This is an invalid argument about eigenvalues, but why?

[nomadreid]

TL;DR

The following argument is obviously wrong somewhere. If a is an eigenvector for the matrix M, then for each of its eigenvectors v, Mv=av. But then for any nonzero k, M(k*v)= (a/k)(k*v), so that for M, a/k is an eigenvector for the eigenvector k*v. This would lead to the absurdity that if an eigenvalue exists, then everything is an eigenvalue for the matrix.

The fallacy in the summary is not covered in the sites discussing eigenvalues, so there must be something blindingly and embarrassingly obvious that is wrong. I would be grateful if someone would point it out. Thanks.

 

[ergospherical]

Not sure what you did there. If $v$ is an eigenvector of eigenvalue $a$, then $M(kv) = k(Mv) = k(av) = a(kv)$ which means that $kv$ is also an eigenvector of the same eigenvalue $a$.

 

[nomadreid]

ergospherical, thanks for the quick reply. Yes, your equations are correct showing that kv would also be an eigenvector of a, but this just says that an eigenvalue will have an infinite number of eigenvectors. However, I am saying (somehow incorrectly) that a/k would be another eigenvalue of kv. Let me restate my (faulty) argument: let kv = w, and let a/k = b. Then Mw=bw. (because Mv=av). That is, b is an eigenvalue of M, different to a (and with a different eigenvector, but that is beside the point). But I shouldn't be able to generate eigenvalues like this.

 

[ergospherical]

Let me restate my (faulty) argument: let kv = w, and let a/k = b. Then Mw=bw. (because Mv=av). That is, b is an eigenvalue of M, different to a (and with a different eigenvector, but that is beside the point). But I shouldn't be able to generate eigenvalues like this.

You're just messing up the algebra and introducing an extraneous factor of $k$ for some reason. $Mv=av \implies M(kv) = a(kv) \implies Mw = aw$.

 

[nomadreid]

ergospherical, I thank you very much for your patience. I see my error (and not sure why I didn't see it before -- but that is often the case with silly mistakes,no?). I am grateful that you answered rather than simply dismissing it as silly. Thread can thus be closed and forgotten.

 

[Mark44]

But then for any nonzero k, M(k*v)= (a/k)(k*v), so that for M, a/k is an eigenvector for the eigenvector k*v.

Aside from the error already pointed out, the above should read "an eigenvalue for the eigenvector kv."

 

[nomadreid]

Mark44, oops, thanks for pointing that typo out... more egg on my face...

 

[nuuskur]

You showed that $kv$ is an eigenvector corresponding to the eigenvalue $ak^{-1}$.

 

[nomadreid]

Thank you, nuuskur. That is indeed what I inadvertently proved instead of what I set out to do; very embarrassing basic error. Thankfully, the people at Physics Forums are a nice group that don't rub it in.