1. linear independence: Eigenvectors/values

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

2. Re: linear independence: Eigenvectors/values

Hey bluelavae.

Hint: What is the eigenvalue equation (Remember Mx = λx)?

3. Re: linear independence: Eigenvectors/values

Hopefully this doesn't come off really dumb:

= λ(av + bw)
= λ(av) + λ(bw)
= 2λ + av + bw

?

4. Re: linear independence: Eigenvectors/values

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.

5. Re: linear independence: Eigenvectors/values

Sorry, It was meant to come out λ ≠ µ, rather than λ = 6 µ.

The second part looks a bit confusing.

6. Re: linear independence: Eigenvectors/values

I didn't quite follow Deveno's proof. Why is $\displaystyle \lambda=6\mu$?

For the linear independence, suppose $\displaystyle xv+yw=0$ with x and y in the underlying field (R?), then $\displaystyle \lambda(xv)+\mu(yv)=0$

Set $\displaystyle c=xv$ and $\displaystyle d=yw,$ then the equations become $\displaystyle c+d=0$ and $\displaystyle \lambda c+\mu d=0$. Multiply the first equation by $\displaystyle \lambda$ and subtract from the second. We get $\displaystyle (\mu-\lambda)d=0$ and then d=0, and then also c=0. So xv=0 and yw=0. If $\displaystyle x\ne0$, 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.

7. Re: linear independence: Eigenvectors/values

his post was edited after i posted my response.