Q:

Prove $A^\alpha$ positive semi definite

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:
$$
    \alpha x^T\cdot I\cdot x + 2x^T\cdot \begin{pmatrix}
    0 & 0 & \cdots & 0 \\
    1 & 0 & 0 & 0 \\
    \vdots & 1 & 0 & 0\\
    1 & 1 & 1 & 0\\
    \end{pmatrix}
\cdot x
$$

A:

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 $\alpha \ge 1$.

