Given a $2 \times 2$ matrix $B$ that satisfies $B^2=3B-2I$, find the eigenvalues of $B$

Question

Given a $2 \times 2$ matrix $B$ that satisfies $B^2=3B-2I$, find the eigenvalues of $B$.

My attempt:

Let $v$ be an eigenvector for B, and $\lambda$ it's corresponding eigenvalue. Also, let $T$ be the linear transformation (not that this is exactly necessary for the question, but just added it in for my understanding.) Therefore,

$$T(v) = Bv = \lambda v$$

Now I'm unsure how to incorporate this information into the quadratic equation given above since by matrix / vector arithmetic isn't extremely solid. Thanks!

Answer 1

Suppose $\lambda $ is an eigenvalue for $B$, with eigenvector $v$. Note that $B^2v=BBv=B(\lambda v)=\lambda^2v$. Apply each side of your equation $B^2=3B-2I$ to the vector $v$ to get $\lambda^2 v= (3\lambda -2)v$, or $(\lambda^2-3\lambda +2)v=0$. If a scalar times a nonzero vector is the zero vector, then the scalar is $0$, so $\lambda^2-3\lambda +2=(\lambda -1)(\lambda -2) =0$. This means that the set of eigenvalues of $B$ is a subset of $\{1,2\}.$ It is impossible to determine which subset from the information given, as diagonal matrix examples show.

Answer 2

Define $p(x) = (x-1)(x-2)$. You have $p(B) = 0$. Hence, the minimal polynomial of $B$ divides $p$. So, your eigenvalues are in $\{1,2\}$.