Find eigenvalues/vectors of $A=\mathbf{u}\mathbf{v}^T+\mathbf{v}\mathbf{u}^T$ where $\mathbf{u},\mathbf{v}$ are orthonormal

Question

Our problem is to compute the eigenvalues and eigenvectors of two matrices formed by products of orthonormal vectors, $\mathbf{u}$ and $\mathbf{v}$:

$A=\mathbf{u}\mathbf{v}^T+\mathbf{v}\mathbf{u}^T$

$B=\mathbf{u}\mathbf{v}^T-\mathbf{v}\mathbf{u}^T$

I started with trying to find the eigenvalues/vectors of A, but I am getting stuck. First, I define the vectors and compute A:

$\mathbf{u}=\begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$

$A= \begin{bmatrix} 2u_1v_1 & u_1v_2+v_1u_2 & \dots & u_1v_n+v_1u_n \\ u_2v_1+v_2u_1 & 2u_2v_2 & \dots & u_2v_n+v_2u_n \\ \vdots & \vdots & \ddots & \vdots \\ u_nv_1+v_nu_1 & u_nv_2+v_nu_2 & \dots & 2u_nv_n \end{bmatrix}$

Our eigenvalue/vector problem is $A\mathbf{w}=\lambda \mathbf{w}$ and I know we find the eigenvalues/vectors by setting $|(A-\lambda I)|=0$

I thought that I might be able to figure it out if I started first with a $2\times2$ matrix, but even that was overwhelming. My determinant is $(2u_1v_1-1)(2u_2v_2-1)-(u_2v_1+v_2u_1)(u_1v_2+v_1u_2)$, which I set to zero. Expanding that out gives an even more terrible equation.

I know that A is a symmetric matrix, so all of the eigenvalues are real, but I have no clue how to figure out what they are. I also tried to start by setting $\mathbf{u}, \mathbf{v}$ to be the standard basis vectors $\in \mathbb{R}^2$, but I realized that I would get eigenvalues/vectors specific only to those specific orthonormal vectors. I don't know how to find them in terms of the elements of $\mathbf{u}$ and $\mathbf{v}$.

I have the same problem for B, which is similar to A but with zeros on the diagonal:

$B= \begin{bmatrix} 0 & u_1v_2-v_1u_2 & \dots & u_1v_n-v_1u_n \\ u_2v_1-v_2u_1 & 0 & \dots & u_2v_n-v_2u_n \\ \vdots & \vdots & \ddots & \vdots \\ u_nv_1-v_nu_1 & u_nv_2-v_nu_2 & \dots & 0 \end{bmatrix}$

Thanks!

Answer 1

Hint: Let $K$ be the base field (I believe you take $K:=\mathbb{C}$). Suppose that $u,v\in K^n$ satisfy $u^\top\,v=v^\top u=0$ and $u^\top\,u=v^\top\,v=1$. Then the matrices $A:=u\,v^\top+v\,u^\top$ and $B:=u\,v^\top-v\,u^\top$ are of rank $2$. Note further that $\text{im}(A)$ and $\text{im}(B)$ equal $\text{span}_K\{u,v\}$. Thus, $0$ is the eigenvalue of $A$, or $B$, with multiplicity $n-2$. What are the remaining eigenvalues of $A$, or $B$? (Observe that, if $-1$ is not a square in $K$, then $B$ is not diagonalizable over $K$.)

Answer 2

Hint: Use a coordinate system where the first two basis vectors are $\bf u$ and $\bf v$.

${\bf u}^T {\bf v}$ and ${\bf v}^T {\bf u}$ are just the scalar $0$.