and are almost isospectral
0.1 General case
The next result shows that and are “almost” isospectral, in the sense that their spectra is the same except possibly the value .
Theorem - Let and be as above. We have
, and moreover
and have the same eigenvalues, except possibly the zero eigenvalue.
Proof : Let .
If then is not invertible. By the result in the parent entry, this implies that is not invertible either, hence .
If is an eigenvalue of , then is not injective. By the result in the parent entry, this implies that is also not injective, hence is an eigenvalue of .
A similar argument proves that non-zero eigenvalues of are also eigenvalues of .
Remark : Note that for infinite dimensional vector spaces the spectrum of a linear mapping does not consist solely of its eigenvalues. Hence, 1 and 2 above are two different statements.
0.2 Finite dimensional case
When the vector space is finite dimensional we can strengthen the above result.
Theroem - and are isospectral, i.e. they have the same spectrum. Since is finite dimensional, this means that and have the same eigenvalues.
Proof : By the above result we only need to prove that: is invertible if and only if is invertible.
Suppose is not invertible. Hence, is not invertible or is not invertible.
For finite dimensional vector spaces invertibility, injectivity and surjectivity are the same thing. Thus, the above statement can be rewritten as: is not injective or is not surjective.
Either way is not invertible.
A similar argument shows that if is not invertible, then is also not invertible, which concludes the proof.
This humble result plays an important role in the spectral theory of operator algebras.
|Title||and are almost isospectral|
|Date of creation||2013-03-22 14:44:51|
|Last modified on||2013-03-22 14:44:51|
|Last modified by||asteroid (17536)|