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.

Theorem 1
''If P is the transition matrix from the basis A to the basis Z, and &beta; 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
$$ \beta = b_{1}z_{1} + b_{2}z_{2} + \cdots + b_{n}z_{n}$$ $$ = b_{1}(p_{11}a_{1} + \cdots + p_{1n}a_{n})+ \cdots + b_{n}(p_{n1}a_{1} + \cdots + p_{nn}a_{n})$$

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

$$\therefore $$

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

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

$$ \Box $$