Help to understand a comment in Hoffman and Kunze's linear algebra book I'm reading Hoffman and Kunze's Linear Algebra and on page 52, the authors said:

Let $P$ be the $n \times n$ matrix whose $i,j$ entry is the scalar $P_{ij}$, and let $X$ and $X'$ be the coordinate matrices of the vector $\alpha$ in the ordered bases $\mathscr{B}$ and $\mathscr{B}'$. Then we may reformulate (2-15) as
  $$
X = PX'. \tag{2-16}
$$
  Since $\mathscr{B}$ and $\mathscr{B}'$ are linearly independent sets, $X=0$ if and only if $X'=0$. Thus from (2-16) and Theorem 7 of Chapter 1, it follows that $P$ is invertible. Hence
  $$
X' = P^{-1}X. \tag{2-17}
$$

I didn't find any mention of this result (highlighted in bold) in the book. How can I prove this fact?

Another way to see this is to use the fact that the vector $\alpha$ has unique coordinates with respect to any fixed basis (as discussed on page 50 of the same textbook).
Let $\{ \beta_1,\dots,\beta_n\}$ be an ordered basis for the $n$-dimensional space $V$. We can always express the zero vector as
$$
0 = 0\beta_1 + \dots + 0\beta_n \quad.
$$
By the uniqueness of coordinates, this means that the coordinates of the zero vector in every ordered basis is
$$
[0]_\mathscr{B} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix},
$$
and conversely, the $n \times 1$ matrix on the right is the coordinate matrix of the zero vector in every ordered basis.
In particular, $X = 0 \Leftrightarrow X'=0$ because $X$ and $X'$ are the coordinate matrices of the vector $\alpha$ in the ordered bases $\mathscr{B}$ and $\mathscr{B}'$, respectively, and by the above discussion $X = 0 \Leftrightarrow \alpha = 0 \Leftrightarrow X'=0$.
