=1\iff \cos t=\frac12 \iff t=60^o=\frac{\pi}{3}$$

Answer 2

It's known that the sum of eigenvalues is the trace of the matrix so

$$2 \cos x=1$$

The conclusion is immediate.