Projector of an eigenvector

Question

If I have an eigenvector $v$ and I want to find the associated projector of this eigenvector (orthogonal onto its subspace), am I right to say the projector is $P = vv^T$?

Does $P = vv^T$ only holds true for eigenvectors or does it holds true for any vectors?

Answer 1

You can build such a projector for an arbitrary vector (not necessarily an eigenvector of some matrix) but as Michael Burr points out you want to normalise $|v| = 1$. To see why it works consider the following:

Define $u' = P(v)u = vv^{T}u$ and consider its properties.

1. $P(v)u' = v |v|^{2} v^{T}u = |v|^{2}vv^{T}u = |v|^{2}P(v)u$ which means that $P(v)^{2} = P(v)$ if the vector is normalised to unit length.
2. $ v^{T}u' = v^{T}v\, v^{T}u = |v|^{2} \, v^{T}u$ so with the same normalisation we get $v^{T}u$ which is the projection of $u$ along $v$ as desired.
3. if $w$ is orthogonal to $v$ then $w^{T}u' = w^{T}v \, v^{T}u = 0$ so $u'$ is orthogonal to vectors perpendicular to $v$.

Points 2 and 3 imply that $u'$ is parallel to $v$, and as such $u$ been projected oto $v$.