Eigenvectors of projection from $\mathbb R^3$ to $\mathbb R^2$.

Question

Say I have a linear transformation that projects from $\mathbb R^3$ to $\mathbb R^2$. Do eigenvectors exist for this specific transformation? Does the same apply when I project from $\mathbb R^2$ to $\mathbb R$?

Answer 1

No. The concept of eigenvector only makes sense for linear maps $T$ from a vector space $V$ into itself, since it is a vector $v\in V$ such that $T(v)=\lambda v$, for some scalar $\lambda$.

Answer 2

To follow up on @JoseCarlosSantos's answer:

On the other hand, because $\Bbb R^2$ is easily embedded in $\Bbb R^3$, as the subset consisting of vectors $(x, y, 0)$, such a "projection" is often written with a $3 \times 3$ matrix. In the case of the simplest possible projection (along the $z$-axis) the matrix is $$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0}$$, and it has eigenvalues $1$ (with any vector in the $xy$-plane as eigenvectors) and $0$), with any vector along the $z$-axis as eigenvector.

In this situation, we say that a matrix $M$ represents a "projection" if $M^2 = M$ (i.e., projecting twice is the same as projecting once). That means that $(M^2 - M)v = 0v = 0$ for any $v$. In particular, if $b$ is an eigenvalue, with eigenvector $u$, it tells you that $$ (b^2 - b) u = 0u $$ so that $b^2 - b = 0$, i.e., $b = 0,1$. Hence the only eigenvalues of a projection are $0$ and $1$.

With this definition, the identity is a kind of "degenerate" projection, because it doesn't lower dimension, and the zero-matrix is the other degenerate one: the "projected to" subspace is $0$-dimensional.