Let F be a linear function over the real field such that F(1,0,0)=2 , F(0,1,0)=1 and F(0,0,1)=-1, find an orthogonal set to ker(F) and us it to calculate the Riesz representative of F
Is it necessary to find the general form of F? I thought it may be F(x,y,z)=2x+y-z but i am not sure it is correct, and even so i do not know how to proceed.
Here is the link to where te problem came from, it´s problem 1.2.4
The Riesz Representation Theorem for finite-dim vector spaces is the following:
Denote $V^*$ the vector space of real linear functionals on the real vector space $V$, a.k.a its dual space.
Let $e_1, \dots, e_n$ denote an orthonormal basis for $V$, Then $\forall v \in V, f(v) = f(<v, e_1>e_1 + \dots + <v, e_n>e_n)$ where $<•,•>$ is an inner product on $V$ under which the mentioned basis we have for $V$ is orthonormal.
Continuing : $ f(v) = <v, e_1>f(e_1) + \dots + <v, e_n>f(e_n)$
$= <v, f(e_1)e_1> + \dots + <v,f(e_n)e_n>$ $ = <v, f(e_1)e_1 + \dots + f(e_n)e_n>$ $= <v, u' > $ where $u' \in V$.
(I am ignoring complex conjugage since we are in a real vector space).
Thus there is exists a unique vector, $u' \in V$, (by definition of a basis) such that $\forall v \in V, f(v) = <v ,u '>$.
In your case, let's find $ker(f)$, By linearity we can see $f((1,0,0) - 2(0, 1,0)) = 0$ $f((0,1, 0) + (0,0,1)) = 0$
We can see from this that $ker(f) = span({(1, -2, 0), (0,1, 1)})$, by the rank-nullity theorem we know $dim(ker(f))=2$. But, as in the Riesz representation theorem we are looking for an orthonormal basis of $V$. So replace our basis for $ker(f)$ with an orthonormal one, call the vectors $u, z$ respectively, and $span({u,z}) = ker(f)$
Let's say we have some vector orthogonal to all vectors in $ker(f)$, call it $w$ with norm 1. Then $w, u, z$ must be an orthonormal basis for $V$.
Can you continue from here? Fill in the vectors $w, u, z$ and see why $f(v) = <v ,f(w)w>, \forall v \in V$.