User:IssaRice/Linear algebra/Dual basis: Difference between revisions

Revision as of 05:26, 30 July 2019

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 $β : = (C O_{2}, C O)$ and $β^{'} : = (C, O)$ .

The change of coordinate matrix, from $β$ to $β^{'}$ , is then $[I_{V}]_{β}^{β^{'}} = (\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix})$ .

The change of coordinate matrix, from $β^{'}$ to $β$ , is the inverse of $[I_{V}]_{β}^{β^{'}}$ , and we have $[I_{V}]_{β^{'}}^{β} = ([I_{V}]_{β}^{β^{'}})^{- 1} = (\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix})$ .

@@ Line 5: / Line 5: @@
 Let <math>V</math> be the space of all mixtures of CO2 and CO, and let <math>\beta := (CO_2, CO)</math> and <math>\beta' := (C,O)</math>.
-The change of coordinate matrix, from <math>\beta</math> to <math>\beta'</math>, is then <math>[I_V]_\beta^{\beta'} = \begin{pmatrix}1 & 1 \\ 2 & 2\end{pmatrix}</math>.
+The change of coordinate matrix, from <math>\beta</math> to <math>\beta'</math>, is then <math>[I_V]_\beta^{\beta'} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}</math>.
 The change of coordinate matrix, from <math>\beta'</math> to <math>\beta</math>, is the inverse of <math>[I_V]_\beta^{\beta'}</math>, and we have <math>[I_V]_{\beta'}^\beta = ([I_V]_\beta^{\beta'})^{-1} = \begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}</math>.