Change of Coordinates and property of invertible matrices (Sec 2.4 Theorem 8, Hoffman Kunze, Linear Algebra) I was reading Linear Algebra by Hoffman Kunze, and encountered this in the Theorem 8 of the Chapter Coordinates, the theorem is stated below : 

What I get about is in the most bottom line (uniqueness), it says 
$"... it\: is\: clear\: that $
$$ \alpha'_{i}=\sum_{i=1}^{n} P_{ij}\alpha_{i}." \tag{a}$$
Is there any easy way to see why it is clear? I tried many ways and what I found was that we start from scratch, (starting the same with the proof, with $\scr \overline{B}$ and then we find an invertible matrix, say $Q$, which we don't know equal to $P$ or not, such that the property (a) above holds with $Q_{ij}$ instead of $P_{ij}$. Then now what we are left to show is that $P$ = $Q$, and we currently have:
$$ x_{i}=\sum_{j=1}^{n} P_{ij} x'_{j}\tag{from (i)}$$ 
and,
$$ x_{i}=\sum_{j=1}^{n} Q_{ij} x'_{j}$$
So together, 
$$ \sum_{j=1}^{n} Q_{ij} x'_{j}=\sum_{j=1}^{n} P_{ij} x'_{j}$$
$$ \sum_{j=1}^{n} (Q_{ij} - P_{ij}) x'_{j}= 0$$
The way I showed that this implies that $P$ = $Q$ is by asserting that $P - Q \neq 0^{n\times n}$ and find a contradiction. Suppose $A := P - Q \neq 0^{n\times n}$ then we can choose a row $r$ such that the k-th entry is non-zero, we can plug in $0$ to any other entries other then the k-th and we are left with something non zero equals to $0$ which is a contradiction, so that $A = 0^{n\times n}$ and $P = Q$, and since $(a)$ holds for $Q$ and $P = Q$, it follows that $(a)$ holds for $P$. 
I'm pretty sure that the proof is correct since we can plug in various values to $x'_{1}, ... , x'_{n} \in F$, since $F$ is a field, it sure contains $0$ and $1$, but this proof seems to be lengthy and is not as clear as how it was written to be by Hoffman and Kunze, I think I'm missing something here, and will be very thankful for a good explanation. Thanks!
 A: Representing $\alpha_j'$ in the ordered basis $(\alpha_1',\dots,\alpha_n')$ will just give you the standard basis vector $e_j$ because $\alpha_j' = 0\alpha_1' + \dots + 1\alpha_j' + \dots + 0 \alpha_n'$. Multiplying $P$ by $e_j$ gives you the $j$-th column of $P$. So equation (i) says that
$$ [\alpha_j']_{\mathcal B} = \begin{bmatrix} P_{1,j} \\ \vdots \\ P_{n,j} \end{bmatrix}. $$
By definition of $[\alpha_j']_{\mathcal B}$, this means that
$$ \alpha_j' = P_{1,j}\alpha_1 + \dots + P_{n,j}\alpha_n = \sum_{i = 1}^n P_{i,j} \alpha_i. $$
A: Theorem 8: this theorem actually starts with the set $\mathcal B'$ which is supposed to be our aimed basis. Therefore, Kunze says: let's define a set of vectors $\mathcal B'$ where its elements are defined by 
$$\alpha'_j=\sum_{i=1}^n P_{ij}\alpha_i$$ and then prove that $\mathcal{B}'$ is a basis for $V$. So the goal is to produce a set with $P$ and to prove that it is a basis.

Uniqueness: I believe Kunze's Theorem 7 in this chapter proves that if $\mathcal{B}$ and $\mathcal{B}'$ are two bases for vector space $V$ over $F$, there exists a unique $n\times n$ matrix $P$ which suffices
$$[\alpha]_\mathcal{B}=P[\alpha]_{\mathcal{B}'}$$
and
$$[\alpha]_{\mathcal{B}'}=P[\alpha]^{-1}_{\mathcal{B}}$$
The uniqueness of $P$ could be simply proved by regarding it as a collection of unique scalars. Since every vector in basis $\mathcal B'$ can be written uniquely as a linear combination of vectors in basis $\mathcal B$, the coefficients are unique in the following picture (and thus the uniqueness of $P$ followers immediately):

