# proving that change of basis matrix is invertible.

• Apr 5th 2013, 06:58 AM
Stormey
proving that change of basis matrix is invertible.
Hi guys.

Let say $\displaystyle \left [ I \right ]_{B}^{C}$ is the change of basis matrix from the base B to base C.
I need to prove that $\displaystyle \left [ I \right ]_{B}^{C}$ is invertible.

So that's what i did so far:
I defined:
$\displaystyle B=\left \{b_1, b_2,... b_n \right \}$
$\displaystyle C=\left \{c_1, c_2,... c_n \right \}$

Now I tried to define the $\displaystyle \left [ I \right ]_{B}^{C}$ matrix (and to show its columns are linear independent), but this is where it got a little complicated for me:
since $\displaystyle c_i\in V$ and B is basis for V, every vector $\displaystyle c_i\in C$ can be represented as a linear combination of B's vectors:

$\displaystyle c_1=\beta_{11} b_1+\beta_{12} b_2+...+\beta_{1n} b_n$
$\displaystyle c_2=\beta_{21} b_1+\beta_{22} b_2+...+\beta_{2n} b_n$
.
.
.
$\displaystyle c_n=\beta_{n1} b_1+\beta_{n2} b_2+...+\beta_{nn} b_n$

and then, these equation can be written as:

$\displaystyle \begin{bmatrix}c_1\\c_2\\.\\.\\.\\c_n\end{bmatrix} =\begin{bmatrix}\beta_{11} & \beta_{12} & . & . & . & \beta_{1n}\\\beta_{21} & \beta_{22} & . & . & . & \beta_{2n}\\. & . & & & & .\\. & . & & & & .\\. & . & & & & .\\\beta_{n1} & \beta_{n2} & . & . & . & \beta_{nn}\end{bmatrix}\begin{bmatrix}b_1\\ b_2\\ .\\ .\\ .\\ b_n\end{bmatrix}$

Now, where is this change of basis matrix?
Is it the matrix with the $\displaystyle \beta_{ij}$ entries?
Am I even writing it correctly?