Prove matrix $uu^T+vv^T$ has rank two.

Question

Can somebody help me in proving that for a given two non-zero column vectors $u$ and $v$ in $\mathbb{R}^n$, such that $u$ and $v$ are linearly independent, then the matrix $uu^T+vv^T$ has rank two.

Answer 1

Let $A=uu^\top + vv^\top$.

Apply Gram-Schmidt on $\{u,v\}$ to get an orthonormal basis $\{w_1,\ldots,w_n\}$ where $\operatorname{span}\{u,v\}=\operatorname{span}\{w_1,w_2\}$.

What is $Aw$ when $w \in \operatorname{span}\{w_3,w_4,\ldots,w_n\}$?

The above hints show the rank is at most $2$. To conclude it is exactly $2$, you need to use linear independence of $u$ and $v$.

Answer 2

$$uu^T+vv^T=\begin{bmatrix}u & v \end{bmatrix} \begin{bmatrix} u^T \\ v^T\end{bmatrix}$$

Clearly, $\begin{bmatrix}u & v \end{bmatrix}$ is a rank $2$ matrix.

Let $\begin{bmatrix}u & v \end{bmatrix}=U\Sigma V^T$ be the singular value decomposition where $U\in \mathbb{R}^{n \times n}, \Sigma \in \mathbb{R}^{n \times 2}$ $V^T \in \mathbb{R}^{2\times 2}$.

The last $n-2$ rows of $\Sigma$ are zero matrices and the first two rows form a diagonal matrix. Let $$\Sigma=\begin{bmatrix} D \\ 0_{(n-2) \times 2} \end{bmatrix}$$

$$uu^T+vv^T=U\Sigma \Sigma^T U^T=U\begin{bmatrix} D^2 & 0_{2 \times (n-2)} \\ 0_{(n-2) \times 2} & 0_{(n-2)\times (n-2)}\end{bmatrix}U^T$$

Rank of $uu^T+vv^T$ is equal to the rank of $D^2$ which is $2$.

Answer 3

There are two parts, why the rank is at least $2$ and why it is at most $2$.

By choosing a vector $w$ with $u^T(w)\neq0$ and $v^T(w)=0$, which is possible since $u,v$ are linearly independent, we have $(uu^T+vv^T)(w)=u^T(w)u$, a nonzero multiple of $u$, so $u$ is in the image of $uu^T+vv^T$; by a similar argument $v$ is as well, so the rank is at least$~2$.

The rank of a sum of two linear maps can never exceed the sum of their ranks (its image is contained in the sum of its images), so the rank of $uu^T+vv^T$ is at most$~1+1=2$.