Prove the eigenvectors are linearly independent

[hover]

Homework Statement

Suppose that a matrix A has real entries (which we always assume) and a complex
(non-real) eigenvalue  $$\lambda$$= a + ib , with b not equal to 0. Let W = U + iV be the corresponding
complex eigenvector, having real and imaginary parts U and V , respectively. Show that
U and V are necessarily linearly independent (meaning that one vector is not a scalar
multiple of the other).
HINT: Argue by contradiction: suppose that U and V are dependent, say, V = rU for
some scalar r, and derive a contradiction (that is, a statement that follows logically from
the supposition, but which is false, such as 0 > 1).

Homework Equations

--

The Attempt at a Solution

This question confuses me because I don't even know how to go about it or where to start. How should I do this? Where do I begin??

Thanks!

 

[micromass]

U+irU being an eigenvector of A with eigenvalues a+bi means that

$$A(U+irU)=(a+bi)(U+irU)$$

Work out the parantheses on both sides and try to reach a contradiction...

 

[sabrathos]

Sorry to revive this question, but I have the same assignment and I'm drawing a blank as to the solution.

I see no contradiction. Our teacher gave us that

A(W) = A(U) + iA(V) = (aU - bV) + i(aV + bU),

so, applying this to micromass's equation, I get

A(U + irU) = (a-br)U + i(ar+b)U,

which is consistent with what I get if I multiply everything out in

(a+bi)(U+irU).

What am I doing wrong?

 

[Dick]

Sorry to revive this question, but I have the same assignment and I'm drawing a blank as to the solution.

I see no contradiction. Our teacher gave us that

A(W) = A(U) + iA(V) = (aU - bV) + i(aV + bU),

so, applying this to micromass's equation, I get

A(U + irU) = (a-br)U + i(ar+b)U,

which is consistent with what I get if I multiply everything out in

(a+bi)(U+irU).

What am I doing wrong?

Keep going. So AU+irAU=(a-br)U+i(ar+b)U. Equate real and imaginary parts. So AU=(a-br)U and rAU=(b+ar)U. Can you finish up?