is $(1 + 1/\gamma)I - \dfrac{ee^T}{m}$ positive definte for any $\gamma > 0$?

Question

Is the following matrix definite postive for any $\gamma > 0$?

$$(1 + 1/\gamma)I - \dfrac{ee^T}{m}$$ where $\gamma > 0$, $I$ is the identity matrix in $\mathbb{R^{m \times m}}$, $e \in \mathbb{R^m}$ is a one-vector, i.e., $e = (1, \ldots, 1)$.

My Atttempt: I think it is positive definite, I try the following multiplying by a vector $x^T$ at left side and $x$ at right side we get

$(1 + 1/\gamma)||x||^2 - (x_1 + \ldots + x_m)^2/m $, since $m>1$ it looks positive but I'm not sure.

Thanks

Answer 1

Note that $ee^T$ is rank-one with a single nonzero eigenvalue of $e$ (with eigenvector $m$). So $(1+1/\gamma)I-ee^T/m$ has eigenvalues $1+1/\gamma$ with multiplicity $m-1$ and $1/\gamma$ with multiplicity one.

Answer 2

Cauchy inequality $$(a_1+a_2+\cdots +a_m)^2\leq m(a_1^2+a_2^2+\cdots +a_m^2).$$ Hence $$(1+\frac{1}{\gamma})(x_1^2+\cdots +x_m^2)-\frac{(x_1+\cdots +x_m)^2}{m}\geq \frac{1}{\gamma}|x|^2>0.$$ That is positive definite.