On a math note, if I understand you, the "given" vectors are the row vectors of the matrix {{-3,2},{-15,8}}, that is the vectors {-3, 2} and {-15, 8}. I do not think there is any general relationship between the row vectors and eigenvectors of a matrix.

The first column of a matrix mat is the vector you get from the product mat.{1,0}. The second column is the result of mat.{0,1}. An eigenvector is a (nonzero) linear combination of the columns that happens, when multiplied by mat, to yield a scalar multiple of itself. (Of course, every product mat.{x,y} is a linear combination of the columns of mat, so the scalar-multiple property is key for eigenvectors.)

You could use Ratios to get the 3:1-type analysis or normalize the eigenvectors to have one of its components to be 1:

evecs = Eigenvectors[N@{{-3, 2}, {-15, 9}}];
Ratios /@ evecs
Normalize[#, Min[Abs[Select[#, # != 0 &]]] &] & /@ evecs
Normalize[#, Norm[#, Infinity] &] & /@ evecs
#/First[#] & /@ evecs
(*
{{-0.230336, -0.973111}, {-0.49079, -0.871278}}  <-- e-vectors
{{4.22474}, {1.77526}}                           <-- ratios of components
{{-1., -4.22474}, {-1., -1.77526}}               <-- smallest component ±1
{{-0.236701, -1.}, {-0.563299, -1.}}             <-- largest component ±1
{{1., 4.22474}, {1., 1.77526}}                   <-- first component 1 (beware ÷0)
*)

Happy holidays to you, too!

Thanks for the information. I am still trying to understand exactly what the unit circle with vectors pointing outward is indicating. Looking at vectors in the plane, it is somewhat easy to identify and understand the flows, but the unit circle vectors is a bit of a stretch for me. In any event, thanks for the help, and have a very merry Christmas and a wonderful New Year.

Mitch Sandlin