How to check if a vector is an eigenvector of a matrix using mathematica?

Question

Here is a vector

$$\begin{pmatrix}i\\7i\\-2\end{pmatrix}$$

Here is a matrix

$$\begin{pmatrix}2& i&0\\-i&1&1\\0 &1&0\end{pmatrix}$$

Is there a simple way to determine whether the vector is an eigenvector of this matrix?

Here is some code for your convenience.

h = {{2, I, 0 },
     {-I, 1, 1},
     {0, 1, 0}};

y = {I, 7 I, -2};

Answer 1

You could use MatrixRank. Here is a function that does this:

eigenvectorQ[matrix_, vector_] := MatrixRank[{matrix . vector, vector}] == 1

For your example:

eigenvectorQ[h, y]

False

Answer 2

For problems with exact coordinates, one could code up the definition of eigenvector. The function eigV finds the eigenvalue for a given vector in the form L == value or returns False if there is none; the function eigQ returns True if there exists an eigenvalue for

ClearAll[eigQ, eigV];
eigV[m_, v_] := Reduce@Thread[(m - SparseArray[{i_, i_} :> L, Dimensions[m]]).v == 0];
eigV[m_][v_] := eigV[m, v]; (* operator form *)
eigQ[m_, v_] := Resolve@Exists[L, eigV[m, v]];
eigQ[m_][v_] := eigQ[m, v]; (* operator form *)

Examples:

eigQ[h] /@ {y, {-I (-2 + Sqrt[3]), 1 - Sqrt[3], 1}}
(*  {False, True}  *)

eigV[h] /@ {y, {-I (-2 + Sqrt[3]), 1 - Sqrt[3], 1}}
(*  {False, L == 1 - Sqrt[3]}  *)

Or simply

eigQ[h, y]
(*  False  *)

For approximate problems, one would have to account for rounding error.

Answer 3

Yes! We just check whether $h.y = (u + I v) y$ holds for some real $u, v \in \mathbb{R}$.

h = {{2, I, 0}, {-I, 1, 1}, {0, 1, 0}};
y = {I, 7 I, -2};
expr = Norm[h.y - (u + I v) y, 2]^2 // ComplexExpand;
Minimize[expr, {u, v}]

{623/6, {u -> 17/18, v -> 0}}

Answer: Nope, it's not.

Answer 4

You already have several good answers. An alternative is to use a Rayleigh quotient,

r = First[y.h.ConjugateTranspose[{y}]/Norm[y]];

The vector y is an eigenvector of h if and only if the matrix $$ h-r1_{3\times3} $$ is singular:

MatrixRank[h - IdentityMatrix[Length[y]] R]<Length[y]
(*False*)

or

Det[h - IdentityMatrix[Length[y]] R] == 0
(*False*)

If you are using floating point numbers, you should change this condition into

MatrixRank[h - IdentityMatrix[Length[y]] R,Tolerance->epsilon]<Length[y]

or

Abs[Det[h - IdentityMatrix[Length[y]] R]] < epsilon

where epsilon is some small number.

As a matter of fact, the MatrixRank method is slightly faster than the Det one. It seems to me that it is also faster than the methods suggested by other users, but confirming this would require a more thorough analysis.

Answer 5

Either

Reduce[h . y == x * y, x]

or

Reduce[(h - IdentityMatrix[Length[h]] x) . y == 0, x]

depending on whether you would rather type $y$ once or twice.

Answer 6

Carl Woll's answer seems to be broken in the newest Mathematica. Here is a slight modification that makes it work

EigenvectorQ[matrix_, vector_] := MatrixRank[Join[matrix.vector, vector, 2]] == 1

MatrixRank gives the number of linearly dependent columns. Join[l1, l2, 2] joins the second vector to the first on the right.

Answer 7

MemberQ[myeigens = Normalize/@Eigenvectors[h], Normalize[y]]|| MemberQ[myeigens, -Normalize[y]]

(* False *)

Answer 8

How about this:

eigenVectorQ[mat_, vec_] := Abs[Dot[#1\[Conjugate], #2]] == Norm[#1] Norm[#2] &[mat.vec, vec]

Then eigenVectorQ[h, y] returns False.