$E$ is the $n$ by $n$ all-ones matrix, and $I$ is $n$ by $n$ identity matrix. $a$ is a real number. What is the condition on $a$ that this matrix $aE+(1-a)I$ is positive semi-definite?
Thanks.
My guess is $ -1/(n-1) \leq a \leq 1$. Can some one help me with the proof?
The symmetric matrix $aE$ has eigenvalue $an$ with multiplicity one and eigenvalue zero with multiplicity $n-1$.
So adding $aE + (1-a)I$ gives us single eigenvalue $an + (1-a) = 1 + a(n-1)$ and eigenvalue $1-a$ with multiplicity $n-1$.
Then the matrix is positive semi-definite if and only if both $1 + a(n-1) \ge 0$ and $1-a\ge 0$. In other words, as you suspected (omitting the case $n=1$):
$$ \frac{-1}{n-1} \le a \le 1 $$