Change of variables in vector spaces

In linear algebra, this is identical to change of basis.

As a start, given two bases, we want to be able to convert all vectors from one to the other.

Suppose:

• $V$ is a vector space,
• ${\rm dim \hspace{3pt}} V=n$,
• $B,D$ are two bases of $V$.

We want:

• given a vector $v \in V$, and its coordinate vector $v^D \in {\bf R}^n$ with respect to $D$,
• find its coordinate vector $v^B \in {\bf R}^n$ with respect to $B$.

This conversion is provided by a linear operator (and a matrix):

$$v^B = P_{DB}v^P$$

This may be confusing: $v_B \in {\bf R}^n$, $v_D \in {\bf R}^n$, is it the same ${\bf R}^n$?

One can/should think of them as two different spaces:

$$v_B \stackrel{\varkappa_B}{=} {\bf R}^n_B {\rm \hspace{3pt} and \hspace{3pt}} v_D \stackrel{\varkappa_D}{=} {\bf R}^n_D,$$

as if two different copies of $V$ and of ${\bf R}^n$.

This is better as we can use the theorem about matrices of linear operators:

Theorem: Given $T \colon V \rightarrow U$ linear operator. The $i^{\rm th}$ column of $A_T$ is made of the coefficients of the linear combination of $T(u_i)$ (i.e., $T$'s values on the basis elements) in terms of $v_1,\ldots,v_n$ a basis of $V$.

It follows that to find $P_{DB}$ we only need its values on $D$ (basis of $V_D$); further these values, as columns, form its matrix. This follows from the above theorem:

"$A_T$ is $m \times n$ matrix the $i^{\rm th}$ column of which is $T(e_i)$, $\{e_i\}$ is the standard basis of ${\bf R}^n$."

We use $v^B=P_{DB}v^p$ for the basis elements.

Let $B=\{v_1,\ldots,v_n\}$, $D = \{u_1,\ldots,u_n\}$, in $V$.

Note: $V_1^B=[1,0,0,\ldots,0]^T \in {\bf R}^n$ comes from $v_1=a_1v_1+\ldots+a_nv_n$.

Find $a_1,\ldots,a_n$.

$$\begin{array}{} v_n^B &= [0,\ldots,0,1]^T \\ u_1^D &= [1,0,\ldots,0]^T \\ u_n^D &= [0,\ldots,0,1]^T \end{array}$$

Therefore, $B$ is the standard basis of $V_B$, and $D$ is the standard basis of $V_D$.

Problem: What is $u_1^B, \ldots, u_n^B$?

We need to rewrite basis $D$ in terms of basis $B$.

So, we need only:

$$u_i^B = P_{DB}u_i^D, i=1,\ldots,n.$$

Hence,

Theorem: The matrix of $P_{DB}$ (with respect to $D$) is made of these columns $$u_1^B,u_2^B,\ldots,u_n^B.$$ Here $P_{DB} \colon V_D \rightarrow V_B$.

Theorem: $P_{BD}=P_{DB}^{-1}$ exercise.

We now can convert vectors from $B$ to $D$ and back!