Rank of $xy^{\mathrm T}+yx^{\mathrm T}$

Question

Suppose that $x\in\mathbb R^n$ and $y\in\mathbb R^n$. The matrices $xy^{\mathrm T}$ and $yx^{\mathrm T}$ are both of rank $1$. Is the sum of these two matrices of rank $1$, i.e. is the matrix $xy^{\mathrm T}+yx^{\mathrm T}$ of rank $1$? I think the answer is positive and I am trying to come up with an argument that shows that this is true.

In general, $$ \operatorname{rank}(A+B)\le\operatorname{rank}A+\operatorname{rank}B, $$ which means that $\operatorname{rank}(xy^{\mathrm T}+yx^{\mathrm T})\le 2$, but I think that the rank is actually equal to $1$.

Any help is much appreciated!

Answer 1

Denote $A = xy^T+yx^T$. If $x$ and $y$ are linearly independent then $\operatorname{rank}(A) = 2$.

Proof: Choose nonzero vectors $u, v \in \operatorname{span}(x, y)$ such that $x^T v = 0$ and $y^T u = 0$. Then $$ A v = x (y^T v) \, , \, A u = y (x^T u) $$ with $y^T v \ne 0 \ne x^T u$, so that both $x$ and $y$ are in the image of $A$.

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More general: If $x_1, \ldots, x_k$ are linearly independent and $y_1, \ldots, y_k$ are linearly independent, then $$ A = x_1 y_1^T + \ldots + y_k y_k^T $$ has the rank $k$. This follows from $$ A = \begin{pmatrix} x_1 & \ldots & x_k\end{pmatrix} \begin{pmatrix} y_1^T \\ \vdots \\ y_k^T \end{pmatrix} =: X \cdot Y $$ and Sylvester’s rank inequality: $$ k = \operatorname{rank}(X) + \operatorname{rank}(Y) - k \le \operatorname{rank}(A) \le k \, . $$

Answer 2

Here is a counterexample:

Let $x=\begin{pmatrix} 0\\1 \end{pmatrix} \in \mathbb{R}^2,y=\begin{pmatrix} 1\\0 \end{pmatrix} \in \mathbb{R}^2$.

Then $xy^T = \begin{pmatrix} 0& 0 \\1 & 0 \end{pmatrix}, yx^T = \begin{pmatrix} 0& 1 \\0 & 0 \end{pmatrix}$ which both have rank 1.

Yet $xy^T + yx^T = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ which has rank 2 as it is nonsingular.