To explain^ Why the beta-coordinate vectors of beta = {b_{1}, ... , b_{n}} are the columns e_{1}, ... , e_{n} of the n times n identity matrix.

Question
Alternate coordinate systems
asked 2021-02-19
To explain^ Why the beta-coordinate vectors of \(\beta = {b_{1}, ... , b_{n}}\)
are the columns \(e_{1}, ... , e_{n}\)
of the \(n \times n\) identity matrix.

Answers (1)

2021-02-20
Consider a vector x in V such that, \(x= c_{1} b_{1} + c_{2} b_{2} + ... + c_{n} b_{n}\)
The coordinates of x relative to basis \(\beta = {b_{1}, b_{2}, ... , b_{n}}\) ,
also called beta-coordinates of x are given by, \([x]_{\beta}=\begin{bmatrix}c_{1} \\...\\c_{n} \end{bmatrix}\)
Since \(\beta = {b_{1}, ..., b_{n}}\) from a basis for V.
Thus, the vectors \({b_{1}, ..., b_{n}} are linearly independent.
Therefore, if any vector b_k is to be written in terms of \(b_{1}, ..., b_{n}\)
That is, an arbitrary vector \(b_{k}\)
can be written as \(b_{k} = 0 \cdot b_{1}, ... + 1 \cdot b_{k} + ... + 0 \cdot b_{n}.\)
Here, k varies from 1 to n.
Thus, the beta-coordinates of \(b_{1}, ..., b_{n}\)
in this case are \(\left\{\begin{bmatrix}c_{1} \\...\\c_{n} \end{bmatrix}\right\}\)
The matrix formed by these beta-coordinates is \(\begin{bmatrix}1 & \cdots & 0 \\ \cdots & \cdots & \cdots \\ 0 & \cdots & 1 \end{bmatrix}\)
That is, beta-coordinate vectors of \(\beta = {b_{1}, ..., b_{n}}\)
are the columns \(e_{1}, ..., e_{n}\) of the
n \cdot n identity matrix.
0

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