Eigenvalues of rank 2 perturbation of the identity

Question

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 $v$.

Answer 1

Suppose $u, v$ are linearly independent. Then $$x = \|v\|u + \|u\|v$$ is non-zero, and \begin{align*} Ax &= (uv^\top + vu^\top)(\|v\|u + \|u\|v) \\ &= \|v\|uv^\top u + \|u\|uv^\top v + \|v\|vu^\top u + \|u\|vu^\top v \\ &= \|v\|(u \cdot v) u + \|u\| \|v\|^2 u + \|v\|\|u\|^2 v + \|u\|(u \cdot v) v \\ &= (\|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 $k < 0$, 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 $k > 0$, 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 $2$.

Answer 2

The example of solution for dim $ 3 \times 3$.

Expressed $u$ and $v$ as $u=[1 \ \ 0 \ \ 0]^T$ and $v=[x\ \ y \ \ z ]$ in some frame.

Then $ A=uv^T+vu^T=\begin{bmatrix} 2x & y & z \\ y & 0 & 0 \\ z & 0 & 0 \end{bmatrix} $.

It's rank 2 symmetric matrix with eigenvalues:

$\lambda_1 = x+ \sqrt{x^2 + y^2 + z^2} $

$\lambda_2 = x- \sqrt{x^2 + y^2 + z^2}$
$\lambda_3 = 0 $.

The eigenvalues of $A+2I$ are ${\lambda_1+2 ,\lambda_2+2,2}$.

Answer 3

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 $<u,v>$, that is the inner product of $ u$ and $v$. Can you see this?

Finally minimum eigenvalues will be $$ 2+ min\{0, <u,v>\}$$