Solution EE.C20 Robert Beezer
The characteristic polynomial of $B$ is \begin{align*} \charpoly{B}{x}&=\detname{B-xI_2}&&\knowl{./knowls/definition.CP.knowl}{\text{Definition CP}}\\ &= \begin{vmatrix} -12-x&30\\-5&13-x \end{vmatrix}\\ &=(-12-x)(13-x)-(30)(-5)&&\knowl{./knowls/theorem.DMST.knowl}{\text{Theorem DMST}}\\ &=x^2-x-6\\ &=(x-3)(x+2) \end{align*} From this we find eigenvalues $\lambda=3,\,-2$ with algebraic multiplicities $\algmult{B}{3}=1$ and $\algmult{B}{-2}=1$.

For eigenvectors and geometric multiplicities, we study the null spaces of $B-\lambda I_2$ (Theorem EMNS). \begin{align*} \lambda&=3&B-3I_2&= \begin{bmatrix} -15&30\\-5&10 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & -2 \\ 0 & 0 \end{bmatrix}\\ &&\eigenspace{B}{3}&=\nsp{B-3I_2}=\spn{\set{\colvector{2\\1}}} \end{align*} \begin{align*} \lambda&=-2&B+2I_2&= \begin{bmatrix} -10&30\\-5&15 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & -3 \\ 0 & 0 \end{bmatrix}\\ &&\eigenspace{B}{-2}&=\nsp{B+2I_2}=\spn{\set{\colvector{3\\1}}} \end{align*} Each eigenspace has dimension one, so we have geometric multiplicities $\geomult{B}{3}=1$ and $\geomult{B}{-2}=1$.