Linear Algebra – Understanding P in Theorem 2.4.8 by Hoffman and Kunze

change-of-basislinear algebramatricesnotationproof-explanation

Code borrowed from here

Suppose $P$ is an $n\times n$ invertible matrix over $F.$ Let $V$ be an $n$-dimensional vector space over $F,$ and let $\cal B$ be an ordered basis of $V.$ Then there is a unique ordered basis $\cal \overline{B}$ of $V$ such that

$$(i)[\alpha]_{\mathcal{B}} = P [\alpha]_{\mathcal{ \overline{B}}}$$
$$(ii) P^{-1}[\alpha]_{\mathcal{B}} = [\alpha]_{\mathcal{ \overline{B}}}$$

for every vector $\alpha \in V$

proof : Let $\cal B$ consist of the vectors $\alpha_1,\dots,\alpha_n.$ Let $\cal \overline{B}= \{\overline{\alpha_1},\dots ,\overline{\alpha_n}\}$ be an ordered set in which $[\alpha]_{\mathcal{B}} = P [\alpha]_{\mathcal{ \overline{B}}}$
, $\color{red}{\text{it is clear that $\overline{\alpha_j}=\sum P_{i,j}\alpha_i$}}$

Thus we need only show that the vectors $\overline{\alpha_j},$ defined by these equations form a basis. Let $Q=P^{-1}.$ Then
\begin{align}
\sum_j Q_{jk}\overline{\alpha_{j}}&=\sum_j Q_{jk}\sum_i P_{ij}\alpha_i\\
&=\sum_j \sum_iP_{ij}Q_{jk}\alpha_i\\
&=\sum_i\Big( \sum_jP_{ij}Q_{jk}\Big)\alpha_i\\
&=\alpha_k
\end{align}

Thus the subspace spanned by the set $\cal \overline{B}$ contains $\cal B$ and hence spans $V.$ Thus $(1)$ is valid due to a previous theorem.

I want to ask what is $P$? Is it any $n \times n$ invertible matrix or a unique one that relate $\alpha_k'$ to $\alpha_k$?

In page 51 $\overline{\alpha} _j = \sum\limits_{i=1}^n P_{ij} \alpha_i $ does it have the same meaning in this theorem or it is just any general matrix ?
for example in the exercise there is this question

  1. Let $\mathcal {B}$ =$\{a_1, a_2,a_3\}$ as be the ordered basis for $R^3$ consisting of
    $a_1 = (1, 0, – 1)$, $a_2 = (1, 1, 1),$ $a_3 = (1, 0, 0)$.
    What are the coordinates of the vector (a, b, c) in the ordered basis $ \cal B$?

in the answer

$$P=\begin{pmatrix}
1 & 1 & 1
\\\\ 0 & 1 & 0
\\\\ -1 & 1 & 0
\end{pmatrix}$$
If not then how to prove $\overline{\alpha_j}=\sum_{i=1}^n P_{ij}\alpha_i$?

another question is what is $[\alpha]_{\mathcal{B}}$ is it the base vector $\alpha_1,\dots,\alpha_n $ or a coordinate of a vector $\alpha $ according to base $ \cal B $? (using the same notation for general vector and the base confuses me a lot)

I apologise if the question seemed elementary or stupid but I am a beginner in this subject.

Best Answer

I think you've got the causality mixed up. The theorem content is the following:

Hypotheses: Given the invertible matrix $P$ and the basis $\mathcal{B}$ .

Conclusions: There exists a unique basis $\overline{\mathcal{B}}$ such that the stated equation holds (equivalently, such that $P_{\mathcal{B}}^{\overline{\mathcal{B}}}=\mathrm{Id}_n$).

Whereas you seemed to be worrying about this:

Hypotheses: Given the bases $\mathcal{B},\overline{\mathcal{B}}$.

Conclusions: There exists an unique invertible matrix $P$ such that the stated equation holds.

This is also true but it has a different proof and is the converse of what the text is showing. The linked "question $3$" you show is an example of the converse: we're given the two bases and then we need to find the change of basis matrix.

So "what is $P$?" It is indeed 'anything' (invertible and of the right dimension). You ask how to prove $\overline{\alpha_j}=\sum P_{i,j}\alpha_i$. You don't. It's true by definition! The authors defined $\overline{\alpha_j}$ by that equation.

Related Question