Eigenvectors and Eigenvalues
Most recent answer: 12/17/2011
Q:
What's an eigenvalue? What's an eigenvector? Why is there eigen- in front of so many words?
- Richard Ong (age 14)
Winston-Salem, NC, USA
- Richard Ong (age 14)
Winston-Salem, NC, USA
A:
If you try Googling eigenvector and eigenvalue, you'll turn up a number of nice short articles. I hope that the little elementary description below will serve as an intuitive introduction to those more formal descriptions.
Draw an arrow on a piece of stretchy plastic.(Call the arrow a vector.) Stretch the plastic out uniformly, and the arrow just gets longer, still pointing the same way. The new vector is just some number times the old vector. That's what we mean by an "eigenvector" of the stretch. Say the stretch was 10%. The new vector is 1.1 times the old vector. The 1.1 is the "eigenvalue". For this simple uniform stretch, every vector is an eigenvector and they all share the same eigenvalue.
Now say you just stretch the plastic sideways, leaving it the same size up-down. An up-down arrow is unchanged, so it's an eigenvector of that stretch with eigenvalue 1.0. A sideways arrow still points sideways, so it's also an eigenvector, with eigenvalue set by how much the stretch was. However, any arrow that was part sideways and part up-down now points in a new direction, more sideways. That means it's not an eigenvector of that stretch, because the result isn't just a multiple of the original.
Eigenvectors and eigenvalues are defined for any linear operation on vectors. A linear operation is one where you can get what the output is on the sum of two vectors by just summing the separate outputs on each of them. So, for example, rotations are linear operations. In two dimensions, the only vector that still points the same way after a rotation (other than full-circle) is 0. So there aren't any interesting eigenvectors for that rotation. In three dimensions, the rotation is around some axis in this 3D space, and any vector along the rotation axis is unchanged, which means it's an eigenvector with eigenvalue 1.0.
As for the German root "eigen" it's related to the English word "own", connected with the meaning of "self". An eigenvector of an operation is one that's mapped to a multiple of itself by the operation.
Mike W.
Draw an arrow on a piece of stretchy plastic.(Call the arrow a vector.) Stretch the plastic out uniformly, and the arrow just gets longer, still pointing the same way. The new vector is just some number times the old vector. That's what we mean by an "eigenvector" of the stretch. Say the stretch was 10%. The new vector is 1.1 times the old vector. The 1.1 is the "eigenvalue". For this simple uniform stretch, every vector is an eigenvector and they all share the same eigenvalue.
Now say you just stretch the plastic sideways, leaving it the same size up-down. An up-down arrow is unchanged, so it's an eigenvector of that stretch with eigenvalue 1.0. A sideways arrow still points sideways, so it's also an eigenvector, with eigenvalue set by how much the stretch was. However, any arrow that was part sideways and part up-down now points in a new direction, more sideways. That means it's not an eigenvector of that stretch, because the result isn't just a multiple of the original.
Eigenvectors and eigenvalues are defined for any linear operation on vectors. A linear operation is one where you can get what the output is on the sum of two vectors by just summing the separate outputs on each of them. So, for example, rotations are linear operations. In two dimensions, the only vector that still points the same way after a rotation (other than full-circle) is 0. So there aren't any interesting eigenvectors for that rotation. In three dimensions, the rotation is around some axis in this 3D space, and any vector along the rotation axis is unchanged, which means it's an eigenvector with eigenvalue 1.0.
As for the German root "eigen" it's related to the English word "own", connected with the meaning of "self". An eigenvector of an operation is one that's mapped to a multiple of itself by the operation.
Mike W.
(published on 12/17/2011)