How can I find the eigenvalues of this $2n \times 2n$ matrix? The matrix I am dealing with is of the form below.
$$\begin{pmatrix} 1-\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}}\\ -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& \cdots & -\frac{1}{\sqrt{2^n}} \\ -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& \cdots & -\frac{1}{\sqrt{2^n}}\\-\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}}\\ \vdots & \vdots& \vdots& \vdots & \ddots & \vdots\\ -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}} \end{pmatrix}$$
It's a $2n \times 2n$ matrix whose $(1,1)$-th entry is $1 - \frac{1}{\sqrt{2^n}}$ and all the others $- \frac{1}{\sqrt{2^n}}$. How can I solve this eigenvalue problem?

Here's a similar approach to what KBS suggests. Begin with the observation that we can write $M = e_1e_1^T - \alpha 11^T$ for some $\alpha \in \Bbb R$. From there, we can write $M = AB$, with
$$
A = \pmatrix{e_1 & \alpha \mathbf 1}, \quad B = \pmatrix{e_1 & - \mathbf 1}^T.
$$
From the fact that $AB$ and $BA$ have the same non-zero eigenvalues, conclude that every eigenvalue of $M$ is either equal to $0$ or is an eigenvalue of
$$
BA = \pmatrix{e_1 & - \mathbf 1}^T\pmatrix{e_1 & \alpha \mathbf 1} = \pmatrix{1 & \alpha\\ -1 & -\alpha n}.
$$
