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!

  • 2
    $\begingroup$ Two words: Cayley and Hamilton. $\endgroup$ – Rodrigo de Azevedo Oct 31 '17 at 13:07
  • $\begingroup$ @RodrigodeAzevedo Haven't heard of Cayley ever, and the only Hamiton I know is a Hamilton path from graph theory $\endgroup$ – Programmer Oct 31 '17 at 13:08
  • $\begingroup$ @RodrigodeAzevedo Oh, wow, I just did some research - so the $P(\lambda) = det(A-\lambda I)=0$ I've been doing this whole time is called the Cayley-Hamilton theorem? $\endgroup$ – Programmer Oct 31 '17 at 13:11
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    $\begingroup$ @Programmer: Not really, but it is related to that. What you have been doing in solving the characteristic equation is finding the eigenvalues. The Cayley-Hamilton theorem is something else, saying that if you plug your whole matrix (not just eigenvalues) into the characteristic equation, it gives the zero matrix. For example, if you had a characteristic equation for $A$ of $\lambda^2 -3\lambda +2=0$, then the Cayley-Hamilton theorem implies that $A^2-3A+2=0$. This fact isn't really used in solving this problem; we can't know what the characteristic equation is from the given info. $\endgroup$ – Jonas Meyer Oct 31 '17 at 18:53

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.

  • $\begingroup$ Hi Jonas, how does $B^2v=BBv=B(\lambda v)=\lambda^2v$? $\endgroup$ – Programmer Nov 1 '17 at 4:51
  • $\begingroup$ @Programmer: Which of those $3$ equations do you not understand? $\endgroup$ – Jonas Meyer Nov 1 '17 at 5:08
  • $\begingroup$ A bit confused how $B(\lambda v)=\lambda^2 v$, since $B=\lambda v$, doesn't $B (\lambda v)=\lambda^2 v^2$? But of course, I don't really understand how arithmetic operations work with vectors/matrices. Could you please instruct me on the rules, thanks! $\endgroup$ – Programmer Nov 1 '17 at 5:12
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    $\begingroup$ @Programmer: There is no $v^2$ that makes sense in this context, and $B=\lambda v$ is wrong. It is correct that $Bv =\lambda v$. It is also the case that in general $A(\lambda v) = \lambda Av$. Being able to "pull out scalars" like this is one of the properties of linearity, and multiplying by a matrix is a linear transformation. So $B(\lambda v) = \lambda Bv = \lambda \cdot \lambda v = \lambda^2 v$. $\endgroup$ – Jonas Meyer Nov 1 '17 at 5:20

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 $\{1,2\}$.


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