Eigenvalue of $B=uv^\text{T}+wz^\text{T}$

Question

We have $u,v,w,z \in R^\text{n}$, how can we express the eigenvectors and eigenvalues of $B=uv^\text{T}+wz^\text{T}$ by analyzing over $u,v,w$ and $z$?

Answer 1

It is easy to obtain the eigenvalues. The matrix $B$ is at most rank-two (as a sum of two (at most) rank-one matrices) and can be expressed as $$ B=CD^T, \quad C:=[u,w], \quad D:=[v,z]. $$ Since $CD^T$ has the same nonzero eigenvalues as $D^TC$ (note that a few answers prove exactly this in contrast to the original question), the (at most two) nonzero eigenvalues of $B$ are the eigenvalues of $$ D^TC=\pmatrix{v^Tu&v^Tw\\z^Tu&z^Tw}. $$

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Obtaining eigenvectors can be tricky. For example, if $D^TC$ is diagonalizable, it does not mean that $CD^T$ is diagonalizable as well. Adding a few assumptions fixes the problem: assume that

• $C$ and $D$ have full column rank, that is, the two vectors forming their columns are linearly independent,
• $\mathrm{Im}(C)\cap\mathrm{Ker}(D^T)$ is trivial,
• $D^TC$ is diagonalizable.

Then if $x$ is an eigenvector of $D^TC$, $Cx\neq 0$ is an eigenvector of $CD^T$. The eigenvectors corresponding to the remaining zero eigenvalues can be chosen from the nullspace of $D^T$.