Help on a proof of dimension of a vector space The proof shows that two bases have the same number of elements, and I can't understand one step. The proof goes:
As $v_1, . . . , v_n$ is a basis of $V$, each $w_k$ can be expressed as a linear combination of the $v_j$ ; thus for each $k$ there are scalars $λ_{1k}, . . . , λ_{nk}$ such that
$$w_k=\sum_{j=1}^nλ_{jk}v_j,~~~k=1,...,m$$

Likewise, for each $j$ there are scalars $μ_{1j}, . . . , μ_{mj}$ such that
$$v_j=\sum_{i=1}^m μ_{ij}w_i,~~~j=1,...,n$$
These give:
$$w_k=\sum_{j=1}^n\sum_{i=1}^mλ_{jk}μ_{ij}w_i,~~~k=1,...,m$$
As the $w_j$ form a basis, we may equate coefficients of $w_k$ on each side of this
equation and obtain
$$1=\sum_{j=1}^nλ_{jk}μ_{kj},~~~k=1,...,m$$
Could anybody shed light on this last step? As rest of the proof is easy. Thank you!
 A: Since $\{w_1,\dots,w_m\}$ is a basis for $V$, each $v\in V$ has a unique representation in the form 
$$v=\alpha_1w_1+\ldots\alpha_mw_m\;:$$
if $v$ had two different representations of this form, their difference would be a non-trivial linear combination of $w_1,\dots,w_m$ equal to $0$, contradicting the linear independence of the basis.
Here we have 
$$w_k=\sum_{j=1}^n\sum_{i=1}^mλ_{jk}μ_{ij}w_i\tag{1}$$
for $k=1,\dots,m$. The two sides of $(1)$ are two representations of $w_k$ as a linear combination of $w_1,\dots,w_n$: on the left the coefficient of $w_k$ is $1$, and the coefficients of the other $w_i$’s are all $0$. On the right the coefficient of $w_i$ is
$$\sum_{j=1}^n\lambda_{jk}\mu_{ij}\;,$$
so it must be $1$ when $i=k$ and $0$ otherwise. In particular, taking $i=k$, we must have
$$\sum_{j=1}^n\lambda_{jk}\mu_{kj}=1\;.$$
A: Since the $w_i$ form a basis, each element of $V$ can be uniquely expressed as a linear combination the $w_i$. In the double sum $w_k$ is expressed in two ways, which therefore has to be the same, that is 
$$
\sum _{j=1}^n \lambda _{jk} \mu _{ij} = 0 \quad \text{for }i\ne k 
$$
while for $i=k$ you obtain what you mention. 
So just think of 
$$
w_k = 0\cdot w_1 + 0\cdot w_2 +\cdots +1\cdot w_k +\cdots +0\cdot w_m.
$$
A: Since the $w_i$'s are a basis, whenever we have scalars $a_1, \dots a_m$ and $b_1, \dots b_m$ such that 
$$ a_1 w_1 + a_2 w_2 \cdots + a_m w_m = b_1 w_1 + \cdots + b_m w_m, $$
then 
$$ a_1 = b_1, \ a_2 = b_2 , \ \ \dots , a_m = b_m. $$
This is what is meant by "equating the coefficients", and it follows since the $w_i$'s are linearly independent.
So your second-last equation can be written
$$ 0 \cdot w_1 + 0 \cdot w_2 + \cdots + 0 \cdot w_{k-1} + 1 \cdot w_k + 0 \cdot  w_{k+1} + \cdots + 0 \cdot w_m= \\ 
\left( \sum_{j=1}^n \lambda_{jk} \mu_{1j} \right) \cdot w_1 + \cdots + \left( \sum_{j=1}^n \lambda_{jk} \mu_{mj} \right) \cdot w_m
$$
Comparing the coefficients of $w_k$ on both sides gives your last equation.
