Let $A$ be a $n\times n$ matrix such that $a_{ij}=ij$,find the eigenvalues of the matrix $A$. Let $A$ be a $n\times n$ matrix such that $a_{ij}=ij$.
I want to find the eigenvalues of the matrix $A$. 
Efforts: $n=2$, I found that eigenvalues are equal to $0,5$.
For $n=3$, I found eigenvalues are equal to $0,0,14$.
For $n=4$ I found eigenvalues are equal to $0,0,0,30$.
For $n=5$ I found eigenvalues are equal to $0,0,0,0,0,55$.
It looks like there is a pattern here. Eigenvalues of $n\times n$ matrix seems to be $\sum_{i=1}^n i^2, \underbrace{0,0,0,..0}_{n-1\text{ times}}$.
But how can I prove it? I mean whether there are some beautiful arguments that makes things easy, and ideas clear. 
Thanks for reading and helping out.
 A: First off, all columns are multiples of one another, so the matrix has rank $1$ and therefore eigenvalue $0$ with multiplicity $n-1$.
For the final eigenvalue, some experimenting gives a strong hint: For $n = 2$, we get that the eigenvector of the non-zero eigenvalue is $(1, 2)^T$, and for $n = 3$ we get $(1, 2, 3)^T$. This is a quite suspicious pattern. Let's check whether that holds in general:
Let $A$ be our $n\times n$ matrix, and $v$ be the vector where the $i$th entry is $i$. We get
$$
Av = \begin{bmatrix}1\cdot 1 + 2\cdot 2 + \cdots + n\cdot n\\
2\cdot 1 + 4\cdot 2 + \cdots + 2n\cdot n\\
\vdots\\
n\cdot 1 + 2n\cdot 2 + \cdots + n^2\cdot n\end{bmatrix} = (1^2+2^2+\cdots + n^2)v
$$
A: $\{x:\sum jx_j=0\}$ is an $n-1$ dimensional subspace. Any basis for this gives you $n-1$ independent eigen vectors with eigen value $0$ and the only other eigen value is $1+2^{2}+3^{2}+\cdots +n^{2}$ as seen easily from definition of eigen values and eigen vectors. 
Use the fact that $\sum\limits_{j=1}^{n}  ijx_j=\lambda x_i$ implies that $x$ is a multiple of $(1,2,\cdots,n)$
