Let $e = \{a,b,c\}$ be an orthonormal basis of some inner product $\circ$ on vector space $\Bbb{R}^3$. Linear map $L:\Bbb{R^3}\to \Bbb{R^3}$ given by $$L x=(a\circ x)a-(c\circ x)b-(b\circ x)c$$ The first two questions were
a) Show that $a-b+c$ is an eigenvector of linear map $L$.
b) Find the matrix of linear map $L$ with respect to basis $e$
c) Find the matrix of $L^n,n\in\Bbb{N}$, with respect to basis $e$
I've included the parts of a) and b) because I think they may be relevant.
Now if I'm correct $a\circ b= 0,a\circ c=0$ because the vectors are orthogonal, and $a\circ a=1$ because $a$ is a unit vector from additivity it follows that $$L(a-b+c)=a-b+c$$ since we have $L(x)=x$ for $x=a-b+c$ it means $a-b+c$ is an eigenvector. However I feel this part should've been done with the matrix of $L$ instead. For b) I've tried $$L(a)=a\\L(b)=-c\\L(c)=-b$$ So I think if $a=(a_1,a_2,a_3),b=(b_1,b_2,b_3),c=(c_1,c_2,c_3)$ the matrix should look something like, $$L =\begin{bmatrix}a_1 & a_2 & a_3 \\-c_1 & -c_2 & -c_3\\-b_1& -b_2 & -b_3\end{bmatrix}$$ However I'm not sure if this is correct or how could I use the matrix $L$ to find $L^n$.
No it isn’t correct because you are found that
$L(a)=a$
$L(b)=-c$
$L(c)=-b$
but with respect to base $e$ you have that the matrix associate to $L$ is that Matrix that has columns the vectors $L(a),L(b),L(c)$ written in the base $e$ and so the matrix is
$M=\left[\begin{matrix}1& 0& 0\\ 0&0&-1\\0&-1&0\end{matrix}\right]$
because,for example, $L(b)$ in the base $e$ is
$L(b)=-c=0 a+0 b+(-1)c$
You can observe that $M^2=I$ and so $M^{2n}=I$ for every $n>1$ and $M^{2n+1}=M$ for every $n>1$