eigen value of the $n \times n $ Matrix $x y^T$

Question

Let $x,y$ be two non zero $n \times 1$ vectors. If $y^T$ denotes the transpose of the matrix $y$, then what are the eigen value of the $n \times n$ matrix $x y^T$?

Attempt: Denoting $x= \big(x_1,\cdots,x_n \big)^T, y = (y_1,\cdots,y_n)^T$

Then, To find the eigen value of $xy^T,$ we solve the system : $xy^T v = \lambda v~:~v=\big(v_1,\cdots,v_n)^T \in \Bbb R^n$

Expanding, we obtain the following equation system:

$$x_1(y_1v_1+\cdots+y_nv_n)=\lambda v_1 \\x_2(y_1v_1+\cdots+y_nv_n)=\lambda v_2 \\.\\.\\.\\x_n(y_1v_1+\cdots+y_nv_n)=\lambda v_n \\$$

How does one move ahead and solve this system?

Answer 1

The case $n=1$ is trivial. Let's see what happens for $n=2$ just to get a flavor: $$ \begin{cases} x_1(y_1v_1+y_2v_2)=\lambda v_1 \\[6px] x_2(y_1v_1+y_2v_2)=\lambda v_2 \end{cases} $$ We want to identify a nonzero eigenvalue $\lambda$. The system becomes $$ \begin{cases} (x_1y_1-\lambda)v_1+x_1y_2v_2=0 \\[6px] x_2y_1v_1+(x_2y_2-\lambda)v_2=0 \end{cases} $$ and we know from elementary arguments that we need $$ (x_1y_1-\lambda)(x_2y_2-\lambda)-(x_2y_1)(x_1y_2)=0 $$ which easily yields $$ \lambda^2-(x_1y_1+x_2y_2)\lambda=0 $$ So a nonzero eigenvalue, if any, has to be $\lambda=x_1y_1+x_2y_2$.

Now the extension should be evident, isn't it?

If we take $v=x$, then your system is satisfied with $\lambda=x_1y_1+x_2y_2+\dots+x_ny_n$. Since $x\ne0$, we're done.

Can there be other eigenvalues? Certainly so, for $n>1$, but with the restriction that they're $0$, because $xy^T$ has rank $1$ and so nullity $n-1$.

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In a faster way: for three column vectors $u,v,w$, it holds $(uv^T)w=(v^Tw)u$, where $v^Tw$ is considered as a scalar. This identity is proved easily by direct computation. Take $u=w=x$ and $v=y$.