Mixing
show norm of self-adjoint operator is maximum of abs value of eigenvalue
$begingroup$
$M: V to V$ linear operator
show
$|M| = max{|text{eigenvalue}|}$
linear-algebra
asked Apr 8 '14 at 1:48
user128616user128616
147213
$endgroup$
add a comment |
$begingroup$
$M: V to V$ linear operator
show
$|M| = max{|text{eigenvalue}|}$
linear-algebra
asked Apr 8 '14 at 1:48
user128616user128616
147213
$endgroup$
add a comment |
$begingroup$
$M: V to V$ linear operator
show
$|M| = max{|text{eigenvalue}|}$
linear-algebra
asked Apr 8 '14 at 1:48
user128616user128616
147213
$endgroup$
$M: V to V$ linear operator
show
$|M| = max{|text{eigenvalue}|}$
linear-algebra
linear-algebra
asked Apr 8 '14 at 1:48
user128616user128616
147213
asked Apr 8 '14 at 1:48
user128616user128616
147213
edited Apr 8 '14 at 4:06
user128616
asked Apr 8 '14 at 1:48
user128616user128616
147213
asked Apr 8 '14 at 1:48
user128616user128616
147213
asked Apr 8 '14 at 1:48
user128616user128616
147213
147213
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $mathbb{C}^n$ such that $Ae_i=lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $lambda=text{max} , {|lambda_1|,...,|lambda_2|}$. Then for any $x in mathbb{C}^n$ such that $|x| leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n in mathbb{C}$, and thus we get that
begin{equation}
begin{split}
|Tx|
& =|T(a_1e_1+...+a_ne_n)|\
& =|a_1 T(e_1)+...+a_nT(e_n)|\
& =|a_1 lambda_1e_1+...+a_nlambda_ne_n|\
& = sqrt{|a_1 lambda_1|^2+...+|a_nlambda_n|^2}\
& leq sqrt{|a_1 lambda|^2+...+|a_nlambda|^2}\
& = lambda sqrt{|a_1|^2+...+|a_n|^2}\
& = lambda |x|\
& leq lambda\
end{split}
end{equation}
Since $x$ was an arbitrary element in $mathbb{C}^n$ such that $|x| leq 1$, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| leq lambda$$
Conversely, notice that $|lambda_i|=lambda$ for some $i$. Then, for that $i$, we have
begin{equation}
begin{split}
|Te_i|
& =|lambda_i e_1|\
& =|lambda_i||e_i|\
& =lambda\
end{split}
end{equation}
Since $|e_i|=1$, this implies that
$$|T|=sup_{|x| leq 1} |Tx| geq lambda$$
Thus, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| = lambda$$
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
$endgroup$
1
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your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
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– Peyman mohseni kiasari
Jan 13 at 7:03
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@Peyman Thank you. I removed the link.
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– Pawel
Jan 13 at 17:26
add a comment |
$begingroup$
The norm of an operator (as derived from the norm on the space) is given by
$$
|M| = max_{x neq 0} frac{|Mx|}{|x|} = max_{|x| = 1} |M x|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
|M| = max_{|x| = 1} sqrt{(Mx)^*Mx} = sqrt{max_{|x| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$. It may help to set $y = U^*x$, noting that $|y| = |x|$.
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
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add a comment |
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This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,ldots,e_n$, (let $lambda_i$ be the evalues arranged so that $|lambda_1| geq cdots |lambda_n|$). For $v in V$ let $a^v_1,ldots,a_n^v$ be the coefficients satisfying
$v = sum_{i=1}^n a_i^v e_i$. Then
begin{align}
|M |^2
&= sup{ langle Mv,Mvrangle mid |v| = 1 } \
&= supleft{ leftlangle Msum_{i=1}^n a^v_i e_i,
M sum_{i=1}^n a^v_i e_irightrangle mid |v| = 1 right} \
&= supleft{ leftlangle sum_{i=1}^n a^v_i lambda_i e_i,
sum_{i=1}^n a^v_i lambda_i e_irightrangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2 |lambda_i|^2 langle e_i,
e_irangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2|lambda_i|^2 mid |v| = 1 right}
end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $|v|=1$. The max occurs when all mass is put at the largest component, $|lambda_1|^2$. Therefore
$$
|M|^2
= |lambda_1|^2.
