Equivalence of Definitions of Change of Basis Matrix

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $R$ be a ring with unity.

Let $G$ be a finite-dimensional unitary $R$-module.

Let $A = \sequence {a_n}$ and $B = \sequence {b_n}$ be ordered bases of $G$.


The following definitions of the concept of Change of Basis Matrix are equivalent:

Definition 1

The matrix of change of basis from $A$ to $B$ is the matrix whose columns are the coordinate vectors of the elements of the new basis $\sequence {b_n}$ relative to the original basis $\sequence {a_n}$.

Definition 2

Let $I_G$ be the identity linear operator on $G$.

Let $\sqbrk {I_G; \sequence {a_n}, \sequence {b_n} }$ be the matrix of $I_G$ relative to $\sequence {b_n}$ and $\sequence {a_n}$.


Then $\sqbrk {I_G; \sequence {a_n}, \sequence {b_n} }$ is called the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence {b_n}$.


Proof

It will be shown that the two matrices defined are equal column-wise.


Let $\ds b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$ for $i$ ranging from $1$ to $n$, where $c_{i j}$'s are scalars.

The uniqueness of the above expression is justified by Expression of Vector as Linear Combination from Basis is Unique.


Then by definition of coordinate vector, the $i$th column of the matrix defined in definition 1 is:

$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{in} \end {bmatrix}^\intercal$


We also have:

$\ds \map {I_G} {b_i} = b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$



So by definition of relative matrix, the $i$th column of the matrix defined in definition 2 is:

$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{i n} \end {bmatrix}^\intercal$


The two matrices are equal, so the two definitions are equivalent.

$\blacksquare$