left inverse operator and the spectrum

Question

Let $D$ be the left inverse bounded operator from $Y$ to $X$ (Banach spaces) of a bounded operator $A$ from $X$ to $Y$, Iam trying to find the spectrum of the operator $AD$ .. My attempt is to work on matrices and to see if I can generalize the result for any bounded operators. I tried different types of matrix and they all gave me different eigenvalues so I don’t know if I can say that the spectrum of $AD$ is the set of complex numbers?

I tried for operators generally, I know that $D$ is left inverse so that $DA=I_X$ but I cant permute $D$ or $A$ since I don’t know if $D$ is invertible or not and that similarly for $A$.. For the spectrum of $DA$ is it the set ${1}$ because $DA=I_X$ ? Any help would be appreciated ..

Martin Argerami · Answer 1 · 2018-01-02T18:00:14.853

1

If $AD$ is invertible, then $ADR=I$ for some $R$, and now $D=DADR=DR$, so $AD=I$. In this case, $$\sigma(AD)=\sigma(I)=\{1\}.$$

If $AD$ is not invertible (so, in particular, $0\in\sigma(AD)$), then we still know that $$\{0\}\cup\sigma(DA)=\{0\}\cup\sigma(AD).$$ If follows, since $\sigma(DA)=\{1\}$, that $$\sigma(AD)=\{0,1\}.$$

The above has nothing to do with operators nor Banach spaces: the argument holds in any unital ring (see here for a purely algebraic argument).

edited Jan 02 '18 at 18:00

answered Jan 01 '18 at 23:25

Martin Argerami

205,756

Thank you Mr Martin for you help, so I think at the end the spectrum of AD is equal to 1, I don't understand that why 0 is included here? – S.N.A Jan 02 '18 at 00:47
Having $0$ in the spectrum means precisely "not invertible". Or, if you want, if $0\not\sigma(AD)$, then $AD$ is invertible. – Martin Argerami Jan 02 '18 at 02:44
(sorry, I meant to type "$0\not\in\sigma(AD)$" above) – Martin Argerami Jan 02 '18 at 02:52
very clear, thank you very much Mr Martin.. – S.N.A Jan 02 '18 at 14:34

left inverse operator and the spectrum

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