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 basis 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 $[I_{V^{\prime }}{]}_{({\phi }_1,{\phi }_2)}^{({\psi }_1,{\psi }_2)}=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)$. This matrix is the transpose of $[I_V{]}_{{\beta }^{\prime }}^{\beta }$. This is not a coincidence.

In Tao's yards and feet example on pages 224-225 of https://terrytao.files.wordpress.com/2011/06/blog-book.pdf we writes that

$1\text{ yard}=3\text{ feet}$

$\text{yards}(L)=\text{feet}(L)/3$

The more general statement of this is that $[I_{V^{\prime }}{]}_{({\phi }_1,{\phi }_2)}^{({\psi }_1,{\psi }_2)}=(([I_V{]}_{\beta }^{{\beta }^{\prime }}{)}^{-1}{)}^{\mathrm{\top }}$. In the one-dimensional case, the transpose does not change the matrix, so we just invert it, going from $(3)$ to $(1/3)$.

Why did I go through all of this? When reading Tao's "Coordinates" entry, the yards/feet example confused me: I knew that the matrix of the dual transformation is just the transpose, so in the one-dimensional case it shouldn't change anything. And yet, we had to take the reciprocal! The way out of this is to realize that in going to the dual space and taking the transpose, we also swap the order of the bases. So Tao could have written the two equations as follows:

$1\text{ yard}=3\text{ feet}$

$\text{feet}(L)=3\times \text{yards}(L)$

Now the matrices are indeed just transposes of each other!

To see what was really going on, I had to first escape the too-simple one-dimensional case, so I went to the two-dimensional example that I remembered from Tao's linear algebra notes.

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https://www.abstractmath.org/MM/MMUsefulBehaviors.htm#selfmonitoring.htm

The students and professors example is like the yards and feet example that tao gives in the coordinates entry of "compactness and contradiction".