Linear Algebra/OLD/Change of Basis

Change of Basis

It was shown earlier that a square matrix can represent a linear transformation of a vector space into itself, and that this matrix is dependent on the basis chosen for the vector space. We will now show how to change the basis of a given vector space

Suppose we have some vector space whose basis is given by the set $A=a_1,a_2,\dots ,a_n$ and we would like to change it to the set $B=b_1,b_2,\dots ,b_n$. The basis B still belongs to vector space aforementioned, so its vectors can be expressed as a linear combination [Eq. 1]

$b_1=p_{11}{a}_1+p_{12}{a}_2+\dots +p_{1n}{a}_n$,

$b_2=p_{21}{a}_1+p_{22}{a}_2+\dots +p_{2n}{a}_n$,

$\dots$

$b_n=p_{n1}{a}_1+p_{n2}{a}_2+\dots +p_{nn}{a}_n$

Each vector of the set B has a coordinate matrix with respect to the basis we started off with, namely the set designated as A. We represent this as

${\left[\begin{array}{c}{b}_1\end{array}\right]}_A=\left[\begin{array}{c}{p}_{11}\\ p_{12}\\ ⋮\\ p_{1n}\end{array}\right]{\left[\begin{array}{c}{b}_2\end{array}\right]}_A=\left[\begin{array}{c}{p}_{21}\\ p_{22}\\ ⋮\\ p_{2n}\end{array}\right]\cdots {\left[\begin{array}{c}{b}_n\end{array}\right]}_A=\left[\begin{array}{c}{p}_{n1}\\ p_{n2}\\ ⋮\\ p_{nn}\end{array}\right]$

Setting these coordinate matrices as the columns of a matrix P gives us a transition matrix. This transition matrix transforms the original basis A to a new basis B of some vector space. The transition matrix is actually the transpose of [Eq. 1] that we saw earlier

$P=\left[\begin{array}{c}{\left[\begin{array}{c}{b}_1\end{array}\right]}_A{\left[\begin{array}{c}{b}_2\end{array}\right]}_A\cdots {\left[\begin{array}{c}{b}_n\end{array}\right]}_A\end{array}\right]=\left[\begin{array}{ccc}{p}_{11}& \cdots & p_{n1}\\ p_{12}& \cdots & p_{n2}\\ ⋮& \ddots & ⋮\\ p_{1n}& \cdots & p_{nn}\end{array}\right]$

To summarize, in order to find the transition matrix from some basis F to some basis G, we must compute the coordinate vector for each element of our original basis F with respect to the other basis G. The matrix whose columns are formed by the coordinate vectors is the transition matrix.

Proof

Theorem 1

If P is the transition matrix from the basis A to the basis Z, and β is an element of the vector space, then it follows that

$P{\left[\begin{array}{c}\beta \end{array}\right]}_Z={\left[\begin{array}{c}\beta \end{array}\right]}_A$

Proof

$\beta =b_1{z}_1+b_2{z}_2+\cdots +b_n{z}_n$ $=b_1(p_{11}{a}_1+\cdots +p_{1n}{a}_n)+\cdots +b_n(p_{n1}{a}_1+\cdots +p_{nn}{a}_n)$

$=(b_1{p}_{11}+\cdots +b_n{p}_{n1})a_1+\cdots +(b_1{p}_{1n}+\cdots +b_n{p}_{nn})a_n$

$\therefore$

${\left[\begin{array}{c}\beta \end{array}\right]}_A=\left[\begin{array}{c}(b_1{p}_{11}+\cdots +b_n{p}_{n1})\\ ⋮\\ (b_1{p}_{1n}+\cdots +b_n{p}_{nn})\end{array}\right]$

$=\left[\begin{array}{ccc}{p}_{11}& \cdots & p_{n1}\\ ⋮& & ⋮\\ p_{1n}& \cdots & p_{nn}\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right]=P{\left[\begin{array}{c}\beta \end{array}\right]}_Z$

$◻$