[SOLVED] Hermitian Function Proof

[Sudharaka]

Hi everyone,

Here's a problem I encountered. I think there's a mistake in this problem.

Problem:

Let \(f:\,V\times V\rightarrow\mathbb{C}\) be a Hermitian function (a Bilinear Hermitian map), \(q:\, V\rightarrow\mathbb{C}\) be given by \(q(v)=f(v,\,v)\). Prove that following,

\[4f(x,\,y)=q(x+y)-q(x-y)+iq(x+iy)-iq(x-iy)\]

When I find the answer I get,

\[q(x+y)-q(x-y)+iq(x+iy)-iq(x-iy)=0\]

I would be really grateful if someone can confirm my answer.

 

[Fantini]

Sudharaka, here's what I tried:

$$
q(x+y) - q(x-y) +i q(x+iy) -i q(x-iy)
$$

$$
= f(x+y,x+y) - f(x-y,x-y) +if(x+iy, x+iy) - if(x-iy, x-iy)
$$

$$
= f(x,x) + f(x,y) + f(y,x) + f(y,y) - (f(x,x) - f(y,x) - f(x,y ) +f(y,y)) +i(f(x,x) + if(y,x) + if(x,y) + i^2 f(y,y)) -i(f(x,x) -if(y,x) -if(x,y) -i^2f(y,y))
$$

$$
= f(x,y) + f(y,x) + f(x,y) + f(y,x) - f(x,y) - f(y,x) - f(x,y) - f(y,x) = 0.
$$

I searched a bit in Wikipedia and found Sesquilinear forms. If we use this instead, meaning that the function $f$ is antilinear in the second coordinate, we'd get this on the third line instead:

$$
= f(x,x) + f(x,y) + f(y,x) + f(y,y) - (f(x,x) - f(y,x) - f(x,y) +f(y,y)) +i(f(x,x) + if(y,x) - if(x,y) - i^2 f(y,y)) -i(f(x,x) -if(y,x) +if(x,y) -i^2f(y,y))
$$

$$=f(x,y) +f(y,x) +f(y,x) +f(x,y) -f(y,x) +f(x,y) -f(y,x) +f(x,y) = 4f(x,y).$$

Just as asked. Possibly what he meant as Hermitian function is really the Sesquilinear?

 

[Opalg]

Possibly what he meant as Hermitian function is really the Sesquilinear?

In the context of a vector space over the complex numbers, Hermitian always means sesquilinear.

 

[Sudharaka]

Sudharaka, here's what I tried:

$$
q(x+y) - q(x-y) +i q(x+iy) -i q(x-iy)
$$

$$
= f(x+y,x+y) - f(x-y,x-y) +if(x+iy, x+iy) - if(x-iy, x-iy)
$$

$$
= f(x,x) + f(x,y) + f(y,x) + f(y,y) - (f(x,x) - f(y,x) - f(x,y ) +f(y,y)) +i(f(x,x) + if(y,x) + if(x,y) + i^2 f(y,y)) -i(f(x,x) -if(y,x) -if(x,y) -i^2f(y,y))
$$

$$
= f(x,y) + f(y,x) + f(x,y) + f(y,x) - f(x,y) - f(y,x) - f(x,y) - f(y,x) = 0.
$$

I searched a bit in Wikipedia and found Sesquilinear forms. If we use this instead, meaning that the function $f$ is antilinear in the second coordinate, we'd get this on the third line instead:

$$
= f(x,x) + f(x,y) + f(y,x) + f(y,y) - (f(x,x) - f(y,x) - f(x,y) +f(y,y)) +i(f(x,x) + if(y,x) - if(x,y) - i^2 f(y,y)) -i(f(x,x) -if(y,x) +if(x,y) -i^2f(y,y))
$$

$$=f(x,y) +f(y,x) +f(y,x) +f(x,y) -f(y,x) +f(x,y) -f(y,x) +f(x,y) = 4f(x,y).$$

Just as asked. Possibly what he meant as Hermitian function is really the Sesquilinear?

In the context of a vector space over the complex numbers, Hermitian always means sesquilinear.

Thanks very much for the reply. I am learning more from MHB than I learn in class.

So to confirm, in the case of bilinear maps over complex fields, Hermitian is the same as sesquilinear with the additional constraint, \(f(v,\,u)=\overline{f(u,\,v)}\) for all \(u,\,v\in V\). Am I correct?

 

[Opalg]

So to confirm, in the case of bilinear maps over complex fields, Hermitian is the same as sesquilinear with the additional constraint, \(f(v,\,u)=\overline{f(u,\,v)}\) for all \(u,\,v\in V\). Am I correct?

Yes.

 

[Sudharaka]

Yes.

Thanks very much, now I understand this perfectly.