Spectrum of operator A
$begingroup$
Let $H$ be a Hilbert space and $A in B(H)$. How to prove that if
$|lambda| = |A|$ and $lambda in W(A)$, then $lambda in sigma_p(A)$?
Here $W(A)$ is the numerical range of the operator $A$:
$$W(A):= big{(Ax,x): |x| = 1big}$$
and $sigma_p(A)$ is the point spectrum of $A$.
I have no idea how to prove this. Can you help me with this?
functional-analysis eigenvalues-eigenvectors hilbert-spaces
edited Jan 16 at 18:49
Hanno
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space and $A in B(H)$. How to prove that if
$|lambda| = |A|$ and $lambda in W(A)$, then $lambda in sigma_p(A)$?
Here $W(A)$ is the numerical range of the operator $A$:
$$W(A):= big{(Ax,x): |x| = 1big}$$
and $sigma_p(A)$ is the point spectrum of $A$.
I have no idea how to prove this. Can you help me with this?
functional-analysis eigenvalues-eigenvectors hilbert-spaces
edited Jan 16 at 18:49
Hanno
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space and $A in B(H)$. How to prove that if
$|lambda| = |A|$ and $lambda in W(A)$, then $lambda in sigma_p(A)$?
Here $W(A)$ is the numerical range of the operator $A$:
$$W(A):= big{(Ax,x): |x| = 1big}$$
and $sigma_p(A)$ is the point spectrum of $A$.
I have no idea how to prove this. Can you help me with this?
functional-analysis eigenvalues-eigenvectors hilbert-spaces
edited Jan 16 at 18:49
Hanno
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
$endgroup$
Let $H$ be a Hilbert space and $A in B(H)$. How to prove that if
$|lambda| = |A|$ and $lambda in W(A)$, then $lambda in sigma_p(A)$?
Here $W(A)$ is the numerical range of the operator $A$:
$$W(A):= big{(Ax,x): |x| = 1big}$$
and $sigma_p(A)$ is the point spectrum of $A$.
I have no idea how to prove this. Can you help me with this?
functional-analysis eigenvalues-eigenvectors hilbert-spaces
functional-analysis eigenvalues-eigenvectors hilbert-spaces
edited Jan 16 at 18:49
Hanno
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
edited Jan 16 at 18:49
Hanno
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
edited Jan 16 at 18:49
Hanno
2,276628
edited Jan 16 at 18:49
Hanno
2,276628
edited Jan 16 at 18:49
Hanno
2,276628
2,276628
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
asked Jan 9 at 15:43
Gera SlanovaGera Slanova
483
483
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is $x$ with $|x|=1$ such that
$$
lambda = langle Ax,xrangle.
$$
By the Cauchy-Schwarz inequality,
$$
|langle Ax,xrangle| le |A|= |lambda|
$$
with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $xne0$, it follows $c_1ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies
$c_1 lambda = c_2$, and $lambda$ is an eigenvalue.
answered Jan 10 at 12:56
dawdaw
24.6k1645
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
There is $x$ with $|x|=1$ such that
$$
lambda = langle Ax,xrangle.
$$
By the Cauchy-Schwarz inequality,
$$
|langle Ax,xrangle| le |A|= |lambda|
$$
with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $xne0$, it follows $c_1ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies
$c_1 lambda = c_2$, and $lambda$ is an eigenvalue.
answered Jan 10 at 12:56
dawdaw
24.6k1645
$endgroup$
add a comment |
$begingroup$
There is $x$ with $|x|=1$ such that
$$
lambda = langle Ax,xrangle.
$$
By the Cauchy-Schwarz inequality,
$$
|langle Ax,xrangle| le |A|= |lambda|
$$
with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $xne0$, it follows $c_1ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies
$c_1 lambda = c_2$, and $lambda$ is an eigenvalue.
answered Jan 10 at 12:56
dawdaw
24.6k1645
$endgroup$
add a comment |
$begingroup$
There is $x$ with $|x|=1$ such that
$$
lambda = langle Ax,xrangle.
$$
By the Cauchy-Schwarz inequality,
$$
|langle Ax,xrangle| le |A|= |lambda|
$$
with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $xne0$, it follows $c_1ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies
$c_1 lambda = c_2$, and $lambda$ is an eigenvalue.
answered Jan 10 at 12:56
dawdaw
24.6k1645
$endgroup$
There is $x$ with $|x|=1$ such that
$$
lambda = langle Ax,xrangle.
$$
By the Cauchy-Schwarz inequality,
$$
|langle Ax,xrangle| le |A|= |lambda|
$$
with equality if and only if $Ax$ and $x$ are linearly dependent. This implies that $Ax$ and $x$ are linearly dependent: $c_1 Ax = c_2 x$. Since $xne0$, it follows $c_1ne0$, and $x$ is an eigenvector to the eigenvalue $c_2/c_1$. Multiplying the equation with $x$ implies
$c_1 lambda = c_2$, and $lambda$ is an eigenvalue.
answered Jan 10 at 12:56
dawdaw
24.6k1645
answered Jan 10 at 12:56
dawdaw
24.6k1645
answered Jan 10 at 12:56
dawdaw
24.6k1645
answered Jan 10 at 12:56
dawdaw
24.6k1645
24.6k1645
add a comment |
add a comment |
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