# Raising a square matrix to the k'th power: From real through complex to real again - how does the last step work?

On page 69 it gives an example where a real, square matrix $A=[(a,-b),(b,a)]$ is raised to the k'th power: $$A^k.(1,0)^T$$ The result must be a real vector. Nevertheless it seems easier to do the calculation via the complex numbers:$$=((a+bi)^k+(a-bi)^k).(1,0)^T/2-i((a+bi)^k-(a-bi)^k).(0,1)^T/2$$ At this stage the result seems to be complex. But then comes the magic step and everything gets real again:$$=Re[(a+bi)^k].(1,0)^T+Im[(a+bi)^k].(0,1)^T$$ Now I did some experiments and made two observations: First, this step seems to yield the correct results - yet I don't know why. Second, the raising of this matrix to the k'th power even confuses CAS (e.g. WolframAlpha aka Mathematica, see e.g. the plots here) because they most of the time seem to think that the results are complex.

My question
Could you please give me a proof/explanation for the correctness of the last step. Perhaps you will even know why CAS are confused too (perhaps it is because their algorithms also go through the complex numbers and they too have difficulties in seeing that the end result will be real?)

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$A$ is of specific form $A=[(a,b)',(-b,a)']$, this may give you a clue. –  mpiktas Jun 14 '11 at 13:54
@mpiktas: Could you please elaborate - thank you! –  vonjd Jun 14 '11 at 13:56
see my answer. The result in that book is not for general matrix $A$. I tried to write all the details, so the answer is self-contained. –  mpiktas Jun 14 '11 at 14:18

What you are using is that for a given complex number $z=a+bi$, we have $\frac{z+\overline{z}}{2}=a={\rm Re}(z)$ and $\frac{z-\overline{z}}{2}=ib=i{\rm Im}(z)$ (where $\overline{z}=a-bi$). Also check that $\overline{z^k}=\overline{z}^k$ for all $k \in \mathbb{N}$.

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Thank you, this clarifies quite a bit. The step seems even to hold when k is real - how does this work? –  vonjd Jun 14 '11 at 14:46
@vonjd, Euler formula: $z^k$=$(|z|\exp(iArgz))^k=|z|^k\exp(ikArgz)$ –  mpiktas Jun 14 '11 at 15:36
So this implies $\frac{z^k+\overline{z}^k}{2}={\rm Re}(z^k)$ and $\frac{z^k-\overline{z}^k}{2}=i{\rm Im}(z^k)$, right? –  vonjd Jun 14 '11 at 17:08

We have that

\begin{align} A=\begin{pmatrix}a & -b\\ b &a \end{pmatrix} \end{align}

with $a$ and $b$ real numbers. In space $L(\mathbb{C}^2)$ this matrix has 2 eigen-values $a+bi$ and $a-bi$ with eigen vectors $\mathbf{b}_1=(1,-i)$ and $\mathbf{b}_2=(1,i)$.

We have $(1,0)=\mathbf{e}_1=(\mathbf{b}_1+\mathbf{b}_2)/2$. So

\begin{align} A^k\mathbf{e}_1=A^k(\mathbf{b}_1+\mathbf{b}_2)/2=A^k\mathbf{b}_1/2+A^k\mathbf{b_2}/2 \end{align}

Now we use the fact that $\mathbf{b}_1$ and $\mathbf{b}_2$ are eigenvectors, hence

$$A^k\mathbf{b}_1=(a+bi)^k\mathbf{b}_1$$

$$A^k\mathbf{b}_2=(a-bi)^k\mathbf{b}_2$$

Now using the fact that

$$\mathbf{b}_1=\mathbf{e}_1-i\mathbf{e}_2$$ $$\mathbf{b}_2=\mathbf{e}_1+i\mathbf{e}_2$$

we arrive at the equation in question:

$$A^k\mathbf{e}_1=((a+bi)^k+(a-bi)^k)/2\mathbf{e}_1-i((a+bi)^k-(a-bi)^k).(0,1)^T/2\mathbf{e}_2$$

We rewrite this as

$$A^k\mathbf{e}_1=c_1\mathbf{e}_1+ic_2\mathbf{e}_2,$$

where $c_1$ and $c_2$ are complex numbers. The vector on the left hand side is real, so the vector on the right-hand side should be real too. The complex vector $\mathbf{c}$ is real if

$$\mathbf{c}=Re (\mathbf{c})$$

So the following must hold

\begin{align} c_1\mathbf{e}_1+ic_2\mathbf{e}_2&=Re(c_1\mathbf{e}_1+ic_2\mathbf{e}_2)\\ &=Re(c_1)\mathbf{e_1}+Re(ic_2)\mathbf{e_2}\\ &=Re(c_1)\mathbf{e_1}-Im(c_2)\mathbf{e_2} \end{align}

and we arrive at the required result.

Update Actually we do not, since I missed the last step. @Dennis provides the ending in its answer. His answer contains the essential trick, my answer just gives the context to the question, i.e. what is the matrix $A$, etc.

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Thank you anyway. I agree, the last step is the crucial one. Btw: you have an extra "\$" in the following line: "We rewrite this as..." –  vonjd Jun 14 '11 at 14:40