Relationship between $[T]_{\mathcal{B}}$ and $[T]_{\mathcal{B'}}$.

Question

Let $T$ be a linear operator on a finite-dimensional space $V$. Let $\mathcal{B}=\{\alpha_1, \dots, \alpha_n \}$ and $\mathcal{B'} = \{\alpha'_1, \dots, \alpha'_n\}$ be two basis for $V$.

How are $[T]_{\mathcal{B}}$ and $[T]_{\mathcal{B'}}$ related?

First, there exists a unique invertible $n \times n$ matrix $P$ such that $$[\alpha]_{\mathcal{B}}=P [\alpha]_{\mathcal{B'}}$$

What exactly is $P$?

I would appreciate a concrete example of this statement.

Note, that $[\alpha]_{\mathcal{B}}$ denotes the coordinates of the vector $\alpha$ with respect to the basis $\mathcal{B}$.

$P$ called a transition matrix or change of basis matrix. — , Jan 05 '15 at 22:02
for an example related to the matrix $P$, look an answer I gave at http://math.stackexchange.com/questions/1068862/covariant-and-contravariant-components-and-change-of-basis/1076807#1076807 — janmarqz, Jan 05 '15 at 22:34

janmarqz · Answer 1 · 2015-01-05T23:02:17.177

The relation you are asking for can be explained as follow:

If you know that $[\alpha]_{\mathcal{B}}=P[\alpha]_{\mathcal{B}'}$ then $$[\alpha]_{\mathcal{B}'}=P^{-1}[\alpha]_{\mathcal{B}},$$ that is, one needs to multiply with $P^{-1}$ the $\alpha$'s old-components to get the new ones.

So, if you want the new components for $T[\alpha]$ you calculate: $$T[\alpha]=PP^{-1}TPP^{-1}[\alpha],$$ and from here you get: $$P^{-1}T[\alpha]=(P^{-1}TP)P^{-1}[\alpha].$$

This last relation tell you that the new components of both $\alpha$ and $T[\alpha]$ are connected by the matrix $P^{-1}TP$, which will be your $[T]_{\mathcal{B}'}$.

Relationship between $[T]_{\mathcal{B}}$ and $[T]_{\mathcal{B'}}$.

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