Find the eigenvalues of $\alpha I +u1^{T}+1u^{T}$

Question

Find the eigenvalues with their algebraic and geometric multiplicities for the matrix

$$B=\alpha I + u 1^T + 1 u^T$$

All I can say that $B$ is symmetric and has real eigenvalues. Now we look at it's characteristic equation and get a determinant of the form $|A+uv^T|=0$ which is standard but it is difficult to solve it.

Answer 1

Hint:

Let $$A= \underline{u} \ \underline{1^T}+\underline{1} \ \underline{u^{T}}$$

Since $\mbox{rank} (u 1^T) \leq 1$ and $\mbox{rank} (1u^{T})\leq 1$ you get that $$\mbox{rank}(A) \leq 2$$

Therefore, $0$ is an eigenvalue of $A$ of geometric multiplicity at least $n-2$. Therefore, it's algebraic multiplicity is at least $n-2$. [note in passing that since the matrix is symmetric, its algebraic and geometric multiplicities are equal].

Let $\lambda_1, \lambda_2$ be the remaining two eigenvalues. Then $$\lambda_1+\lambda_2= tr(A) = 2(u_1+u_2+...+u_n)\\ \lambda_1^2+\lambda_2^2= tr(A^2)$$

Solving, you get the last two eigenvalues.

Now, it is easy to argue that the eigenvalues of $B$ are simply $\alpha+$ the eigenvalues of $A$.