why $ Rank(xy^T+yx^T)=2$ for nonzero and non-collinear $x,y \in R^n$

Question

Suppose we have nonzero $x,y \in R^n$ and $x,y$ are not collinear.

How to prove $Rank(xy^T+yx^T)=2$ ?

Answer 1

Let $v$ be a nonzero vector in $\text{span}\{x,y\}$ that is orthogonal to $x$. (Why does such a $v$ exist?) Then $(xy^\top + yx^\top)v = x(y^\top v)$, so $x$ is in the column space of $x y^\top + y x^\top$.

By a similar argument, one can show that $y$ is in the column space of $x y^\top + y x^\top$ (by choosing $v$ to be a nonzero vector in $\text{span}\{x,y\}$ that is orthogonal to $y$).

So far we have shown $$\text{span}\{x,y\} \subseteq \text{colspace}(x y^\top + y x^\top).\tag{1}$$

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Any $n$-dimensional vector can be written [uniquely] in the form $$ax+by+v$$ where $a$ and $b$ are scalars, and $v$ is orthogonal to both $x$ and $y$. Then any vector in the column space of $xy^\top + yx^\top$ can be written as $$(xy^\top + yx^\top)(ax + by + v) = x(a(y^\top x) + b (y ^\top y)) + y(a (x^\top x) + b(x^\top y)),$$ which shows that $$\text{colspace}(xy^\top + yx^\top) \subseteq \text{span}\{x, y\}.\tag{2}$$

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Combining (1) and (2) above yields $$\text{colspace}(xy^\top + yx^\top) = \text{span}\{x, y\}.$$

Taking the dimensions of both sides yields $$\text{rank}(xy^\top + yx^\top) = 2.$$

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Remark: non-collinearity of $x$ and $y$ is used in the proof of (1) as well as in the final step. It is not needed in the proof of (2).

Answer 2

Observe that $$\mathbf{x} \mathbf{y}^T+\mathbf{y}\mathbf{x}^T=\mathbf{x} \mathbf{y}^T+(\mathbf{x}\mathbf{y}^T)^T.$$ The following are fairly well-known result (see this forum):

1. $\text{rank}(A)=\text{rank}(A^T)$.
2. $\text{rank}(A+B) \leq \text{rank}(A)+\text{rank}(B)$. From these it follows that $$\text{rank}(\mathbf{x} \mathbf{y}^T+\mathbf{y}\mathbf{x}^T)=\text{rank}(\mathbf{x} \mathbf{y}^T+(\mathbf{x}\mathbf{y}^T)^T) \leq 2 \, \text{rank}(\mathbf{x}\mathbf{y}^T).$$

Let $\mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix}$ and $\mathbf{y}=\begin{bmatrix}y_1\\y_2\\ \vdots \\y_n\end{bmatrix}$. Then $$\mathbf{x} \mathbf{y}^T=\begin{bmatrix}x_1y_1 & x_1y_2& \ldots x_1y_n\\x_2y_1 & x_2y_2& \ldots x_2y_n\\\\\vdots& \vdots &\vdots\\x_ny_1 & x_ny_2& \ldots x_ny_n\end{bmatrix}.$$ It is clear that if $\mathbf{x},\mathbf{y} \neq \mathbf{0}$, then this matrix has rank $1$. Therefore, $$\text{rank}(\mathbf{x} \mathbf{y}^T+\mathbf{y}\mathbf{x}^T) \leq 2.$$ Now use non-collineraity to claim why it will be $=2$.

Answer 3

You mentioned that rank being at most 2 is clear to you so all that is needed is to show that it is at least two.

Since x and y are not parallel, take the component of x perpendicular to y and apply the matrix to it. Then reverse the roles of x and y in this procedure. You will find that the two outputs generated this way are proportional to x and y and are thus not colinear. So the rank is at least two.