Inner Product Space and Linear Transformation

Let V be an inner product space, and let w be a fixed vector in V.
Let L:V -> R be defined by L(v) = (v,w) for v in V. Show that L is a linear transformation.

I'm just confused on how I would show this. I attempted to try it and I was hoping for some feedback, and suggestions. Thanks a lot guys!

I know that to be a linear transformation, then
L(u + v) = L(u) + L(v)
L(cu) = cL(u)

And since V is an inner product space that means
(u,u) > 0
(v,u) = (u,v)
(u+v,w) = (u,w) + (v,w)
(cu,v) = c(u,v)

So I have a vector v = [a1...an] and w = [b1...bn]
If L(v)= (v,w) then it = a1b1 + .... + anbn
L(w) = (w,v) (I assume) = b1a1 + .... + bnan

Then to figure out L(v+w) = L(v) + L(w) wouldn't be too hard. I would just have to separate the 2 equations so that it would be L(v) and L(w).

The second part, I would just multiply some c into L(v) which would give me
L(cv) = [ca1b1 + ca2b2 +...+canbn] = c[a1b1 + ... + anbn] = cL(v)

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Originally Posted by Brokescholar

Let V be an inner product space, and let w be a fixed vector in V.
Let L:V -> R be defined by L(v) = (v,w) for v in V. Show that L is a linear transformation.

I'm just confused on how I would show this. I attempted to try it and I was hoping for some feedback, and suggestions. Thanks a lot guys!

I know that to be a linear transformation, then
L(u + v) = L(u) + L(v)
L(cu) = cL(u)

And since V is an inner product space that means
(u,u) > 0
(v,u) = (u,v)
(u+v,w) = (u,w) + (v,w)
(cu,v) = c(u,v)

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You already have it. L(u+v)= (u+v,w)= (u,w)+ (v,w)= L(u)+ L(v)
L(cu,v)= (cu, w)= c(u,w)= cL(u).

There is no need for any of the rest of this:

Quote:

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So I have a vector v = [a1...an] and w = [b1...bn]
If L(v)= (v,w) then it = a1b1 + .... + anbn
L(w) = (w,v) (I assume) = b1a1 + .... + bnan

Then to figure out L(v+w) = L(v) + L(w) wouldn't be too hard. I would just have to separate the 2 equations so that it would be L(v) and L(w).

The second part, I would just multiply some c into L(v) which would give me
L(cv) = [ca1b1 + ca2b2 +...+canbn] = c[a1b1 + ... + anbn] = cL(v)

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