Congruent diagonal matrix

Question

For two days I reflect on this question without an answer:

If $A=(i+j-1)_{1\le i,j\le n}$ is matrix in $\mathcal M_n(\mathbb R)$, the question is to find basis in which $A$ is congruent to diagonal matrix. I know that since I look for a congruent matrix I'll consider $A$ as matrix of a quadratic form and since $A$ is symmetric then it's diagonalizable in an orthonormal basis but how I can find it? Any help would be appreciated.

Answer 1

step1:Find the characteristic polynomial of $A$, then characteristic values of it;
step2:Find the characteristic vectors belonging to each characteristic value;
step3:Schimidt orthogonalizate these values and then unit them.
Then you will get the orthonormal basis you want.

Answer 2

I wanted to write this as a comment but it got extremely long. This is not a complete answer to your question.

Since $A$ is symmetric it is diagonalizable. For $n=2$,

$A = \left[\begin{matrix} 1& 2\\ 2 & 3\end{matrix}\right]$

The characteristic polynomial is $p_A(t) = x^2-4x-1 = (x-(2-\sqrt{5}))(x-(2+\sqrt{5}))$

For $n=3$,

$A = \left[\begin{matrix} 1& 2&3\\ 2 & 3&4\\ 3&4&5\end{matrix}\right]$

The characteristic polynomial is $p_A(x) = -x(x^2+9x+6) = -1/4x(x-(9-\sqrt{105}))(x-(9+\sqrt{105}))$.

In general your polynomial is $p_A(x) = (-1)^nx^{n-2}((-1)^nx^2-bx-c)$ where $b=\text{trace}(A) = n^2$ and $c$ is the sum of the 2x2 minors of $A$ lying along the diagonal. So in partictular you will have an $n-2$ dimension kernel and two other $1-$dimensional eigenspaces.

Once you know an easy formula/"closed form" for $c$ the bases should be straight-forward to find. This is what I came up with:

$ c = \displaystyle \sum_{i=1}^{n-1} \sum_{j=1}^{n-i} \left|\begin{matrix} 2i-1& 2i-1+j\\ 2i-1+j & 2i-1+2j\end{matrix}\right| = \sum_{i=1}^{n-1}\sum_{j=1}^{n-i}-j^2 = \dfrac{n^4-n^2}{12}$.

So $p_A(x) = (-1)^nx^{n-2}\left((-1)^nx^2 - n^2x -\frac{n^4-n^2}{12}\right)$

As for the eigenvectors, they will look messy, no doubt.