Q:

Finding eigenvalues

This is the question from my homework that was due Monday.  So right now i just want to understand how to do them in order to be ready for the final exams. 
For b), c), d), e) i found three vectors that lie on the plane and from there i solved b), c), d), e).  And it is not the right way, since the question asks to do them all without calculating matrix [f]  
He said the for question b, there is going to two eigenvalues since it is reflection.  And he wrote a really complicated solution for b). 
Can someone help me with that?  It is really frustrating, since i understood this topic pretty well and when i saw these question, i did even know from where to start.  

A:

(a) The unit normal vector to the plane is $(3, -2, 1)/\sqrt{14}$. We could find a vector within the plane by choosing an arbitrary vector and subtracting its projection onto the unit normal vector, but it is easier to just eyeball it: $$(1,1,-1)\cdot(3,-2,1) = 0$$
To get a 2nd vector, an easy method (since we are in $\Bbb R^3$) is to just take the cross product:
$$(3,-2,1)\times(1,1,-1) = (1,4,5)$$
So an orthonormal basis would be $\{(1,1,-1)/\sqrt 3, (1,4,5)/\sqrt{42}\}$.
(b) Let $$\begin{align}v_1 &= \left(\frac 3{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac 1{\sqrt{14}}\right)\\v_2 &= \left(\frac 1{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac {-1}{\sqrt{3}}\right)\\v_3 &=\left(\frac 1{\sqrt{42}}, \frac{4}{\sqrt{42}}, \frac 5{\sqrt{42}}\right)\end{align}$$
Then $B = \{v_1, v_2, v_3\}$ is an orthonormal basis for $\Bbb R^3$. Since $f$ is reflection in the plane, it leaves $v_2, v_3$ unchanged and reverses $v_1$. I.e.:$$f(v_1) = -1v_1$$ $$f(v_2) = 1v_2$$ $$f(v_3) = 1v_3$$
Obviously $-1$ and $1$ are eigenvalues of $f$, $\operatorname{span}\{v_1\}$ is a subspace of the eigenspace of $-1$, while $\operatorname{span}\{v_2,v_3\}$ is within the eigenspace of $1$. Since these together span all of $\Bbb R^3$, they must be the entire eigenspaces. 
(c) $Q$ is constructed from the three eigenvectors $v_1,v_2, v_3$ (I'll leave it to you to figure out if you need need them as rows or columns), $D$ is the matrix of eigenvalues $$D=\begin{bmatrix}-1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$
(d) $$(QDQ^T)^T = (Q^T)^TD^TQ^T = QDQ^T$$
(e) The determinant is the product of the eigenvalues $= (-1)*1*1 = -1$

