Hey bluelavae.
Hint: What is the eigenvalue equation (Remember Mx = λx)?
Recall that two vectors v and w are linearly independent if the only solution to the vector equation av + bw = 0
is the solution a = b = 0.
Let v and w be eigenvectors of a matrix M with corresponding nonzero eigenvalues λ and µ such that λ ≠ µ.
(i) Given scalars a and b, ﬁnd an expression for, M(av + bw), that does not contain M.
(ii) Hence or otherwise, prove that the eigenvectors v and w must be linearly independent.
Thanks
it's kind of on the right track, but still a miss.
first, we use the linearity of M, to write:
M(av + bw) = aM(v) + bM(w).
next, we use the fact that v,w are eigenvectors with eigenvalues λ,μ to write:
aM(v) + bM(w) = a(λv) + b(μw) <--this expression does not involve M.
we can also write it as 6a(μv) + b(μw), given that λ = 6μ.
now, suppose av + bw = 0. clearly, M(0) = 0, so since M(av + bw) = 6a(μv) + b(μw), we have 6a(μv) + b(μw) = 0. since μ ≠ 0, we can multiply this by 1/μ to get:
6av + bw = 0. together with av + bw = 0, this gives us 5av = 0. since v ≠ 0 (it is an eigenvector, and eigenvectors are non-zero), we have 5a = 0, that is: a = 0. b = 0 follows quickly as a consequence, so {v,w} is a linearly independent set.
EDIT: if we just stipulate that λ ≠ μ, we have:
av + bw = 0
λ(av) + μ(bw) = 0
equivalently:
λ(av) + λ(bw) = 0
λ(av) + μ(bw) = 0
thus (λ-μ)(bw) = 0, so that either λ-μ = 0, or bw = 0. since λ ≠ μ, λ-μ ≠ 0, so we must have bw = 0, so b = 0, and thus a = 0.
I didn't quite follow Deveno's proof. Why is ?
For the linear independence, suppose with x and y in the underlying field (R?), then
Set and then the equations become and . Multiply the first equation by and subtract from the second. We get and then d=0, and then also c=0. So xv=0 and yw=0. If , we get v=0, but an eigen vector by definition is not 0. Thus x=0 and similarly y=0.
Of course this generalizes to n eigenvectors with all different eigenvalues. If you know about the van der Monde matrix, it's easy to prove.