Multilinear Algebra

The set of all linear maps from $V\to W$ where $V,W$ are vector spaces, is called $\text{Hom}(V,W)$. And is itself a vector space, where the zero function $\mathbf{O}(\mathbf{v})={\mathbf{0}}_{\mathbf{w}}$ for each $\mathbf{v}\in V$ acts as the identity vector, and the map $f+g$ is defined as the map that $x\mapsto f(x)+g(x)$ and the map $\lambda f$ is defined as the one $x\mapsto \lambda f(x)$.
Now, given $\text{Hom}(V_1,V_2)$ and $\text{Hom}(W_1,W_2)$ are vector spaces, we can have linear maps between them. This is the "multilinear nature"

if $V$ is a vector space over $\mathbb{R}$, then $\text{Hom}(V,\mathbb{R})=V^{\ast }$ is called the dual space of $V$. An element of $V^{\ast }$ is called a "co-vector".
For example if $P$ is the vector space of all polynomials with real entries and real coefficients, the definite integral of an element of $P$ from $0$ to $1$ is an operator call it $I$: $I(p)={\int }_0^{1}p(x)dx$, then $I$ is a linear map from $P$ to $\mathbb{R}$, hence $I\in P^{\ast }$, so $I$ is a co-vector of $P$.
In general if $V$ is a vector space over $F$, $\text{Hom}(V,F)$ is called the dual of $V$. An element of $V^{\ast }$ is often called a functional

if $V$ is a vector space (maybe of infinite dimension), then $B$ is a Hamel basis for $V$ if for all $\mathbf{v}\in V$, there exist a unique finite subset $\{v^1,v^2,\dots v^n\}$ of $B$, and $n$ non zero scalars $f_1,f_2,\dots f_n$ such that $\mathbf{v}=\sum _{i=1}^n{v}^i{f}_i$.

If $V$ is finite dimensional, with basis $<{\mathbf{e}}_{\mathbf{i}}{>}_{i=1}^n$, Then the $j$'th projection map $p_j:V\to F$ is the one that gives the scalar co-efficient of the $j$'th basis vector when the input vector is written as a linear combination of the basis vectors. that is $$ p_{j}(\lambda_{1}\mathbf{e_{1}}+ \lambda_{2}\mathbf{e_{2}} + \dots \lambda_{n}\mathbf{e_{n}}) = \lambda_{j}$$
If $V$ is infinite dimensional, with Hamel basis $B$, then $p_b:V\to F$ gives the scalar coefficient of the basis vector $b\in B$, when the input vector is written as a unique linear combination of a subset of $B$. that is,

In both cases, the co-ordinate projection maps are linear maps.
In the finite dimensional case $$p_{j}\left( a\sum_{i=1}^n\lambda_{i}\mathbf{e_{i}} +b\sum_{i=1}^n\mu_{i}\mathbf{e_{i}} \right) = a\lambda_{j} + b\mu_{j}$$ and for the infinite dimensional case, I wont write the case-wise breakdown, but okay. Moreover, in both cases the set of all co-ordinate projection maps (also called co-ordinate functionals) are linearly independent vectors in $V^{\ast }$, for the infinite dimension case, consider the functional which is a linear combination of co-ordinate maps being equal to the zero functional. $$ L =\sum_{b \in B}f_{b}p_{b} = \mathbf{0}$$
For any basis $b\in B$, $L(b)=0\in F$ (as L is the zero functional) but $L(b)=f_b$ Hence $f_b=0$, running through all $b\in B$, we notice that each $f_b=0\in F$. Hence $L$ is linearly independent
Similarly, for the finite dimensional case let $$ T = \sum_{i=1}^n\lambda_{i}p_{i} = \mathbf{0}$$, then $T({\mathbf{e}}_{\mathbf{j}})={\lambda }_j=0\in F$, hence running through all ${\mathbf{e}}_{\mathbf{i}}$,we have ${\lambda }_i=0\in F$ for each $i=1\to n$. so $<p_i{>}_{i=1}^n$ is a list of $n$ linearly independent vectors in $V^{\ast }$. let $f\in V^{\ast }$, for any $\mathbf{v}=\sum _{i=1}^n{\lambda }_i{\mathbf{e}}_{\mathbf{i}}\in V$, Then $f(\mathbf{v})=\sum _{i=1}^n{\lambda }_{i}f({\mathbf{e}}_{\mathbf{i}})$ That is $f$ is completely determined by what it does to each of the basis vectors.
Hence we see (beautifully!) that $f=\sum _{i=1}^{n}f({\mathbf{e}}_{\mathbf{i}})p_i$. Notice that $(\sum _{i=1}^{n}f({\mathbf{e}}_{\mathbf{i}})p_i)({\mu }_k{\mathbf{e}}_{\mathbf{k}})=\sum _{i=1}^n(f({\mathbf{e}}_{\mathbf{i}}){\mu }_k)\cdot (p_j({\mathbf{e}}_{\mathbf{k}}))=f({\mathbf{e}}_{\mathbf{k}}){\mu }_k=f({\mu }_k{\mathbf{e}}_{\mathbf{k}})$.
So the co-ordinate projection functionals are a basis for $V^{\ast }$ when $V$ is finite dimensional.
Moreover, even if $V$ is infinite dimensional, for any $\mathbf{v}\ne \mathbf{0}\in V$, there exists a functional $g$ for which $g(\mathbf{v})\ne 0$. Notice $\mathbf{v}=\sum _{i=1}^n{v}^i{f}_i$. then simple set $g=p_{v^k}$ then $g(\mathbf{v})=f_k\ne 0\in F$
Also, $dim(V)=dim(V^{\ast })$