Show that there exists a matrix $A$ such that $R_a(x)=Ax$ for all $x\in \mathbb R^n$.

Let $a(\neq 0)\in \mathbb R^n$. Let us define the reflection operator $R_a=x-2\frac{x.a}{a.a}a$. Show that there exists a matrix $A$ such that $R_a(x)=Ax$ for all $x\in \mathbb R^n$ and prove that $\det A=-1$.

My try: I chose the standard basis vectors of $\mathbb R^n$ as $e_1,e_2,\dots,e_n$. Then I took $R_a(e_i)$, expressed it in terms of $e_i$ and got the matrix $A$.

My questions are twofold:

Is there any anlternate way to find the matrix $A$.
How to find $\det A$. I find that it is easy to calculate $\det A$ for $n=2,3$ but what is the way out for higher dimensions? Can someone please help me out?

Well one way to do this is without explicitly finding the matrix at all! I think this is probably the most elegant: note that linear maps are in correspondence with with matrix operators, so all you need to do is check that $R_a$ is a linear map! And this is easy: $$ R_a(\lambda x + y) = \lambda x + y - 2\frac{(\lambda x + y) \cdot a}{a \cdot a} a = \lambda x + y - \lambda \cdot 2 \frac{x \cdot a}{a \cdot a} a - 2\frac{y \cdot a}{a \cdot a} a = \lambda R_a(x) + R_a(y). $$

As for the determinant calculation, let's avoid matrices for that too. Recall that the determinant of a linear map (or equivalently a corresponding matrix) is the product of its generalized eigenvalues with multiplicity. So, we just need to find the eigenvalues of $R_a$.

First note that if $x$ any vector orthogonal to $a$ then $R_a(x) = x - 2\frac{x \cdot a}{a \cdot a} a = x$, so $x$ is an eigenvector with eigenvalue $1$. Since the orthogonal complement of the span $S = \operatorname{span}(a)$ in $\mathbb{R}^n$ is $n - 1$ dimensional, this means that we have found $(n - 1)$ eigenvalues which are all $1$. On the other hand the vector $a$ is not in the orthogonal complement $S^\perp$, and we have $R_a(a) = a - 2\frac{a \cdot a}{a \cdot a} a = -a$, so $a$ is an also an eigenvector, this time with eigenvalue $-1$!

Thus we have (without doing any matrix calculations) pinned down the $n$ generalized eigenvalues of $R_a$: they are all $1$, except for a single $-1$. Thus $$ \det(R_a) = \underbrace{1 \cdot 1}_{\text{$n-1$ times}} \cdot (-1) = -1, $$ as desired.

As asked in comments: If we want to compute the $n \times n$ matrix which explicitly represents the action of $R_a$ on $\mathbb{R}^n$ the recipe is just that it will have $i$th column equal to $R_a(e_i)$, where $\{e_i\}$ is the standard basis of $\mathbb{R}^n$. If $a \in \mathbb{R}^n$ is arbitrary (let's write $a = \sum_{j = 1}^n a_j e_j$) then this is perhaps little bit yuck: $$ R_a(e_i) = e_i - 2\frac{e_i \cdot a}{a \cdot a} a = e_i - 2\frac{a_i}{a \cdot a} \sum_{j = 1}^n a_j e_j. $$ So in general this is going to look like: $$ [R_a]_{\{e_j\}} = \begin{bmatrix} 1 - 2\frac{a_1^2}{a \cdot a} & - 2\frac{a_1 a_2}{a \cdot a} & \cdots & -2 \frac{a_1 a_n}{a \cdot a} \\ - 2\frac{a_1 a_2}{a \cdot a} & 1 - 2\frac{a_2^2}{a \cdot a} & \cdots & - 2\frac{a_2 a_n}{a \cdot a} \\ -2 \frac{a_1 a_3}{a \cdot a} & - 2\frac{a_2 a_3}{a \cdot a} & \cdots & - 2\frac{a_3 a_n}{a \cdot a} \\ \vdots & \vdots & \ddots & \vdots \\ - 2\frac{a_1 a_n}{a \cdot a} & - 2\frac{a_2 a_n}{a \cdot a} & \cdots & 1 - 2\frac{a_n^2}{a \cdot a} \end{bmatrix}. $$