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> [!proposition] Proposition. ([[eigenvalues of outer product]])
> Let $x,y \in \mathbb{F}^{N}$. The [[matrix]] $xy'$ has a *single* possibly nonzero [[eigenvalue]] equal to $y'x$. All other [[eigenvalue]]s are $0$.
> \
> As a corollary, the [[eigenvalue]]s of $xy'$ are all equal to $0$ whenever $x$ and $y$ are [[orthogonal]].
> [!proposition] Lemma 1.
> $x,y \in \mathbb{F}^{N}$, the [[characteristic polynomial]] $\det(I-xy')$ is equal to $1-y'x$.
\
**Proof of Lemma 1.** We have $\begin{align}
\det (I-xy')= & (\det 1)\det(I-x (1)^{-1} y') \\
= & \det \begin{bmatrix}
1 & y' \\ x & I
\end{bmatrix} \ \ (*) \\
= & (\det 1) \det(1-y' I^{-1}x) \ \ (**) \\
= & \det(1-y'x) \\
= & 1-y'x
\end{align}$
where in $*$ and $**$ we apply the two det-matrix-inversion lemmas and the final equality follows as the [[determinant]] of a scalar.
> [!proposition] Lemma 2.
> As a corollary of **lemma 1**, we get that for $\lambda \neq 0$: $\begin{align}
\det (\lambda I_{N} - xy')=&\det \big( (\lambda I_{N})\left( I_{N}-\frac{xy'}{\lambda} \right) \big) \\
=& \det(\lambda I_{N}) \det (I_{N}- (\frac{1}{\lambda}x)y') \\
= & \lambda^{N} (1-y'(\frac{1}{\lambda}x)) \\
= & \lambda^{N} - \lambda^{N-1}y'x,
\end{align}$
and for $\lambda=0$: $\begin{align}
\det (\lambda I_{N}-xy')= & \det(-xy') \\
= & (-1)^{N} \det (xy') \\
= & \begin{cases}
0 & \text{ if } N > 1 \\
-xy & \text{ if } N=1,
\end{cases}
\end{align}$
\
where the last equality follows because $xy'$ has [[rank]] $1$ as an [[outer product]].
> [!proof]- Proof. ([[eigenvalues of outer product]])
We know that the [[eigenvalue]]s of the [[matrix]] $xy'$ may be computed via the [[characteristic polynomial]] $\det (xy' - \lambda I_{N})=0, $
which is $0$ when and only when $\det (\lambda I_{N}-xy')=0.$
Using **lemma 2** we see that the solutions to this polynomial in lambda are the solutions to $\lambda^{N} - \lambda^{N-1}y'x=0.$
Rearranging yields $\lambda^{N}=\lambda^{N-1}y'x,$
canceling then gives us $\lambda=y'x$ as one solution and $\lambda=0$ as the other $N-1$ solutions. This is what we wanted to show.
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#### References
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