Eigenvalues and Eigenspaces of a Projection

Question

Let $P$ be the orthogonal projection onto a subspace $E \subset V$ ($V$ being an inner product space) with $\mathrm{dim(V)}=n$, $\mathrm{dim(E)}=r$. Obtain the eigenvalues and eigenspaces, along with their algebraic/geometric multiplicities.

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I tried to first let $B = (b_{1}, ..., b_{r})$ be an orthonormal basis of $E$. By the projection formula, we will have (let $T$ be the projection mapping):

$T(v) = \displaystyle\sum_{i=1}^{r} (v,v_{i})v_{i}$

Then I tried finding the transformation matrix $T$ such that $T(v) = Tv$ (say in the standard basis), but I haven't had any luck finding the transformation matrix (which would then allow us to compute the respective eigenvalues and eigenspaces). I was thinking of perhaps choosing the respective basis to $B$ and then completing it to a basis of $V$ (say Steinitz' Lemma), but I'm not sure if this is the way to proceed.

What would be the way to approach this exercise?

Answer 1

Notice that $Pv=v$ if $v \in E$.

Also notice that $Pv=0$ if $v \in E^{\perp}$, i.e, if $v$ is orthogonal to every vector in $E$.

Then notice that $V=E \oplus E^{\perp}$ (in the informal sense), and use this to show that the only eigenvalues are $0$ and $1$, and the respective eigenspaces are $E^{\perp}$ and $E$.

Answer 2

You could approach this on a conceptual rather than a computational level. First, remember that to say a vector $\def\v{{\bf v}}\v$ is an eigenvalue of the transformation $T$, with eigenvalue $\def\l{\lambda}\l$, means that $$T(\v)=\l\v\ .$$ Next, draw a diagram to illustrate a projection. Projection of $\Bbb R^3$ onto a plane will do as a schematic diagram, no matter the number of dimensions in your question. Make sure your diagram shows a plane $P$, a vector $\v$ and the projection $T(\v)$.

Now, is there any way that the projection could be a scalar times $\v$? For example, can you draw a case in which the projection is twice $\v$, that is, $$T(\v)=2\v\ ?$$ With a bit of thought, you should be able to see (hint) that there are only two values of $\l$ for which it is possible that $T(\v)=\l\v$, and you should be able to find the corresponding possibilities for $\v$ in each case.

Answer 3

orthogonal projection $P$ from $V$ onto the subspace $E$ of $V$ decomposes into an identity on $E$ and $0$ on $E^\perp$ so the eigenvalues of $P$ are $1$ repeated dimension of $E$ times and the rest are $0$ repeated. the corresponding eigenvectors are the basis of $E$ and $E^\perp$ respectively.

Answer 4

The included images illustrate the orthogonal projection(the general case), the orthogonal projection of a vector $v\in E$ ($v$ is an eigenvector corresponding to the eigenvalue 1), respectively the orthogonal projection of a vector in $E^\perp$ ($v$ is an eigenvector corresponding to 0)