2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

Hint: every vector $v \in V$ determines a linear transformation $V^* \to \mathbb{R}$ by the mapping $f \mapsto f(v)$ (called evaluation at $v$).

$\endgroup$
  • 1
    $\begingroup$ @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 $(e_i)''(e_j)=0$ $i\neq j$ $\endgroup$ – Arcane1729 Nov 14 '15 at 19:31
  • 1
    $\begingroup$ 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. $\endgroup$ – Arcane1729 Nov 14 '15 at 19:44
  • 1
    $\begingroup$ 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? $\endgroup$ – Arcane1729 Nov 14 '15 at 19:49
  • 1
    $\begingroup$ @Arcane1729 Looks right to me! The way to say it is that "a finite-dimensional vector space is naturally isomorphic to its double dual". The "naturally" means that it doesn't matter which basis you're working in. On the other hand, a finite-dimensional space is isomorphic to its dual space, but not naturally. $\endgroup$ – Eric Auld Nov 14 '15 at 21:57
  • $\begingroup$ I understand . :) I will now go and try the original problem. $\endgroup$ – Arcane1729 Nov 14 '15 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.