Given a matrix $A$ with eigenvalue...

Question

Given a matrix $A$ with the eigenvalue $ \lambda $ and eigenvector $v$. Let $b$ be some vector. Show that the vector $v$ is also an eigenvector for the matrix $B = A-v b^T$ and construct a formula for the eigenvalues.

I have no idea how to begin this problem. Hints needed!

Answer 1

Act on $v$ by $B$ and expand using the distributive property. You get $Bv = Av - vb^Tv = \lambda v - vb^Tv$. What is an alternative way of writing $vb^T$? (Hint: it's a scalar that often crops up in the context of two vectors...)

Answer 2

$$(A-vb^{T})(v)=Av-vb^{T}v=(\lambda-b^{T}v)v$$ So the eigenvalue is $$\lambda -b^{T}v$$ This same formula holds for all eigenvalues $\lambda$ and eigenvectors $v$.