Question

# 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.

Alternate coordinate systems
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.

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.