1st year linear algebra: linearly independent eigenvectors

Can one eigenvalue for a matrix produce more than one linearly independent eigenvector?

There is a theorem in my textbook that says, "if v1,...,vr are eigenvectors that correspond to distinct eigenvalues s1,...,sr then the set {v1,...,vr} is linearly independent"

Does this mean that a matrix can only have as many linearly independent eigenvectors as distinct eigenvalues? In other words does the implication go the other way as well (making the set of lin. ind. eigenvectors bigger makes the list of distinct eigenvalues bigger too)?

One of the questions on a practice exam is, "This matrix (it is 3 by 3) has two distinct eigenvalues s1=2 and s2=3. Determine whether A is diagonalisable and explain how you arrived at your answer."

I didn't think one eigenvalue could produce more than one linearly independent eigenvector, so it seems obvious from the definition of diagonalisable, "A nXn matrix A is diagonalisable iff A has n lin. ind. eigenvectors" so the question is too easy, would take two lines quoting the theorem and definition of diagonalisable, but there is a whole page worth of room ... plus it gives you the whole matrix which you wouldn't need to answer the question, so I know I am misunderstanding something.

The matrix in the question is {[3, -1, -3]T, [-1, 1, -7]T, [1 0 4]T} if that helps at all.

Thanks in advance for your replies, my exam is on wednesday :-/

My linear algebra is a little rusty but... here's what I remember.

I'm a very visual person, so I tend to visualize linear algebra. I visualize an eigenvalue by imagining a (2-d, for simplicity) plane, and first picking a point (this will be easily converted to an eigenvector; just draw an arrow from the origin to the point you picked). When you apply any linear transformation to this point, the resulting point (a vector) must lie in a line from the origin and the original point.

That is, the resulting vector must be a scaled version of the original vector: algebraically, Ax = Lx where A is the linear transformation matrix, x is the original vector you picked (which turns out to be an eigenvector), and L is a constant of proportionality (I don't want to type lambda; it's an eigenvalue).

Now, if you pick any point on the line from the origin through the original point, you'll have chosen a distinct eigenvector, with an associated scaling factor (eigenvalue) of L. The thing is, these eigenvectors all correspond to the same eigenvalue and any 3 of them (we're dealing with 3 dimensions with your matrix) are very clearly linearly dependent. In essence, you'll only get one "good" eigenvector from the eigenvalue of L.

Is this true in all cases? Given a scaling factor L, will you only get one "good" eigenvector? I'd say NO, since I can pretty easily imagine two lines going through the origin that happen to have the same scaling factor, but which produce two different "good" eigenvectors.

So I'd say the answer to your question is "yes", since a scaling factor L could produce more than one "good" eigenvector. However, I don't have an algebraic proof.


Actually, I'll try to algebraically describe my visual stuff. That should produce a counterexample.

Let (x1, x2) = the original vector. Let L*(x1, x2) = the scaled vector. Let A = the transformation matrix (2x2 in this case; I'm going for simplicity), ((a11, a12), (a21, a22)). Let (y1, y2) = a linearly independent vector to (x1, x2). Let L*(y1, y2) = the scaled linearly independed vector.

Then Ax = Lx (a11*x1 + a12*x2, a21*x1 + a22*x2) = (Lx1, Lx2), or a11*x1 + a12*x2 = Lx1, and a21*x1 + a22*x2 = Lx2.

Also, Ay = Ly (a11*y1 + a12*y2, a21*y1 + a22*y2) = (Ly1, Ly2), or a11*y1 + a12*y2 = Ly1, and a21*y1 + a22*y2 = Ly2.

Rearranging the order of these equations,
a11*x1 + a12*x2 = Lx1
a11*y1 + a12*y2 = Ly1

and

a21*x1 + a22*x2 = Lx2
a21*y1 + a22*y2 = Ly2

Since you're in a linear algebra course, you know that a11, a12; a21, a22 are thus uniquely identified for a given choice of x1, x2, y1, y2, and L. Since it is very possible to choose two linearly independent vectors x and y [(x1, x2), (y1, y2), respectively], you can contruct a linear transformation matrix A from (any) two linearly independent eigenvectors [x and y], but with 1 eigenvalue [L].

So yup, there's a counterexample. Hope that helps to answer your question.