# 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 $A=a_{1},a_{2},\ldots ,a_{n}$  and we would like to change it to the set $B=b_{1},b_{2},\ldots ,b_{n}$ . The basis B still belongs to vector space aforementioned, so its vectors can be expressed as a linear combination [Eq. 1]

$b_{1}=p_{11}a_{1}+p_{12}a_{2}+\ldots +p_{1n}a_{n}$ ,

$b_{2}=p_{21}a_{1}+p_{22}a_{2}+\ldots +p_{2n}a_{n}$ ,

$\ldots$

$b_{n}=p_{n1}a_{1}+p_{n2}a_{2}+\ldots +p_{nn}a_{n}$

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

${\begin{bmatrix}b_{1}\\\end{bmatrix}}_{A}={\begin{bmatrix}p_{11}\\p_{12}\\\vdots \\p_{1n}\\\end{bmatrix}}{\begin{bmatrix}b_{2}\\\end{bmatrix}}_{A}={\begin{bmatrix}p_{21}\\p_{22}\\\vdots \\p_{2n}\\\end{bmatrix}}\cdots {\begin{bmatrix}b_{n}\\\end{bmatrix}}_{A}={\begin{bmatrix}p_{n1}\\p_{n2}\\\vdots \\p_{nn}\\\end{bmatrix}}$

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

$P={\begin{bmatrix}{\begin{bmatrix}b_{1}\\\end{bmatrix}}_{A}{\begin{bmatrix}b_{2}\\\end{bmatrix}}_{A}\cdots {\begin{bmatrix}b_{n}\\\end{bmatrix}}_{A}\end{bmatrix}}={\begin{bmatrix}p_{11}&\cdots &p_{n1}\\p_{12}&\cdots &p_{n2}\\\vdots &\ddots &\vdots \\p_{1n}&\cdots &p_{nn}\\\end{bmatrix}}$

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

$P{\begin{bmatrix}\beta \\\end{bmatrix}}_{Z}={\begin{bmatrix}\beta \\\end{bmatrix}}_{A}$

#### Proof

$\beta =b_{1}z_{1}+b_{2}z_{2}+\cdots +b_{n}z_{n}$  $~~~=b_{1}(p_{11}a_{1}+\cdots +p_{1n}a_{n})+\cdots +b_{n}(p_{n1}a_{1}+\cdots +p_{nn}a_{n})$

$~=(b_{1}p_{11}+\cdots +b_{n}p_{n1})a_{1}+\cdots +(b_{1}p_{1n}+\cdots +b_{n}p_{nn})a_{n}$

$\therefore$

${\begin{bmatrix}\beta \\\end{bmatrix}}_{A}={\begin{bmatrix}(b_{1}p_{11}+\cdots +b_{n}p_{n1})\\\vdots \\(b_{1}p_{1n}+\cdots +b_{n}p_{nn})\\\end{bmatrix}}$

$={\begin{bmatrix}p_{11}&\cdots &p_{n1}\\\vdots &~&\vdots \\p_{1n}&\cdots &p_{nn}\\\end{bmatrix}}{\begin{bmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\\\end{bmatrix}}=P{\begin{bmatrix}\beta \\\end{bmatrix}}_{Z}$

$\Box$