Change of basis formula - intuition/is this true? Let $E$ and $F$ be two bases of the same $n$-dimensional vector space $U$. 
Does the change of basis matrix from $E$ to $F$ have size $n \times n$? Give a counterexample or brief justification. 
My thoughts:
The change of basis matrix from $E$ to $F$ is formed by expressing the coordinates of vector in $U$ with respect to $E$ as a coordinate with respect to $F$ (Although, I'd also like someone to help me make this statement clearer). 
Since $E$ and $F$ both have dimension $n$, this matrix will have $n$ columns (as there are $n$ basis vectors). However, I'm not sure if it'll need $n$ rows. 
Any help with this would be much appreciated!
 A: Call $P$ the change of basis matrix of dimension $p \times q$, and $\mathbf{v}_E$ be a vector wrt the E basis and $\mathbf{v}_F$ the vector wrt the F basis.
$\mathbf{v}_E$ is $n \times 1$ so for compatibility of multiplication: $P \,\mathbf{v}_E$, we require $q=n$.
Given $\mathbf{v}_F$ is also $n \times 1$ then $p=n$ so $P$ is $n \times n$
Explaining the last line:
$\mathbf{v}_F \in U$ and must have $n$ components.
The $(i^{th}$ component of $\mathbf{v}_F) = (i^{th} \text{row of P}) \times \,\mathbf{v}_E$ 
Therefore $P$ must have the same number of rows as $\mathbf{v}_F$ has components.
Therefore $P$ has $n$ rows.
A: If $E=\{e_1,...,e_n\}$ is the first basis and $F=\{f_1,...,f_n\}$ is the second then we can write each
$$f_1=P_{11}e_1+\cdots+P_{n1}e_n$$
$$f_2=P_{12}e_1+\cdots+P_{n2}e_n$$
$$\cdots$$
$$f_n=P_{1n}e_1+\cdots+P_{nn}e_n$$
where you need a set of $n\times n$ coefficients $P_{ij}$. 
And to warranty linear independence of the $f_i$, it is better that the matrix
$$P=[P_{ij}],$$
have determinant non-zero. 
