TITLE: How to prove that following is rank-r modification of identity?
QUESTION [1 upvotes]: I was reading about Householder's reflection in Chapter 5 in Golub's Matrix Computation. Given that v is some n-dimensional vector, and $\mathbb{I}$ is $n \times n$ identity matrix,
$$Q = \mathbb{I}-\beta v v^{T}$$
where $\beta$ can be thought of a scalar, is rank-$1$ modification of identity. Now I could explicitly check that $v v^{T}$ has rank 1 because each row (or column) is proportional to every other row (or column). But when I have some general situation, like
$$Q = \mathbb{I} - W Y^{T}$$
where W,Y $\in \mathbb{R^{n\times r}}$, how to prove that $Q$ is now a rank-r modification of identity?

REPLY [1 votes]: Technically, this is a "rank-$\leq r$ modification" of identity, since the columns of $W,Y$ might not be linearly independent.
In the case where these are linearly independent, this is indeed a rank-$r$ modification. $Y^T$ has $r$ linearly independent rows, and since $W$ has full column rank we will have $\operatorname{rank}(WM) = \operatorname{rank}(M)$ for any matrix $M$.
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