Given the transformation expression: $\;\;\boldsymbol A\cdot\,\boldsymbol b\ = \boldsymbol b - 2(\boldsymbol b\,\cdot\,\hat{\boldsymbol n})\hat{\boldsymbol n}\;\;\text{where}\; \boldsymbol A\cdot\,\boldsymbol b$ is the reflected vector of $\boldsymbol b$ from the reflection plane and $\hat{\boldsymbol n}$ is the unit vector normal to the reflection plane.
How do I get the transformation matrix $\boldsymbol A\;$ if for example, the unit vector $\hat{\boldsymbol n}$ is expressed as $2(\hat{\boldsymbol i} + \hat{\boldsymbol j} + \hat{\boldsymbol k} )$
I have tried expressing $\boldsymbol b$ in terms of it's components $b_{1}\hat{\boldsymbol i} + b_{2}\hat{\boldsymbol j}+b_3\hat{\boldsymbol k}\;$ but I get stuck when I open up the term $\;(\boldsymbol b\,\cdot\,\hat{\boldsymbol n})\hat{\boldsymbol n}\;$ as $\frac{ (b_{1}\hat{\boldsymbol i} + b_{2}\hat{\boldsymbol j}+b_3\hat{\boldsymbol k})\,\cdot 2(\hat{\boldsymbol i} + \hat{\boldsymbol j} + \hat{\boldsymbol k})}{12} \cdot 2(\hat{\boldsymbol i} + \hat{\boldsymbol j} + \hat{\boldsymbol k} )$
From the transformation expression $$\boldsymbol A\cdot\,\boldsymbol b\ = \boldsymbol b - 2(\boldsymbol b\,\cdot\,\hat{\boldsymbol n})\hat{\boldsymbol n}$$
One can simply derive that:
$$\boldsymbol A = \boldsymbol I - 2 \, \hat{\boldsymbol n}\,\hat{\boldsymbol n}^T$$
where $\boldsymbol I$ is the identity matrix
If, for example,
$$ \hat{\boldsymbol n} = \frac{1}{\sqrt 3}(\hat{\boldsymbol i} + \hat{\boldsymbol j} + \hat{\boldsymbol k} )$$
Then
$$\boldsymbol{A}_{i,j} = \delta(i-j) - \frac{2}{3}$$