7
$\begingroup$

How can I find the eigenvalues of the sum of rank-one matrices $vv^T + ww^T$?

I know that the eigenvalues of each term is $v^Tv$ and $w^Tw$, respectively.

  • 0
    To say things properly, the **only nonzero eigenvalue** of $vv^T$ is $v^Tv$.2012-02-22

3 Answers 3

0

You cannot expect to be able to express the eigenvalues of $vv^T + ww^T$ from the eigenvalues of each of the two summands.

For instance, given $t\in[0,1]$ let $$ v=\begin{bmatrix}1\\ 0\end{bmatrix}, \ \ w=\begin{bmatrix}t\\ \sqrt{1-t^2}\end{bmatrix}. $$ Then $v^Tv=1$, $w^Tw=1$, which "forgets" about $t$. But $$ vv^T+ww^T=\begin{bmatrix}1+t^2&t\sqrt{1-t^2}\\ t\sqrt{1-t^2}&1-t^2\end{bmatrix}, $$ which has eigenvalues $1+t,1-t$.

10

Let $A=vv^T+ww^T$, then $A$ has at most rank $2$. So we want to know the potentially non-zero eignevalues. We have $\operatorname{Tr}(A)=|v|^2+|w|^2$ and $\operatorname{Tr}(A^2)=|v|^4+|w|^4+2\langle v,w\rangle^2$ so these eigenvalue must satisfy $\lambda_1+\lambda_2=|v|^2+|w|^2$ and $\lambda_1\lambda_2=|v|^2|w|^2-\langle v,w\rangle^2$. So $\lambda_1$ and $\lambda_2$ are the roots of $X^2-(|v|^2+|w|^2)X+|v|^2|w|^2-\langle v,w\rangle^2$. The discriminant is $$\Delta=(|v|^2+|w|^2)^2+4\langle v,w\rangle^2-4|v|^2|w|^2=(|v|^2-|w|^2)^2+4\langle v,w\rangle^2>0$$ (or equal to $0$ if $|v|=|w|$ and $v$ and $w$ are orthogonal; in this case we have only one root) so $$\lambda_1=\frac{|v|^2+|w|^2-\sqrt{(|v|^2-|w|^2)^2+4\langle v,w\rangle^2}}2$$ and $$\lambda_2=\frac{|v|^2+|w|^2+\sqrt{(|v|^2-|w|^2)^2+4\langle v,w\rangle^2}}2,$$ the other eigenvalues being $0$.

  • 0
    can you explain better the implication $Tr(A^2)=|v|^4+|w|^4+2(v,w)^2 \Rightarrow \lambda_1\lambda_2=|v|^2+|w|^2-(v,w)^2$?? I understand that $\lambda_1^2+\lambda_2^2=Tr(A^2)$... Is it a straightforward computation or am I forgotting some theorem? Thanks in advance S.2014-02-13
  • 0
    We can expand $A^2$ as $vv^Tvv^T+2vv^Tww^T+ww^Tww^T$ then compute the trace. Then $2\lambda_1\lambda_2=(\lambda_1+\lambda_2)^2-\lambda_1^2-\lambda_2^2$.2014-02-13
  • 0
    This is excellent!2014-11-14
7

Let $u_1$ and $u_2$ be orthonormal vectors such that $v = a u_1$ and $w = b u_1 + c u_2$ for some scalars $a,b,c$. Thus in an orthonormal basis starting with $u_1, u_2$, the matrix of $v v^T + w w^T$ has first two rows and columns $\pmatrix{a^2 + b^2 & bc \cr bc & c^2\cr}$ and everything else $0$. The nonzero eigenvalues of your matrix are the eigenvalues of this.

  • 0
    I should have mentioned, $v,w$ could have any $n>1$ elements.2012-02-22
  • 0
    It doesn't matter, Robert's answer still applies. As he says, the "first two rows and columns" are the matrix mentioned. What is not completely clear to me (at least without writing it!) is how easy it is to find $u_2$ and $b,c$ in general.2012-02-22
  • 1
    To find $u_2$, use Gram-Schmidt. As for $b$ and $c$, they are $u_1^T w$ and $u_2^T w$ respectively.2012-02-23