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?
$\mathcal{B} = \{b_1, b_2, \dots, b_n\}$ and $\mathcal{B}' = \{b_1', b_2', \dots, b_n'\}$ are bases, in particular linearly independent sets. Given a vector $\alpha$ in the vector space, the coordinate matrix of $\alpha$ with respect to $\mathcal{B}$ has coordinates corresponding to the unique coefficients
$$
\alpha = x_1b_1 + x_2b_2 + \dots + x_nb_n,
$$
and its coordinate matrix with respect to $\mathcal{B}'$ has coordinates corresponding to the unique coefficients
$$
\alpha = x_1'b_1' + x_2'b_2' + \dots + x_n'b_n'.
$$

Suppose $X = 0$, then clearly $\alpha = 0$. However, since $\mathcal{B}'$ is linearly independent, we must have $X'=0,$ since $X'\neq 0$ would imply there exists a nontrivial linear combination of the basis vectors in $\mathcal{B}'$ equal to the zero vector, contradicting the linear independence of a basis. Similarly if we assume $X' = 0$.
It is actually related to the matrix of basis change.

P is actually the matrix represents the basis change from the basis of X' to that of X.
Now we know that the basis change matrix is always invertible because of the fact that B and B ' are independent  .
It is easy to see.
You can proove  by inequality of the ranks of matrices 
Or from the scratch that are required to proove the inequalities .

Hope that it is of some use.