Resolvent Set of Adjoint is Complex Conjugate
$begingroup$
Given a linear operator $A$ from a Hilbert space $mathscr{H}$ to itself, we define the resolvent set as
$$rho (A) = {lambda in mathbb Cmid A-lambda Itext{ is invertible}}$$
I would like to show that the resolvent set of the adjoint is given by
$$rho(A^*)={lambdainmathbb Cmid A^* -lambda Itext{ is invertible}}= overline{rho(A)} = {lambdainmathbb Cmid bar{lambda} in rho (A)}$$
Here the bar denotes the complex conjugate. To show this, we have:
$$begin{align*}
A^* -lambda I:text{ is invertible} \[1ex]
& iff (A^* -lambda I)^*:text{ is invertible} \[1ex]
& iff A - bar{lambda}I:text{ is invertible} \[1ex]
&iff bar{lambda} in rho(A)
end{align*}$$
Hence $lambdainbarrho (A)ifflambdainrho (A^*)$.
Is this solution correct?
I would like to make sure, as the solution someone else gave me seems more complicated (rather needlessly) than this.
operator-theory hilbert-spaces spectral-theory
edited Dec 11 '18 at 12:10
amWhy
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
$endgroup$
add a comment |
$begingroup$
Given a linear operator $A$ from a Hilbert space $mathscr{H}$ to itself, we define the resolvent set as
$$rho (A) = {lambda in mathbb Cmid A-lambda Itext{ is invertible}}$$
I would like to show that the resolvent set of the adjoint is given by
$$rho(A^*)={lambdainmathbb Cmid A^* -lambda Itext{ is invertible}}= overline{rho(A)} = {lambdainmathbb Cmid bar{lambda} in rho (A)}$$
Here the bar denotes the complex conjugate. To show this, we have:
$$begin{align*}
A^* -lambda I:text{ is invertible} \[1ex]
& iff (A^* -lambda I)^*:text{ is invertible} \[1ex]
& iff A - bar{lambda}I:text{ is invertible} \[1ex]
&iff bar{lambda} in rho(A)
end{align*}$$
Hence $lambdainbarrho (A)ifflambdainrho (A^*)$.
Is this solution correct?
I would like to make sure, as the solution someone else gave me seems more complicated (rather needlessly) than this.
operator-theory hilbert-spaces spectral-theory
edited Dec 11 '18 at 12:10
amWhy
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
$endgroup$
2
$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53
add a comment |
$begingroup$
Given a linear operator $A$ from a Hilbert space $mathscr{H}$ to itself, we define the resolvent set as
$$rho (A) = {lambda in mathbb Cmid A-lambda Itext{ is invertible}}$$
I would like to show that the resolvent set of the adjoint is given by
$$rho(A^*)={lambdainmathbb Cmid A^* -lambda Itext{ is invertible}}= overline{rho(A)} = {lambdainmathbb Cmid bar{lambda} in rho (A)}$$
Here the bar denotes the complex conjugate. To show this, we have:
$$begin{align*}
A^* -lambda I:text{ is invertible} \[1ex]
& iff (A^* -lambda I)^*:text{ is invertible} \[1ex]
& iff A - bar{lambda}I:text{ is invertible} \[1ex]
&iff bar{lambda} in rho(A)
end{align*}$$
Hence $lambdainbarrho (A)ifflambdainrho (A^*)$.
Is this solution correct?
I would like to make sure, as the solution someone else gave me seems more complicated (rather needlessly) than this.
operator-theory hilbert-spaces spectral-theory
edited Dec 11 '18 at 12:10
amWhy
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
$endgroup$
Given a linear operator $A$ from a Hilbert space $mathscr{H}$ to itself, we define the resolvent set as
$$rho (A) = {lambda in mathbb Cmid A-lambda Itext{ is invertible}}$$
I would like to show that the resolvent set of the adjoint is given by
$$rho(A^*)={lambdainmathbb Cmid A^* -lambda Itext{ is invertible}}= overline{rho(A)} = {lambdainmathbb Cmid bar{lambda} in rho (A)}$$
Here the bar denotes the complex conjugate. To show this, we have:
$$begin{align*}
A^* -lambda I:text{ is invertible} \[1ex]
& iff (A^* -lambda I)^*:text{ is invertible} \[1ex]
& iff A - bar{lambda}I:text{ is invertible} \[1ex]
&iff bar{lambda} in rho(A)
end{align*}$$
Hence $lambdainbarrho (A)ifflambdainrho (A^*)$.
Is this solution correct?
I would like to make sure, as the solution someone else gave me seems more complicated (rather needlessly) than this.
operator-theory hilbert-spaces spectral-theory
operator-theory hilbert-spaces spectral-theory
edited Dec 11 '18 at 12:10
amWhy
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
edited Dec 11 '18 at 12:10
amWhy
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
edited Dec 11 '18 at 12:10
amWhy
192k28225439
edited Dec 11 '18 at 12:10
amWhy
192k28225439
edited Dec 11 '18 at 12:10
amWhy
192k28225439
192k28225439
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
asked Dec 8 '18 at 2:25
rubikscube09rubikscube09
1,199718
1,199718
2
$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53
add a comment |
2
$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53
2
2
$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53
$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53
add a comment |
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$begingroup$
Yes, your proof is correct.
$endgroup$
– Alonso Delfín
Dec 8 '18 at 2:53