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, $\lim\_{n\rightarrow\infty}x^n = \lim\_{n\rightarrow\infty}P^n\vec{x}=\vec{x}$