Estabilish wheter there exists a basis of $R^3$ made of eigenvectors of P or not, and if so, wheter such a can be made of mutually orthogonal vectors $Pv=v-(w\cdot v)\cdot w$
Where $w=(1,−1,0)$
i.Write the matrix $A$ that represnts $P$ in the canonical bases of $\mathbb{R}^3$
ii.Find the eigenvalues and eigenvectors of $P$
iii. Estabilish wheter there exists a basis of $\mathbb{R}^3$ made of eigenvectors of $P$ or not, and if so, wheter such a can be made of mutually orthogonal vectors .
vi.Give the geometrical interpretation of the action of $P$
What i have done so far:
$Pv=(y,x,z)$ => $$\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$
eigenvalues and eigenvectors:
$$\begin{pmatrix}λ &1&0\\1&λ &0\\0&0&λ -1\end{pmatrix}$$
λ = 1
λ = -1
For λ = 1:
$$\begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}$$
X2=s X3=t
X=$S*\begin{pmatrix}-1\\1\\0\end{pmatrix}+t*\begin{pmatrix}0\\0\\1\end{pmatrix}$
For λ = -1:
$$\begin{pmatrix}-1&1&0\\1&-1&0\\0&0&-2\end{pmatrix}$$
X2=s 
$X=S*\begin{pmatrix}1\\1\\0\end{pmatrix}$
Finally: $$\begin{pmatrix}-1&0&1\\1&0&1\\0&1&0\end{pmatrix}$$
Here i stopped, i`m not sure in my computations and i do not understand the third question at all, can someone Explain simpler. Thx for attention
 A: I don't know what procedure have you to follow to calculate the eigenvectors, whether the standard or can be accepted some more "ad hoc", using the particularities of $P$. Anyway, I first use the second then the first.
The vectors $u$ orthogonal to $w$, having $u\cdot w=0$ are clearly eigenvectors of $P$ with eigenvalue $1$: $Pu=u-(u\cdot w)w=u$. These eigenvectors form a subspace of dimension two:
$(x,y,z)\cdot(1,-1,0)^T=0\implies x-y=0\implies u=(k,k,l),\;k,l\in\mathbb R$,
letting us to choose $f_1=(1,1,0)^T$ and $f_2=(0,0,1)^T$, orthogonals, $f_1\cdot f_2=0$
Further, $u=\lambda w$ is too an eigenvector: $P\lambda w=\lambda w-(\lambda w\cdot w)w=(1-w^2)\lambda w$, with eigenvalue $1-w^2=-1$. We choose the same $w$.
So, the answers to ii) and ii) are yet given: $f_1,f_2$ and $w$ are eigenvectors with eigenvalues $1$ and $-1$, form a basis and they are mutually orthogonal.
Geometrically: Consider the vector $v$ decomposed in the new basis: $v=a_1f_1+a_2f_2+a_3w$ and the action of $P$ on $v$
$Pv=P(a_1f_1+a_2f_2+a_3w)=P(a_1f_1+a_2f_2)+P(a_3w)=a_1f_1+a_af_2-a_3w$
The resulting vector has its third component in this basis reversed: $Pv$ is the mirror image of $v$ respect to the plane $x-y=0$.

$\vert P-\lambda I\vert=\begin{vmatrix}
 -\lambda&1&0\\
 1&-\lambda&0\\
 0&0&1-\lambda
\end{vmatrix}
=(1-\lambda)(\lambda^2-1)=0$
$\lambda=\pm1$
$\lambda=1$) $Pv=v$
$\begin{cases}
 y=x\\
 x=y
\end{cases}\;$, etc.
$\lambda=-1$) $Pv=-v$
$\begin{cases}
 y=-x\\
 x=-y\\
 z=-z
\end{cases}\;$, etc.
