Change of Base matrices 
Let $E$ and $F$ be two bases of the same n dimensional vector space $U$
  $\bullet$ if $P$ is the change of base matrix from $E$ to $F$ and Q the change of base matrix from $F$ to $E$ then $P=Q^{-1}$  

My Attempt
I have tried this with a few trial bases for $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ , but how would you go about proving this??  
Any help wold be very much appreciated.
 A: Let $E = \{e_1 , \ldots , e_n\}$ and $F = \{f_1, \ldots ,f_n\}$. Let $p_{ij}$ denote the element of $P$ in the $i^{th}$ column and  $j^{th}$ row, and similarly $q_{ij}$ for $Q$. Then writing out the matrix multiplication for each $1 \le i \le n$ we have
$$
f_i = \sum_{i=1}^{n} p_{ij} e_j =  \sum_{i=1}^{n} p_{ij} \sum_{j=1}^n q_{jk} f_k =  \sum_{i=1}^{n}\sum_{j=1}^n p_{ij}  q_{jk} f_k .
$$
Since $F$ is a basis each of the $f_k$ are linearly independant, and so the only term in the above sum which can non-zero is when $i = k$ in which case it is $1$, as then we get $f_i = f_i$. Now $\sum_{j=1}^n p_{ij}  q_{jk} $ is just the expression for the $(i,k)$-th element of the product $PQ$, and since it is $0$ for $i \ne k$ and $1$ for $i = k$ it follows that $PQ = Id$. You can do this the other way to show that $QP = Id$ and so $P = Q^{-1}.$
A: Remember these functions are bijections, so this follows basically from the very definition of inverse function:
$$ \begin{cases}E=\{e_i\}\\F=\{f_i\}\end{cases}\implies\;\;\;Pe_i=f_i\iff Qf_i=P^{-1}f_i=e_i$$
