Orthogonally diagonalizing a matrix given a linear mapping, without calculating the original matrix Let f:R^3 ---> R^3 be the linear mapping which reflects over the plane 3x-2y+z=0. The goal of this question is to orthogonally diagonalize the matrix [f] without calculating [f].
(a) Determine an orthanormal basis for the plane
(b) Determine all eigenvalues of [f]. For each eigenvalue, describe it's eigenspace, and find an orthanormal basis for it. 
(c) Find an orthogonal matrix Q and a diagonal matrix D such that Q^T[f]Q=D
(d) explain why [f] must be symmetric
(e) determine det[f]
I can get part (a), but I have no idea how to go about getting the eigenvalues...  
 A: Hint
Take two linearly independent vectors of the given plane and construct from it an orthonormal basis of the plane $(v_1,v_2)$ using the Gram-Schmidt process. Let $v_3=v_1\wedge v_2$. Notice that $f(v_1)=v_1$ and $f(v_2)=v_2$ and $f(v_3)=-v_3$ and the eigenvalues are $1,1,-1$. The matrix $Q=(v_1\; v_2\;v_3)$ and $D=\operatorname{diag}(1,1,-1)$. Notice that $Q$ is orthogonal i.e. $Q^T=Q^{-1}$
so $[f]$ is symmetric. The determinant of $[f]$ is the product of the eigenvalues.
A: i think you can find an orthonormal basis without gram-schmidt. here is how:
first find three orthogonal vectors two of which are in the plane. pick $v_1 = [3, -2, 1]^T$. now a vector orthogonal to this is $v_2 = [0,1,2]^T$ finding a vector orthogonal to these two vectors is the null space of a matrix made up these two rows. from the selection of the rows this matrix is already in row echelon form with $z$ being the free variable.  so $v_3 = [5, 6,-3]^T$. the reflection $f$ has this effect on these vectors: $f(v_1) = -v_1, f(v_2) = v_2$ and $f(v_3) = v_3.$ with $q = [v_1/|v_1|, v_2/|v_2|, v_3/|v_3|],$ we have $T[f]q = q\  diag(-1, 1, 1)$ and $q^Tq = qq^T = I_3.$    
