Question

Someone explains me, please Every vector on C has the form Aab=az+bz¯ for some a,b in C Let u= (u1,u2), v =(v1,v2) (u, v)=((u1,u2),(v1,v2)) = u1v1+u2v2 I got it. What I didn't get is (u,v) = Re(u,v¯)=Re(u¯,v) Give me example, please

Answer 1

If I understand correctly, you're asking why
\[ (u,v) = \mathcal{Re}(u,\bar v) = \mathcal{Re}(\bar u,v) \]
Where \( (u,v) \) is the inner product of u and v, and the bar denotes complex conjugation. Is that right?

Answer 2

Yes, sir

Answer 3

Well then right off the bat, it doesn't make sense. \( (u,\bar v) \) is already real, so I suspect you really mean to say something else.

Answer 4

Nope, I don't have anything else. The lecture note start the logic by that equation with the note that Re(z) = Re (z bar)
However, I don't get it (the very first sentence of the logic)
If I accept it without knowing what it is, the rest of the logic is OK to me

Answer 5

Oh. Well that is true by definition. For the complex conjugate, you just flip the sign of the imaginary part - the real part stays the same.

Answer 6

Please, give me an example

Answer 7

If \( z = 3 + 4i\), then \( \bar z = 3 - 4i \).

Answer 8

yes

Answer 9

Is that not what you're asking?

Answer 10

the part above (u,v) = Re(u, v bar)= Re (u bar, v)

Answer 11

Again - that doesn't make sense. (u, v bar) and (u bar, v) are already real, under the definition you've given above.

Answer 12

Can you unfold that corner?

Answer 13

this part is next

Answer 14

I have to copy down and follow the prof's lecture during 2hours. Sometimes, I just copy without understanding it. He goes so fast.

Answer 15

Well,
\[ \mathcal{Re}(u \bar v) = \mathcal{Re}(\bar u v) = (u,v)\]

Perhaps that's what you meant?

Answer 16

oooh, I got it. hahaha... you see, (u, v) = u1v1 + u2v2
that's all real
let's take \(u = \left[\begin{matrix}u_1\\iu_2\end{matrix}\right]\)
and \(v = \left[\begin{matrix}v_1\\iv_2\end{matrix}\right]\) so, \(\bar u = \left[\begin{matrix}u_1\\-iu_2\end{matrix}\right]\) and \(\bar v = \left[\begin{matrix}v_1\\-iv_2\end{matrix}\right]\)
now, take \(Re(u,\bar v)= u_1v_1 +u_2v_2\) Am I right?

Answer 17

There should not be any i's in the column vectors.