# Double Dual Spaces linear maps of linear maps

Hi I am stuck on a problem about dual spaces which I've spent hours on but I just cant grasp the idea of functions of functions- The problem in mind uses the vector space $V$ of polynomials of degree $\leq 2$ over the field $\mathbb{R}$ and I have been given $3$ linear maps from $V$ to $\mathbb{R}$- e.g a map $\phi_1$ which integrates the polynomial between two limits and so outputs a real number. So these maps are my basis for $V'$ the dual space of $V$. The question requires me to find the basis of $V''$. I could get nowhere with this after hours. So I thought I'll go back and find the basis of $V$ that this dual basis of $V'$ $\{\phi_1,\phi_2,\phi_3\}$ corrsponds to. So I got three linearly independent polynomials to be the basis of $V$. (They must be unique right?) And that's as far as I got.

So I didnt really want to spoil this problem- and thought I'd try understand this for an easier one and try again later. If we now take $V=\mathbb{R^3}$ with basis vectors $e_1=\begin{pmatrix} 1 \\0 \\ 0 \end{pmatrix}$, $e_2=\begin{pmatrix} 0 \\ 1 \\0 \end{pmatrix}$, $e_3=\begin{pmatrix} 0 \\0 \\1 \end{pmatrix}$- the standard basis. The dual basis vectors are $(e_i)'$ where $(e_i)'(e_i)=0$ and $(e_i)'(e_j)=0$ for $i\neq j$ So we have The dual basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$ which are elements of $(\mathbb{R^3})^T$ How would we find the the basis corresponding to this for $V''$ - without using any extra theorems or artillery just directly from this problem?

Hint: every vector $v \in V$ determines a linear transformation $V^* \to \mathbb{R}$ by the mapping $f \mapsto f(v)$ (called evaluation at $v$).
• @So I can pick a single vector $v \in R^3$ say $\begin {pmatrix} 1 \\ 1\\ 0 \end {pmatrix}$ and pick any $f \in \mathbb{R^3}^*$ say $(1,3,1)$ and get the evaulation at $v =\begin {pmatrix} 1 \\ 1\\ 0 \end {pmatrix}$ , $f(v)= 4$ but we can find the dual basis for the dual basis $\mathbb{R^3}^*$ in the same way as we found it for for $\mathbb{R^3}$- by seeing what linear maps $(e_i)''$ send the linear maps $\{(1,0,0),(0,1,0),(0,0,1)\}$ we obtained to $0$'s and $1$'s i.e we need the double dual basis $\{(e_1)'',(e_2)'',(e_3)''\}$ s.t $(e_i)''(e_i)=1$ and $(e_i)''(e_j)=0$ $i\neq j$ – Arcane1729 Nov 14 '15 at 19:31
• But if$e_i$ is sent to the value which it sends stuff to- then the basis we need is precisely the one of V that we started with? E.g need to pick v= $\begin {pmatrix} 1 \\ 0\\0 \end {pmatrix}$ for $(e_1)'= (0,0,1)$ since that's what is sent to $1$ by $(e_1)'$ and similarly $\begin {pmatrix} 0 \\ 1\\0 \end {pmatrix}$ and $\begin {pmatrix} 0 \\ 0\\1 \end {pmatrix}$ are sent to $0$ by $(e_1)'$. It works out for $(e_2)'$ and $(e_3)'$. So the $v$'s in $V$ that determine $V^* \rightarrow \mathbb{R}$ are precisely the basis of $\mathbb{R^3}$ we started with. – Arcane1729 Nov 14 '15 at 19:44
• But what are the linear maps of $V^{**}$ that I need? $e_1, e_2,e_3$ the standard basis of $\mathbb{R^3}$? Except you need right multiplication instead of left multiplication to get a real number and not a $3$x$3$ matrix? Is this all correct? – Arcane1729 Nov 14 '15 at 19:49