User:IssaRice/Linear algebra/Dual basis

see p. 224 of https://terrytao.files.wordpress.com/2011/06/blog-book.pdf

the following example is based on p. 115 of https://terrytao.files.wordpress.com/2016/12/linear-algebra-notes.pdf

Let $V$ be the space of all mixtures of CO2 and CO, and let $\beta :=(C{O}_2,CO)$ and ${\beta }^{\prime }:=(C,O)$.

The change of coordinate matrix, from $\beta$ to ${\beta }^{\prime }$, is then $[I_V{]}_{\beta }^{{\beta }^{\prime }}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)$.

The change of coordinate matrix, from ${\beta }^{\prime }$ to $\beta$, is the inverse of $[I_V{]}_{\beta }^{{\beta }^{\prime }}$, and we have $[I_V{]}_{{\beta }^{\prime }}^{\beta }=([I_V{]}_{\beta }^{{\beta }^{\prime }}{)}^{-1}=\left(\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right)$.

Now let us extend this example to discuss dual spaces.

Given some mixture (i.e. linear combination) $a\times C{O}_2+b\times CO$, the dual basis of $\beta$ consists of two linear functionals $({\phi }_1,{\phi }_2)$ such that

${\phi }_1(a\times C{O}_2+b\times CO)=a$

${\phi }_2(a\times C{O}_2+b\times CO)=b$

Similarly, given some mixture $c\times C+d\times O$, the dual space of ${\beta }^{\prime }$ consists of two linear functionals $({\psi }_1,{\psi }_2)$ such that

${\psi }_1(c\times C+d\times O)=c$

${\psi }_2(c\times C+d\times O)=d$

Now we can ask, given ${\phi }_1$, how can we write it in terms of ${\psi }_1,{\psi }_2$?

Given the mixture $a\times C{O}_2+b\times CO$, we can write this as $(a+b)\times C+(2a+b)\times O$. Thus, ${\psi }_1(a\times C{O}_2+b\times CO)=a+b$ and ${\psi }_2(a\times C{O}_2+b\times CO)=2a+b$. In other words, our task is to express $a$ in terms of $a+b$ and $2a+b$. We get $(2a+b)-(a+b)=a$, so ${\phi }_1=-{\psi }_1+{\psi }_2$.

Similarly, we get ${\phi }_2=2{\psi }_1-1{\psi }_2$.

Thus, the matrix is $[I_{V^{\prime }}{]}_{({\phi }_1,{\phi }_2)}^{({\psi }_1,{\psi }_2)}=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)$.