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[SOLVED] Hermitian Function Proof

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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
Feb 7, 2012
2,702
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
Feb 5, 2012
1,621
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. :p

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
Feb 7, 2012
2,702
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
Feb 5, 2012
1,621