Quick proof regarding inner products

[JG89]

Homework Statement

Let V be an inner product space. If <x,y> = <x,z> for all x in V, then prove that y = x.

Homework Equations

The Attempt at a Solution

Since V is a vector space, it follows that (y - z) is an element of V. Since <x,y> = <x,z> for ALL x, we put x = (y-z). Then we have <y - z, y> - <y - z, z> = 0. Since inner products are conjugate linear in the second component, we can write:

<y - z, y> - <y - z, z> = <y -z, y - z> = 0. I know that in general <w,w> = 0 if and only if w = 0. Thus we must have y - z = 0, implying that y = z.

Is this correct?

 

[bartek2009]

I think that you lose generality if you put x=(y-z).
Probably the proof applies then to that particular case.
Anyway, there is my proof:

1. y = z + w
2. (x, z + w) = (x, z)
3. (x, z) + (x, w) = (x, z)
4. (x, w) = 0

now, since x can be any vector, the difference between y and z,
namely w, must be a null vector and so y and z are equal.

P.S. In the question you are asked to prove x=y
but then you prove z=y so I guess there is a typo.

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[Dick]

Homework Statement

Let V be an inner product space. If <x,y> = <x,z> for all x in V, then prove that y = x.

Homework Equations

The Attempt at a Solution

Since V is a vector space, it follows that (y - z) is an element of V. Since <x,y> = <x,z> for ALL x, we put x = (y-z). Then we have <y - z, y> - <y - z, z> = 0. Since inner products are conjugate linear in the second component, we can write:

<y - z, y> - <y - z, z> = <y -z, y - z> = 0. I know that in general <w,w> = 0 if and only if w = 0. Thus we must have y - z = 0, implying that y = z.

Is this correct?

Your proof is just fine.

 

[JG89]

Thanks Dick.