Algebraic multiplicity is the number of times of occurance of an eigenvalue and geometric multiplicity is the number of linearly independent eigenvectors associated with that eigenvalue.
@Gopal,
Good interpretation.
But you need to explain the following points:
The vector space V generated by v1 has dimension 1.
Hence, W has dim = n-1.
If the the multiplicity of λ1 > 1, then you find v2 in W,
you assumed it belongs to the eigenspace of the eigenvalue λ1 .Why?
Also, the following statement page (2).
(In m1 steps we get m1 linearly independent eigenvectors associated with λ1 and nothing more further.) is not clear.
It needs a proof. May be more difficult than the initial question itself.
@Issam
@Issam vectorspece V is generated by a nonzero vector v1. So its of dimension 1. Is it not clear? Then for the next, we know that Rn is of dimension n and so Rn has an orthogonl basis. So the space spanned by the n-1 orthogonalvectors to v1 is W and is o dmension n-1.Is this not clear?
If the the multiplicity of λ1 > 1, then you find v2 in W,
you assumed it belongs to the eigenspace of the eigenvalue λ1 .Why?
If the the multiplicity of λ1 > 1, then the restriction of A to W is a linear operator from W to W. So it has λ1 as an eigenvalue nd the eigenvector then definitely belong to W and where else?
(In m1 steps we get m1 linearly independent eigenvectors associated with λ1 and nothing more further.) is not clear.
In m1 steps the occurence of λ1 as an eigenvalue to the restricted operator A ends. and by very construction the m1 eigevectors associated with λ1 are lin independent too.
Hop that this explanation suffices,
regards
Indulal
@ Indulal,
Read my previous answer once more.
You didn't answer my question yet.
I agree with the assumptions until the end of line five; then my question starts in line 6 :
[you assumed it belongs to the eigenspace of the eigenvalue λ1.Why?]
In other words: v2 is in W,
prove that it belongs to the eigenspace of λ1.
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