0
$\begingroup$

I'm a little unsure where to start this question off and proceed...

Suppose that $C\in M_2(\mathbb{R})$ has a non-real eigenvalue $\lambda=a+bi$, and suppose that the rotation matrix $$R=\frac{1}{\sqrt{a^2+b^2}} \begin{pmatrix} a & b\\ -b & a \end{pmatrix}$$ has an anticlockwise angle of rotation of $\pi/3$. If the top-most entry of $C^3$ is $-64$ find $a$ and $b$.

I'm given the following hint.

Express $C$ in terms of $\lambda$, $R$, and an invertible matrix $Q \in M_2(\mathbb{R})$, and then take cubes. You will not be able to actually find Q however.

I'm also given the following proposition to use.

Let $A$ be a real $2 \times 2$ matrix that has a non-real complex eigenvalue $\lambda=a+bi$. If w is an eigenvector for $\lambda$, then the real matrix $Q=(\text{Re}(\mathbf{w})\hspace 0.5cm \text{Im} (\mathbf{w}))$ is invertible and $$Q^{-1}AQ=\begin{pmatrix} a & b\\ -b & a \end{pmatrix}=s\begin{pmatrix} \frac{a}{s} & \frac{b}{s}\\ \frac{-b}{s} & \frac{a}{s} \end{pmatrix}$$ where $s=|\lambda|=\sqrt{a^2+b^2}.$

Can someone help me figure how how to solve this? I'm not sure what the eigenvalues and its relation to the rotation matrices have to really do with any of this or how to apply the hint or proposition...

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Let $\rho=|\lambda|=\sqrt{a^2+b^2}$. Then $\lambda=\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)$. So, the eigenvalues of $C$ are $\rho\left(\cos\left(-\frac\pi3\right)\pm\sin\left(-\frac\pi3\right)i\right)$. Since they are distinct, $C$ is similar to$$\begin{bmatrix}\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)&0\\0&\rho\left(\cos\left(-\frac\pi3\right)-\sin\left(-\frac\pi3\right)i\right)\end{bmatrix},$$and therefore $C^3$ is similar to $\left[\begin{smallmatrix}-\rho^3&0\\0&-\rho^3\end{smallmatrix}\right]$. But this is a scalar matrix and therefore it is similar to itself and only to itself. So,$$C^3=\begin{bmatrix}-\rho^3&0\\0&-\rho^3\end{bmatrix},$$which implis that $\rho=4$. So $\{a,b\}=\left\{2+2\sqrt3\,i,2-2\sqrt3\,i\right\}$.

$\endgroup$
6
  • $\begingroup$ Another followup. How did you get 0's on the opposite diagonal entries? My calculations don't seem to be giving 0. $\endgroup$
    – Future Math person
    Mar 19, 2021 at 17:39
  • $\begingroup$ Do you mean that, when you compute$$\begin{bmatrix}\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)&0\\0&\rho\left(\cos\left(-\frac\pi3\right)-\sin\left(-\frac\pi3\right)i\right)\end{bmatrix}^3,$$you don't get what I got? $\endgroup$
    – José Carlos Santos
    Mar 19, 2021 at 18:13
  • $\begingroup$ I don't. Should you not be getting: $\begin{bmatrix}\frac{1}{2}&\frac{-\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}^3$ instead? Also $\rho$ is $1$ so is that not redundant to write? $\endgroup$
    – Future Math person
    Mar 19, 2021 at 18:20
  • $\begingroup$ For any two numbers $a$ and $b$, and for any natural number $n$, $\left[\begin{smallmatrix}a&0\\0&b\end{smallmatrix}\right]^n=\left[\begin{smallmatrix}a^n&0\\0&b^n\end{smallmatrix}\right]$. And I have no clue about the reason why is it that you say that $\rho=1$. $\endgroup$
    – José Carlos Santos
    Mar 19, 2021 at 19:26
  • $\begingroup$ No I get that part but how are you getting $\begin{bmatrix}\rho\left(\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)i\right)&0\\0&\rho\left(\cos\left(-\frac\pi3\right)-\sin\left(-\frac\pi3\right)i\right)\end{bmatrix},$ in the first place? Isn't the matrix supposed to be written as $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ $\endgroup$
    – Future Math person
    Mar 19, 2021 at 19:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .