Eigenvalues of rank 2 perturbation of the identity Let $I$ be the identity matrix, and $u, v\in\mathbb{R}^n$. What can we say about the eigenvalues of the following matrix?

$$2I+vu^{\mathsf{T}}+uv^{\mathsf{T}}$$

I'm particularly interested in the minimum eigenvalue of this matrix in terms of $u$ and $v$.
Suppose $u, v$ are linearly independent. Then
$$x = \|v\|u + \|u\|v$$
is non-zero, and
\begin{align*}
Ax &amp;= (uv^\top + vu^\top)(\|v\|u + \|u\|v) \\
&amp;= \|v\|uv^\top u + \|u\|uv^\top v + \|v\|vu^\top u + \|u\|vu^\top v \\
&amp;= \|v\|(u \cdot v) u + \|u\| \|v\|^2 u + \|v\|\|u\|^2 v + \|u\|(u \cdot v) v \\
&amp;= (\|u\| \|v\| + u \cdot v)x.
\end{align*}
Hence $\|u\| \|v\| + u \cdot v$ is an eigenvalue for $A$. Note that, since $u, v$ are linearly independent, this eigenvalue is strictly positive, making it the maximum eigenvalue.

Similarly, if $y = \|v\|u - \|u\|v \neq 0$, then $Ay = (u \cdot v - \|u\| \|v\|)y$, hence we also obtain a strictly negative eigenvalue $u \cdot v - \|u\| \|v\|$.

Since $A$ is the sum of two rank $1$ matrices, it is of rank at most $2$. Hence, the only other eigenvalue is $0$, making $u \cdot v - \|u\| \|v\|$ the minimum eigenvalue.

If $u = 0$ or $v = 0$, then $A = 0$, and the eigenvalues are all $0$. Note that this agrees with the given formula for the minimum eigenvalue.

If $v = ku \neq 0$ for some $k$, then $A$ is rank $1$ with eigenvector $u$ and corresponding non-zero eigenvalue $2k\|u\|^2$. If $k &lt; 0$, then this is a negative number, and thus the minimum eigenvalue. Note that $2k\|u\|^2 = u \cdot v - \|u\|\|v\|$, which is to say that the formula continues to hold. If $k &gt; 0$, then the minimum eigenvalue is $0$, which is again equal to $u \cdot v - \|u\| \|v\|$.

Therefore, in any case, the minimum eigenvalue for $A$ is $u \cdot v - \|u\| \|v\|$, and indeed the eigenvalues are, up to multiplicity, in ascending order,
$$u \cdot v - \|u\| \|v\|, \underbrace{0, \ldots, 0}_{n - 2 \text{ times}}, u \cdot v + \|u\| \|v\|.$$
The eiegenvalues for $A + 2I$ are the same as $A$, but shifted by $2$.
Rank of $vu^T$ is $1$ same is the case for $uv^T$.

Both $n\times n$, so in total $n$ eigenvalues for both.

Now out of those $n$, $n-1$ will be $0$ for both

The last eigenvalue will be $&lt;u,v&gt;$, that is the inner product of $ u$ and $v$. Can you see this?

Finally minimum eigenvalues will be $$ 2+ min\{0, &lt;u,v&gt;\}$$