Sum of two rank $1$ matrices with some property gives rank $2$ matrix

Question

Let $\{u_1,u_2, \ldots,u_n \}$ be a set of $n$ orthonormal vectors.

Similarly let $\{v_1,v_2, \ldots ,v_n\}$ be another set of n orthonormal vectors.

Prove that $u_1v_1^T + u_2v_2^T$ is a rank $2$ matrix.

I know that $u_1v_1^T$ and $u_2v_2^T$ are both rank $1$ matrix since they are outer product of two vectors.

But how to show that their sum is actually rank $2$ matrix?

Answer 1

Hint 1: Produce a linearly independent set of size $n-2$ in its null space. This shows the rank is $\le 2$.

Hint 2: Show that its column space includes $u_1$ and $u_2$. This shows the rank is $\ge 2$.

Answer 2

\begin{align} u_1v_1^T + u_2v_2^T &=\begin{bmatrix}u_1 & u_2 \end{bmatrix} \begin{bmatrix} v_1^T \\ v_2^T \end{bmatrix} \\ &=\begin{bmatrix}u_1 & u_2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} v_1^T \\ v_2^T \end{bmatrix} \\ \end{align}

is the reduced SVD, hence the rank is $2$.

Alternatively, it can be rewritten in the form of

$$\begin{bmatrix} u_1 \ldots u_n \end{bmatrix}\begin{bmatrix} I_2 & \\ & O_{n-2} \end{bmatrix} \begin{bmatrix} v_1^T \\ \vdots \\ v_n^T\end{bmatrix}$$

where the first and third matrix is an orthogonal (and hence non-singular matrices).