# [SOLVED]Hermitian Function Proof

#### Sudharaka

##### Well-known member
MHB Math Helper
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

MHB Math Helper
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

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
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

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
Yes. Thanks very much, now I understand this perfectly. 