Is $\sum_{k=1}^n\sum_{j=1}^nk\cdot j=\sum_{k=1}^nk\cdot\sum_{j=1}^nj$? I have found confronting information on other websites and thus will ask my question here.
Is the following conversion legitimate?
$$\sum_{k=1}^n\sum_{j=1}^nk\cdot j=\sum_{k=1}^nk\cdot\sum_{j=1}^nj$$
I have seen someone doing something similar on Matroids Matheplanet. 
Thank you for your help.
Philipp
 A: Yes.
Since $k$ is a multiplier not related to $j$, it can be "factored" out, resulting in the product of two summations.
Simple example, using $n=3$:
$$\text{LHS}=\sum_{k=1}^3\sum_{j=1}^3 k\cdot j=\sum_{k=1}^3 k+2k+3k=\sum_{k=1}^3 6k=6\sum_{k=1}^3 k=6(1+2+3)=36\\
\text{RHS}=\sum_{k=1}^3k\sum_{j=1}^3 j=(1+2+3)(1+2+3)=36$$
For the general case, 
$$\begin{align}
\text{LHS}&=\sum_{k=1}^n\sum_{j=1}^n k\cdot j\\
&=\sum_{k=1}^n k+2k+3k+\cdots +nk\\
&=\sum_{k=1}^3 (1+2+3+\cdots+n)k\\
&=\sum_{k=1}^n \frac {n(n+1)}2k\\
&=\frac {n(n+1)}2\sum_{k=1}^nk\\
&=\frac {n(n+1)}2\cdot\frac {n(n+1)}2\\
&=\sum_{k=1}^n k\sum_{j=1}^n j \end{align}$$

Note: If $k$ is not a multiplier for the entire summand, then it cannot be "factored" out.   
Example:
$$\sum_{k=1}^n\sum_{j=1}^n j+k\neq \sum_{k=1}^nk+\sum_{j=1}^nj$$
A: $$\sum_{k}\sum_{j}kj=\sum_{k}\left(k(1+2+3+...)\right)$$
$$=(1+2+3...+n)1+(1+2+3...+n)2+...+(1+2+3...+n)n$$
And $$\sum_{k}k\sum_{j}j=(1+2+3...+n)(1+2+3...+n)$$
Clearly, by distribution, the two are equal.
A: Yes 
\begin{eqnarray}
\sum_{k=1}^n\sum_{j=1}^nk\cdot j &=& \sum_{k=1}^n k \cdot (1+2+...+n)\\
&=&1(1+2+...+n)+2(1+2+...+n)+...+n(1+2+...+n )\\
 &=& (1+2+...+n)(1+2+...+n)\\
&=&\sum_{k=1}^nk\cdot\sum_{j=1}^nj
\end{eqnarray}
A: Since you have final sums, there is no problem. See
$$
\sum_{k=a}^n\sum_{j=b}^m c_kd_j=\sum_{k=a}^n\left(\sum_{j=b}^m c_kd_j\right)=
\sum_{k=a}^n\left(c_k\sum_{j=b}^m d_j\right)=\left(\sum_{k=a}^nc_k\right)\left(\sum_{j=b}^m d_j\right)
$$
You just use the distributive law. At the first equation I added the brackets to clarify the meaning. Next $c_k$ is in the sum over $j$ like a constant. So you can exchange them. Finally the sum over j is independent of $k$ and you can exchange them again.
A: 
Trivia: $\Large\sum, \int, \newcommand{\derivative}[3] {\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \derivative{n}{y}{x}$ are linear operations.

What does it mean?
$\Large f(x)$ is linear if and only if
$$\Large f(x+y)=f(x)+f(y)$$
$$\Large f(ax)=a\cdot f(x)$$
Let 
$$\large f(x)=\sum_{j=1}^nx$$
And
$$\large a=\sum_{k=1}^nk$$
So
$$\large f(ax)= \sum_{j=1}^n\left(\sum_{k=1}^nk\right)x$$
