Vtyjkyuk

If someone did not study functional analysis but just studied linear algebra, how to let them understand the idea of Riesz representation theorem for finite-dimensional vector spaces?

The Riesz representation theorem Wikipedia

Let $H$ be a Hilbert space, and let $H^*$ denote its dual space, consisting of all continuous linear functionals from $H$ into the field $mathbbR$ or $mathbbC$. If $x$ is an element of $H$, then the function $varphi_x$, for all $y$ in $H$ defined by:
beginalign*
varphi_x (y) = langle y,x rangle
endalign*
where $langle cdot,cdot rangle$ denotes the inner product of the Hilbert space, is an element of $H^*$. The Riesz representation theorem states that every element of $H^*$ can be written uniquely in this form.

This description is abstract to me. Since linear algebra is sort of the reduced functional analysis, at the very first step, I am thinking to understand the reduced Riesz representation theorem applied to linear algebra.

In linear algebra, we intend to solve the problem of a linear system
beginalign*
A x = b
endalign*
where $A in mathbbR^m times mathbbR^n$ is an $m$ by $n$ matrix, $x in mathbbR^n$ is an $n$ by $1$ column vector and $b in mathbbR^m$ is an $m$ by $1$ column vector. The matrix $A$ transforms vectors in $mathbbR^n$ to vectors in $mathbbR^m$, thus we say $A: mathbbR^n to mathbbR^m$. But the vector $b$ is actually in the column space of $A$, say $C(A) = mathbbR^r subset mathbbR^m$, which has dimension $r$ that denotes the rank of $A$. Thus we can say $A: mathbbR^n to mathbbR^r$. If we have an $m$ by $1$ column vector $y$, then we can write
beginalign
y^T A x = y^T b
endalign
We can rewrite it in the form of inner product
beginalign
langle y,Ax rangle = langle y,b rangle
endalign
And if we consider $b$ as a functional in the dual space of $mathbbR^r$, denoted by $varphi_Ax(cdot) := langle cdot,b rangle$, then
beginalign
varphi_Ax (y) = langle y, A x rangle
endalign
Note that the mapping between $b$ and $varphi_Ax$ is one-to-one. We say, every $b$ in $mathbbR^r$ can be written uniquely in this form. It is very close to the equation in the Riesz representation theorem, but it seems we have to use $Ax$ instead of $x$, unless $A=I$ and $m=n=r$?

I am trying to state the reduced version of the Riesz representation theorem in linear algebra, as follows:

$mathbbR^r$ is a Hilbert space, and its dual space $(mathbbR^r)^*=mathbbR^r$, consisting of all continuous linear functionals from $mathbbR^r$ into the field $mathbbR$. If $Ax$ is an element of $mathbbR^r$, then the function $varphi_Ax$, for all $y$ in $mathbbR^r$ defined by:
beginalign*
varphi_Ax (y) = langle y,Ax rangle
endalign*
where $langle cdot,cdot rangle$ denotes the inner product of the Hilbert space, is an element of $(mathbbR^r)^*$. The Riesz representation theorem states that every element of $(mathbbR^r)^*$ can be written uniquely in this form. That is, every vector $b$ in $mathbbR^r$ can be represented by $langle y,Ax rangle$.

This looks like a connection to the "weak formulation" of $Ax = b$, namely, we can find the solution $x in mathbbR^n$ of $Ax = b$, if for every "test" vector $y in mathbbR^m$ there holds $varphi_Ax (y) = langle y,Ax rangle$.

I am still not fully understand the theorem at this moment, so there might be something wrong stated above. Any comments? Could you provide me with a more clear structure of the reduced Riesz representation theorem in linear algebra?

In addition, the proof of the Riesz representation theorem in textbooks usually take with a nullspace of $varphi$ denoted by $mathrmker(varphi)$ and its orthogonal space $mathrmker(varphi)^perp$. In linear algebra, we know the row space of a matrix is always orthogonal to its nullspace. Is there any connection between these two ideas? In other words, can we prove the reduced Riesz representation theorem for finite-dimensional vector spaces with using only the concepts in linear algebra?

Here is a formulation of Riesz representation theorem from mine lecture notes which you might find helpfull.

Let $V,W$ be vector spaces over a field $K$ with $dim(V), dim(W)< infty$ and $f:Vtimes W rightarrow K$ be a nondegenerate bilinear form. Then for every $pi in W^*$there exists $v in V$ such that $$pi=f(v,cdot)$$

Every non-trivial linear functional $Phi$ on a vector space $V$ is characterized by its null space $N=mathcalN(Phi)$, which is always of co-dimension one in $V$. This is because $Phi(v)=1$ for some $v$, which allows you to write
$$
w = (w-Phi(w)v)+Phi(w)v,
$$
and $w-Phi(w)vin N$ is easily directly verified. So $V=Noplus[v]$. Conversely, every such decomposition defines a unique linear functional $Phi$ such that $N=mathcalN(Phi)$ and $Phi(v)=1$. To see that $Phi$ is unique, suppose $Psi$ is another such linear function. Then $Phi(w-Phi(w)v)=0$ implies $Psi(w-Phi(w)v)=0$ or $Psi(w)=Psi(v)Phi(w)=Phi(w)$ for all $w$. This is true for finite- or infinite-dimensional real or complex linear spaces.

If $Phi$ is a non-zero continuous linear functional on a Hilbert space $V$, then $N=mathcalN(Phi)=Phi^-10$ is closed and, therefore, there is a unique vector $v$ such that $v perp N$. So $V=Noplus [v]$, where the decomposition is orthogonal. And we can assume $Phi(v)=1$ by multiplying $v$ by an appropriate scalar. The linear functional $Psi(w)=fraclangle w,vranglelangle v,vrangle$ has the same null space as $Phi$ and $Psi(v)=Phi(v)=1$. Therefore $Psi=Phi$.

A Hilbert space is needed in infinite dimensions in order to come up with a vector that is orthogonal to the null space of a given continuous linear functional. Continuity ensures that the null space is closed, so that this may be done. In finite-dimensional spaces, only Gram-Schmidt is needed.