Solve $A^k e_1$ for real $A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ using complex eigenvalues

Question

Solve $A^k e_1$ for real $A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ using complex eigenvalues. Where $e_1 = (1,0)^T = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is the standard basis vector.

I believe I can do the problem, but I get a different answer than my textbook on the very last step.

With complex $\lambda_1 = a+bi$, we have corresponding eigenvector $b_1 = (1, -i)^T = \begin{pmatrix} 1 \\ -i \end{pmatrix}$

With complex $\lambda_2 = a-bi$, we have corresponding eigenvector $b_2 = (1, i)^T = \begin{pmatrix} 1 \\ i \end{pmatrix}$

Since $e_1 = (b_1 + b_2) / 2$:

\begin{align*} A^k e_1 &= \frac{1}{2} A^k (b_1 + b_2) \\ &= \frac{1}{2} \left[ (a + bi)^k b_1 + (a - bi)^k b_2 \right] \\ &= \left[ (a + bi)^k + (a - bi)^k \right] e_1/2 - i \left[ (a + bi)^k - (a - bi)^k \right] e_2/2 \\ &= \text{Re} [(a+bi)^k] e_1 + \text{Im} [(a+bi)^k] e_2 \\ \end{align*}

Up to the last step, I completely follow and agree with my textbook.

If we use the fact that ${(z^*)}^k = {(z^k)}^*$ and the facts that $z + z^* = 2 \cdot \text{Re} (z)$ and $z - z^* = 2 \cdot \text{Im} (z)$, that would seem to give a slightly different answer with an added $-i$ term from what the textbook gives:

\begin{align*} &= \text{Re} [(a+bi)^k] e_1 - i \cdot \text{Im} [(a+bi)^k] e_2 \\ \end{align*}

Can anyone tell me which answer is correct and why?

Answer 1

Note that $$z - z^* = 2i \cdot \text{Im} (z)$$

Not $$z - z^* = 2 \cdot \text{Im} (z)$$

Answer 2

It can easily be calculated that $\lambda_{1}=a+bi$ and $\lambda_{2}=a-bi$. Now write

$\lambda_{1}=\sqrt{a^{2}+b^{2}}\,\,$$(cos\theta+isin\theta)$ and

$\lambda_{2}=\sqrt{a^{2}+b^{2}}\,\,$$(cos\theta-isin\theta)$ .

Notice that $A$ is clearly diagonalizable, hence $A=S\Lambda\,S^{-1}$ and clearly $A^{k}=S\Lambda^{k}S^{-1}$.

Therefore $A^{k}=S(\sqrt{a^{2}+b^{2}})^{k}$$\begin{pmatrix} cosk\theta+isink\theta & 0 \\ 0& cosk\theta-isink\theta \\ \end{pmatrix}$$S^{-1}$.

Notice that $\theta=cos^{-1}\dfrac{a}{\sqrt{a^{2}+b^{2}}}$ . Hence

$cosk\theta=cosk(cos^{-1}\dfrac{a}{\sqrt{a^{2}+b^{2}}})$ and

$sink\theta=sink(cos^{-1}\dfrac{a}{\sqrt{a^{2}+b^{2}}})$.

It is easy to calculate $S=\begin{bmatrix} 1 & 1 \\ -i & i \\ \end{bmatrix}$.

And $S^{-1}=\begin{bmatrix} \dfrac{1}{2} & \dfrac{i}{2} \\ \dfrac{1}{2} & \dfrac{-i}{2} \\ \end{bmatrix}$

So everything has been calculated and we just multiply by $e_{1}$.

If my calculations are correct we get the matrix:

$(\sqrt{a^{2}+b^{2}})^{k}$$\begin{pmatrix} cos(kcos^{-1}(\dfrac{a}{\sqrt{a^{2}+b^{2}}}))& -sin(kcos^{-1}(\dfrac{a}{\sqrt{a^{2}+b^{2}}})) \\ sin(kcos^{-1}(\dfrac{a}{\sqrt{a^{2}+b^{2}}}))& cos(kcos^{-1}(\dfrac{a}{\sqrt{a^{2}+b^{2}}})) \\ \end{pmatrix}$ and we multiply by $e_{1}$ to get:

$\begin{pmatrix} cos(kcos^{-1}\dfrac{a}{\sqrt{a^{2}+b^{2}}}) \\sin(kcos^{-1}\dfrac{a}{\sqrt{a^{2}+b^{2}}}) \end{pmatrix}(\sqrt{a^{2}+b^{2}})^{k}$.

The result has been checked for $k=2$ and it is correct!!