Finding eigenvalues of a particular $3\times3$-matrix, i.e. solving a third order polynomial.

Question

Consider the matrix $$\begin{bmatrix}2&1&0\\1&1&0\\0&0&1\end{bmatrix}$$

The characteristic polynomial is of the form $$(1-\lambda)^2(2-\lambda)-1 = 0.$$ Expanding this out, we get $$\lambda^3 - 4\lambda^2 +5 \lambda -1 = 0.$$ How would I go about solving for $\lambda$ in this case? Do I have to do long division to find a factor? If that's the case, do I just guess and check factors until I get a nice division with no remainder?

Thanks.

Answer 1

The characteristic polynomial is not $(1-\lambda)^2(2-\lambda)-1$, it is $$(1-\lambda)^2(2-\lambda)-(1-\lambda),$$ from which it is immediate that $\lambda=1$ is an eigenvalue.

In general, when you are asked to compute eigenvalues of a matrix as an exercise, the divisors of the constant coefficient are good candidates. In this case these are just $1$ and $-1$.

Answer 2

The characteristic polynomial is $$(1-\lambda)^2(2-\lambda)-(1-\lambda)=0$$ $$\implies (1-\lambda)\bigg((1-\lambda)(2-\lambda)-1\bigg)=0$$ $\implies \lambda=1$ or $\lambda^2-3\lambda+1=0$ $\implies \color{red}{\lambda=1}$ or $\color{blue}{\lambda=\cfrac{3+\sqrt5}{2}}$ or $\color{#180}{\lambda=\cfrac{3-\sqrt5}{2}}$