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Topic: Intersection over an index
Replies: 3   Last Post: Oct 18, 2012 2:44 AM

 Messages: [ Previous | Next ]
 Bob Hanlon Posts: 906 Registered: 10/29/11
Re: Intersection over an index
Posted: Oct 18, 2012 2:42 AM

intersectM[m1_?MatrixQ, m2_?MatrixQ] :=
Select[m2, MemberQ[m1, #] &];

intersectEV[m : {__?MatrixQ}] := Module[
{ev = Eigenvectors /@ m},
Fold[intersectM[#1, #2] &, First[ev], Rest[ev]]]

A[1] = {{-1, -3, 1}, {0, -3, 0}, {-1, -1, -1}};

Eigenvectors[A[1]]

{{1, 1, 1}, {-I, 0, 1}, {I, 0, 1}}

A[2] = {{-2, -1, 1}, {0, -1, 0}, {-1, 1, -2}};

Eigenvectors[A[2]]

{{-I, 0, 1}, {I, 0, 1}, {0, 1, 1}}

A[3] = {{-2, -1, -1}, {0, -1, 0}, {1, -1, -2}};

Eigenvectors[A[3]]

{{I, 0, 1}, {-I, 0, 1}, {0, -1, 1}}

A[4] = {{-2, -1, 1}, {0, -1, 0}, {1, -1, -2}};

Eigenvectors[A[4]]

{{-1, 0, 1}, {1, 0, 1}, {0, 0, 0}}

The first three have common eigenvectors

intersectEV[Table[A[k], {k, 3}]]

{{I, 0, 1}, {-I, 0, 1}}

intersectEV[Table[A[k], {k, 4}]]

{}

Bob Hanlon

On Tue, Oct 16, 2012 at 8:12 PM, Geoffrey Eisenbarth
<geoffrey.eisenbarth@gmail.com> wrote:
> Given a set of n many matrices A[k], I'd like to find any common eigenvectors. Using
>
> Intersection[Table[Eigenvalues[A[k]],{k,1,n}] doesn't seem to work. For instance:
>
> A[1] = {{-1, -3, 1}, {0, -3, 0}, {-1, -1, -1}};
> A[2] = {{-2, -1, 1}, {0, -1, 0}, {-1, 1, -2}};
> Intersection[Table[A[p], {p, 1, 2}]]
>
> gives me
> {{{-2, -1, 1}, {0, -1, 0}, {-1, 1, -2}}, {{-1, -3, 1}, {0, -3,
> 0}, {-1, -1, -1}}}
>
>
> Any suggestions?
>

Date Subject Author
10/18/12 Bob Hanlon
10/18/12 Sseziwa Mukasa
10/18/12 Geoffrey Eisenbarth