Find eigenvalues and eigenvectors of the operator $A$ The question is: Find the eigenvalues and eigenvectors of the operator $A$ on $\Bbb{R}^3$ given by $A\mathbf{x}=|\mathbf{a}|^2 \mathbf{x}- (\mathbf{a} \cdot \mathbf{x}) \mathbf{a}$, where $\mathbf{a}$ is a given constant vector. How do you know without any calculations that $A$ must have an orthonormal eigenbasis? 
I have seem examples similar to this question. I'm wondering if there's any systematic way to solve this kind of questions. Someone showed to me that you first get $\mathbf{x}=\lambda\mathbf{a}$. What is the reason behind that and how does this help to solve the problem? 

Notice that $A$ is a symmetric matrix. Because $A_{ij}=(Ae_{j}\cdot e_{i})=|a|^{2}(e_{j}\cdot e_{i})-(a\cdot e_{j})(a\cdot e_{i})=(Ae_{i}\cdot e_{j})=A_{ji}$. So it is diagonalizable and has an orthonormal eigen basis.
Intuition behind the eigen vectors: Notice that $A'x:=(a\cdot x)a$ is a projection matrix projecting every vector along the direction of $a$. Modify a little bit and define $A''x:=\frac{1}{|a|^{2}}(a\cdot x)a$. Then $A''$ is an orthogonal projection, and $x-\frac{1}{|a|^{2}}(a\cdot x)a$ is the vector orthogonal to the direction of $\frac{1}{|a|^{2}}(a\cdot x)a$.
(Geometrically) It is clear that $a$ is an eigen vector of $A''$ corresponding to the eigenvalue $1$. Another eigenvalue of $A''$ is $0$ and eigenvectors are the vectors which are orthogonal to $a$, let's call them $a_{1}^{\bot}$ and $a_{2}^{\bot}$. 
Now if we look at $I-A''$, then it is a projection onto the space generated by $a_{1}^{\bot},a_{2}^{\bot}$. In fact, $(I-A'')a_{i}^{\bot}=a_{i}^{\bot}$ for $i=1,2$ and $(I-A'')a=0$, i.e., $a_{i}^{\bot}$ s are the eigenvectors of $(I-A'')$ corresponding to the eigenvalue $1$, and $a$ is an eigenvector of $(I-A'')$ corresponding to the eigenvalue $0$.
In our case $A=|a|^{2}(I-A'')$. Using the above intuition we find the eigenvalues of $A$ are $|a|^{2}, 0$ and the corresponding eigenvectors are $a_{1}^{\bot}, a_{2}^{\bot}$ and $a$ respectively.  
