# Thread: Rewrite a real linear transformation

1. ## Rewrite a real linear transformation

Hello,

I am trying to prove the following:

Let $I: \mathbb{C} \to \mathbb{C}$ be the identity and $\bar{I}:\mathbb{C} \to \mathbb{C}$ be conjugation. If $T: \mathbb{C} \to \mathbb{C}$ is real linear then there exists unique $w_1,w_2\in \mathbb{C}$ such that $T=w_1I+w_2 \bar{I}$.

The definition I have of real linear is: $T:\mathbb{C} \to \mathbb{C}$ is called real linear if $T(\lambda z + \gamma w)=\lambda T z + \gamma T w$ for all vectors and all real scalars.

I have also that $T$ is real linear if and only if $T \sim M$ with $M=\begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}$ and $T(x+iy)=\alpha x + \beta y + i(\gamma x + \delta y)$

So I wrote out $T=w_1I+w_2 \bar{I}$ and I'm having trouble believing this to be true at all as I would need, $$Tz\sim Mz = \begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}w_1+w_2 & 0 \\ 0 & w_1-w_2\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}=(w_1I+w_2 \bar{I})z$$where $\mathbb{C} \ni z \sim \begin{pmatrix} x \\ y \end{pmatrix}$

This should imply that $w_1+w_2=\alpha$, $0=\beta=\gamma$ and $w_1-w_2=\delta$.

There has to be something really wrong with how I'm working this. Is it something to do with the one matrix having complex entries and the other real?

2. ## Re: Rewrite a real linear transformation

Hey bkbowser.

You will have to show that if you assume w1 and w2 are real then so must be the rest of the numbers on the LHS.

Because it has to be real linear, then only real numbers should exist in the appropriate coefficients.

Match up real and complex parts and show that this occurs and you should be done.

3. ## Re: Rewrite a real linear transformation

So, $T(x+iy)=\alpha x+\beta y +i(\gamma x -\delta y)$ would have to equal $w_1(x+iy)+w_2(x-iy)$. We can rewrite the second equation as $(w_1+w_2)x+i(w_1-w_2)y$ now match up real and complex parts. First, $(w_1+w_2)x=\alpha x+\beta y$ and $(w_1-w_2)iy=\gamma x +\delta y$...

Do you mean to do this kind of thing?

4. ## Re: Rewrite a real linear transformation

You have to show that the complex parts cancel out for real linear if the vectors are all real numbers and the scalars are all real numbers.

If you can show that with your conjugate then you're done.

5. ## Re: Rewrite a real linear transformation

By the definition of a real linear transformation $T(z)=\alpha x + \beta y + i(\gamma x +\delta y)$. Now follow the hint and use $z=x+iy$ with $x=\frac{z+\bar{z}}{2}$, and $y=\frac{z-\bar{z}}{2i}$: \begin{equation*}\begin{split}T(z) &= \alpha \frac{z+\bar{z}}{2} +\beta \frac{z-\bar{z}}{2i} + i \Big(\gamma \frac{z+\bar{z}}{2} + \delta \frac{z-\bar{z}}{2i}\Big) \\ &= \alpha \frac{z+\bar{z}}{2} -i \beta \frac{z-\bar{z}}{2} + i \gamma \frac{z+\bar{z}}{2} + \delta \frac{z-\bar{z}}{2} \\ 2T(z)& = \alpha (z+\bar{z}) -i \beta( z-\bar{z}) + i \gamma (z+\bar{z}) + \delta (z-\bar{z}) \\ &= \alpha z +\alpha \bar{z} - i \beta z + i \beta \bar{z} +i\gamma z + i \gamma \bar{z} +\delta z - \delta \bar{z} \\ T(z) &= \frac{\alpha-i\beta-i\gamma+\delta}{2}z+\frac{\alpha +i\beta+i\gamma - \delta}{2} \bar{z}.\end{split}\end{equation*}

Set $$w_1=\frac{\alpha-i\beta-i\gamma+\delta}{2} \:\:\:\: \text{ and, } \:\:\:\: w_2 =\frac{\alpha +i\beta+i\gamma - \delta}{2}.$$

All that remains to be shown is the uniqueness of $w_1$, and $w_2$. Clearly if the real scalars $\alpha,\beta,\gamma,$ and $\delta$ were uniquely determined by $T$ then this would imply that $w_1$, and $w_2$ would be unique. But then this follows easily from the assumption that there exists $\epsilon,\zeta,\eta$ and $\theta$ such that $T \sim \begin{pmatrix}\epsilon & \zeta \\ \eta & \theta\end{pmatrix}$ since $$\begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}=\begin{pmatrix}\epsilon & \zeta \\ \eta & \theta\end{pmatrix} \iff \begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}- \begin{pmatrix}\epsilon & \zeta \\ \eta & \theta\end{pmatrix} = 0$$implies $\alpha = \epsilon$, $\beta = \zeta$, $\gamma = \eta$, and $\delta=\theta$.

6. ## Re: Rewrite a real linear transformation

$T=w_1 I+w_2 \bar{I}$ implies

$T(1)=w_1+w_2$ and $T(i)=w_1 i-w_2 i$

$w_1=\frac{1}{2} (T(1)-i\text{ }T(i))$ and

$w_2=\frac{1}{2} (T(1)+i\text{ }T(i))$

which proves uniqueness

To complete the proof

$T(z)=T(x+i y)= x T(1) + y T(i)$

$=x\left(w_1+w_2\right)+ y \left(w_1 i-w_2 i\right)$

$=w_1(x+i y)+w_2(x-i y)$

$=w_1I(z)+w_2\bar{I}(z)$