User:IssaRice/Linear algebra/Riesz representation theorem

Let's take the case where $V=\mathbf {R} ^{n}$ and the inner product is the usual dot product. What does the Riesz representation theorem say in this case? It says that if we have a linear functional $T:\mathbf {R} ^{n}\to \mathbf {R}$ then we can write $T$ as $Tv=v\cdot u$ for some vector $u\in \mathbf {R} ^{n}$. But we already know (from the correspondence between matrices and linear transformations) that we can represent $T$ as a 1-by-n matrix. And v can be thought of as a n-by-1 matrix. And now the dot product is the same thing as the matrix multiplication! In other words, the Riesz Representation Theorem (at least in this simple setting) is just saying that multiplying by a row vector is the same thing as dotting with that vector: $u^{T}v=v\cdot u$.

If $\sigma =(e_{1},\ldots ,e_{n})$ is the standard basis of $\mathbf {R} ^{n}$ and $(1)$ is the standard basis of $\mathbf {R}$, then $Tv=[Tv]^{(1)}=[T]_{\sigma }^{(1)}[v]^{\sigma }=[T]_{\sigma }^{(1)}\cdot [v]^{\sigma }=[v]^{\sigma }\cdot [T]_{\sigma }^{(1)}=v\cdot [T]_{\sigma }^{(1)}$.

So in the case $V=\mathbf {R} ^{n}$ we can understand the Riesz representation theorem as saying something we already knew. What Riesz representation theorem does is extend this same sort of "representability" to all finite-dimensional inner product spaces V and all linear functionals $T:V\to \mathbf {F}$.

this video talks about this for $\mathbf {R} ^{2}$