Eigenvalues, eigenvectors, and powers of the projection onto a subspace specified by a basis

Question

Let $U$ be a $3$-dimensional subspace of $\mathbb{R^6}$ with basis vectors $\{\vec{u_1}$,$\vec{u_2}$,$\vec{u_3} \}$ and $P$ be the matrix which projects $\mathbb{R^6}$ onto $U$.

(a) What are the eigenvalues, and eigenvectors of $P$?

(b)Let $\vec{x} \in \mathbb{R^6}$ be an arbitrary vector, describe the limit as $n\rightarrow \infty$ of $\vec{x_n} = P^n\vec{x}$

(a) I tried to understand $U$ by defining it to be:

$U=\{\vec{v} \in \mathbb{R^6} :\vec{v}=\vec{u_1}+\vec{u_2}+\vec{u_3} \}$

More importantly I thought about what are the possible eigenvalues and eigenvectors of a projection matrix.

For every $v \in U$, $Pv=v$ are the eigenvectors with $\lambda=1$.

And, for every $v \bot U$, $Pv=0$ are the eigevectors with $\lambda=0$.

I am not sure what the definite answer is, i think that the basis vectors are linearly independent and should have 3 eigenvalues but not sure if they would all be equal to $\lambda=1$ in this case.

(b)my attempt at N-epsilon proof.

Pf:

Let $\vec{x} \in \mathbb{R^6}$ and $\forall \epsilon > 0$ .

Choose $m$ s.t. $\vec{x}-\epsilon \leq P^m\vec{x}$ and $N$,

s.t. $P^m\vec{x}=P^mP^n\vec{x}\leq P^n\vec{x}$,$\forall n>N$.

Thus $\vec{x}-\epsilon \leq P^n\vec{x} \leq \vec{x}$, for sufficiently large $n$.

Hence, $\lim_{n\rightarrow\infty}x^n = \lim_{n\rightarrow\infty}P^n\vec{x}=\vec{x}$

Answer 1

For (a), you correctly identified the eigenvectors and eigenvalues: anything in $U$ is an eigenvector with eigenvalue $1$ and anything in $U^\perp$ is an eigenvector with eigenvalue $0$. So there are six eigenvectors (up to linear independence) which form a basis for $\mathbb R^6$. (The fact that $U$ is three dimensional is not relevant for counting the eigenvectors!)

For (b), notice that $P^2 = P$. (Try to visualise this - projecting twice is the same as projecting once.) Does this help?

Also notice that you can decompose any $x$ uniquely as $x = u + u^\perp$, where $u \in U$ and $u^\perp \in U^\perp$ - does this help?

Answer 2

Somehow, a mild algebraic point of view would be helpful here:

$P$ is an idempotent operator, meaning that projecting twice is the same as projecting once. In other words, $P^2=P$.

This should handle part $1$. You were right about the eigenvectors, and I think that you identified the relevant information.