So I'm trying to solve this question, and I'm having some trouble.
It seems that I can "guess" the eigenvalues and eigenvectors (trying this out for a few n's I've found the pattern), but there has to be a cleaner way of solving this.
What am I missing?
I've tried:
- Characteristic polynomial - By proving that for all $\lambda_i$s to be positive, $\alpha\geq0$. But I can't seem to get a nice closed solution
- Decomposing this to the following form thinking it might bring me closer to a nice solution: $$ \alpha x^T\cdot I\cdot x + 2x^T\cdot \begin{pmatrix} 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 0 \\ \vdots & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ \end{pmatrix} \cdot x $$
Write $A^\alpha = e e^T + (\alpha-1) I$, with $e$ being a vector of ones.
Note that $A^\alpha$ is real, symmetric.
It is straightforward to compute the eigenvalues by noting that $e$ is an eigenvector, and any $y \bot e$ is an eigenvector.
Now translate this into the requirement that $A^\alpha \ge 0$.
Here is another way:
$x^T A^\alpha x = (\sum_x x_k)^2+(\alpha-1) \sum_k x_k^2 $.
We need $\alpha \|x\|_2^2 \ge \|x\|_2^2 - (e^T x)^2$, or $\alpha \ge 1- {(e^T x)^2 \over \|x\|_2^2}$ (for $x \neq 0$, of course).
Hence we see that $A^\alpha \ge 0$ iff $\alpha \ge 1- (e^T x)^2$ for all unit vectors $x$ iff $\alpha \ge 1$.