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This is my attempt at a solution:
A^2(x) = A(A(x))
A^2(x) = lambda^2(x)
Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The statement for general N>0 follows in the same way.
Am I missing anything?
This is my attempt at a solution:
A^2(x) = A(A(x))
A^2(x) = lambda^2(x)
Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The statement for general N>0 follows in the same way.
Am I missing anything?
Originally Posted by temaire
This is my attempt at a solution:
A^2(x) = A(A(x))
A^2(x) = lambda^2(x)
Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The statement for general N>0 follows in the same way.
Am I missing anything?
Not really: now just round up the proof doing some induction.
Tonio
A proof by induction works like this: first show your statement holds foror
or the first
for which it is true, then assume it to be true for
(this is your induction hypothesis) and then show it to be true for
using your hypothesis. if
is an eigenvalue of
,
and
so
is an eigenvalue of
for
. suppose it is true for
it is essentially what you said but a a bit more rigorous
induction is a really useful tool, you should practice using it
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