In diagonalization, can the eigenvector matrix be any scalar multiple? One can decompose a diagonalizable matrix (A) into A = C D C^−1, where C contains the eigenvectors and D is a diagonal matrix with the eigenvalues in the diagonal positions. So here's where I get confused. If I start with a random eigenvector matrix D
> D
     [,1] [,2]
[1,]    7    0
[2,]    0    5

and a random eigenvector matrix C
> C
     [,1] [,2]
[1,]    4    2
[2,]    1    3

There should be some matrix A with those eigenvectors and eigenvalues. When I compute A by multiplying C D C^-1 I get 
> A<-C%*%D%*%solve(C)
> A
     [,1] [,2]
[1,]  7.4 -1.6
[2,]  0.6  4.6

My understanding is that if I then work backwards and diagonalize A I should get the same matrix C and D, I started with to get A in the first place. But for reasons that escape me, I don't.
> eigen(A)
$values
[1] 7 5

$vectors
      [,1]      [,2]
[1,] 0.9701425 0.5547002
[2,] 0.2425356 0.8320503 

the first eigenvector is a multiple of column 1 of C and the second eigenvector is a multiple of column 2 of C. For some reason it feels strange that I have this relationship:
xC D (xC)^-1 = C D C    where x is a scalar. Did I screw up somewhere or is this true?

Well:
$$(xA)^{-1}=\frac{1}{x}A^{-1}\qquad x\neq 0$$ so the result is actually the same. Eigenvectors are vectors of an eigenspace, and therefore, if a vector is an eigenvector, then any multiple of it is also an eigenvector. When you build a matrix of eigenvectors, you have infinite of them to choose from, that program is calculating two of them that it wants, and they don't have to be the ones you chose at the beginning.