$$
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $mathbb{C}^n$ such that $Ae_i=lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $lambda=text{max} , {|lambda_1|,...,|lambda_2|}$. Then for any $x in mathbb{C}^n$ such that $|x| leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n in mathbb{C}$, and thus we get that
begin{equation}
begin{split}
|Tx|
& =|T(a_1e_1+...+a_ne_n)|\
& =|a_1 T(e_1)+...+a_nT(e_n)|\
& =|a_1 lambda_1e_1+...+a_nlambda_ne_n|\
& = sqrt{|a_1 lambda_1|^2+...+|a_nlambda_n|^2}\
& leq sqrt{|a_1 lambda|^2+...+|a_nlambda|^2}\
& = lambda sqrt{|a_1|^2+...+|a_n|^2}\
& = lambda |x|\
& leq lambda\
end{split}
end{equation}
Since $x$ was an arbitrary element in $mathbb{C}^n$ such that $|x| leq 1$, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| leq lambda$$
Conversely, notice that $|lambda_i|=lambda$ for some $i$. Then, for that $i$, we have
begin{equation}
begin{split}
|Te_i|
& =|lambda_i e_1|\
& =|lambda_i||e_i|\
& =lambda\
end{split}
end{equation}
Since $|e_i|=1$, this implies that
$$|T|=sup_{|x| leq 1} |Tx| geq lambda$$
Thus, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| = lambda$$
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
$endgroup$
1
$begingroup$
your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
add a comment |
$begingroup$
Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $mathbb{C}^n$ such that $Ae_i=lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $lambda=text{max} , {|lambda_1|,...,|lambda_2|}$. Then for any $x in mathbb{C}^n$ such that $|x| leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n in mathbb{C}$, and thus we get that
begin{equation}
begin{split}
|Tx|
& =|T(a_1e_1+...+a_ne_n)|\
& =|a_1 T(e_1)+...+a_nT(e_n)|\
& =|a_1 lambda_1e_1+...+a_nlambda_ne_n|\
& = sqrt{|a_1 lambda_1|^2+...+|a_nlambda_n|^2}\
& leq sqrt{|a_1 lambda|^2+...+|a_nlambda|^2}\
& = lambda sqrt{|a_1|^2+...+|a_n|^2}\
& = lambda |x|\
& leq lambda\
end{split}
end{equation}
Since $x$ was an arbitrary element in $mathbb{C}^n$ such that $|x| leq 1$, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| leq lambda$$
Conversely, notice that $|lambda_i|=lambda$ for some $i$. Then, for that $i$, we have
begin{equation}
begin{split}
|Te_i|
& =|lambda_i e_1|\
& =|lambda_i||e_i|\
& =lambda\
end{split}
end{equation}
Since $|e_i|=1$, this implies that
$$|T|=sup_{|x| leq 1} |Tx| geq lambda$$
Thus, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| = lambda$$
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
$endgroup$
1
$begingroup$
your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
add a comment |
$begingroup$
Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $mathbb{C}^n$ such that $Ae_i=lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $lambda=text{max} , {|lambda_1|,...,|lambda_2|}$. Then for any $x in mathbb{C}^n$ such that $|x| leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n in mathbb{C}$, and thus we get that
begin{equation}
begin{split}
|Tx|
& =|T(a_1e_1+...+a_ne_n)|\
& =|a_1 T(e_1)+...+a_nT(e_n)|\
& =|a_1 lambda_1e_1+...+a_nlambda_ne_n|\
& = sqrt{|a_1 lambda_1|^2+...+|a_nlambda_n|^2}\
& leq sqrt{|a_1 lambda|^2+...+|a_nlambda|^2}\
& = lambda sqrt{|a_1|^2+...+|a_n|^2}\
& = lambda |x|\
& leq lambda\
end{split}
end{equation}
Since $x$ was an arbitrary element in $mathbb{C}^n$ such that $|x| leq 1$, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| leq lambda$$
Conversely, notice that $|lambda_i|=lambda$ for some $i$. Then, for that $i$, we have
begin{equation}
begin{split}
|Te_i|
& =|lambda_i e_1|\
& =|lambda_i||e_i|\
& =lambda\
end{split}
end{equation}
Since $|e_i|=1$, this implies that
$$|T|=sup_{|x| leq 1} |Tx| geq lambda$$
Thus, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| = lambda$$
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
$endgroup$
Recall that any self-adjoint matrix is diagonalizable in some orthonormal basis, i.e. there exists an orthonormal basis $e_1,...,e_n$ of $mathbb{C}^n$ such that $Ae_i=lambda_ie_i$ for $i=1,2,...,n$. Denote the linear transformation associated with matrix $A$ by $T$. Let $lambda=text{max} , {|lambda_1|,...,|lambda_2|}$. Then for any $x in mathbb{C}^n$ such that $|x| leq 1$, $x=a_1e_1+...+a_ne_n$ for some $a_1,...,a_n in mathbb{C}$, and thus we get that
begin{equation}
begin{split}
|Tx|
& =|T(a_1e_1+...+a_ne_n)|\
& =|a_1 T(e_1)+...+a_nT(e_n)|\
& =|a_1 lambda_1e_1+...+a_nlambda_ne_n|\
& = sqrt{|a_1 lambda_1|^2+...+|a_nlambda_n|^2}\
& leq sqrt{|a_1 lambda|^2+...+|a_nlambda|^2}\
& = lambda sqrt{|a_1|^2+...+|a_n|^2}\
& = lambda |x|\
& leq lambda\
end{split}
end{equation}
Since $x$ was an arbitrary element in $mathbb{C}^n$ such that $|x| leq 1$, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| leq lambda$$
Conversely, notice that $|lambda_i|=lambda$ for some $i$. Then, for that $i$, we have
begin{equation}
begin{split}
|Te_i|
& =|lambda_i e_1|\
& =|lambda_i||e_i|\
& =lambda\
end{split}
end{equation}
Since $|e_i|=1$, this implies that
$$|T|=sup_{|x| leq 1} |Tx| geq lambda$$
Thus, we conclude that
$$|T|=sup_{|x| leq 1} |Tx| = lambda$$
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
edited Jan 13 at 17:26
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
answered Dec 13 '16 at 16:20
PawelPawel
3,1641022
3,1641022
1
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your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
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– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
add a comment |
1
$begingroup$
your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
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– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
1
1
$begingroup$
your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
your link is dead: "Error. Page cannot be displayed. Please contact your service provider for more details. (13)"
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 7:03
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
$begingroup$
@Peyman Thank you. I removed the link.
$endgroup$
– Pawel
Jan 13 at 17:26
add a comment |
$begingroup$
The norm of an operator (as derived from the norm on the space) is given by
$$
|M| = max_{x neq 0} frac{|Mx|}{|x|} = max_{|x| = 1} |M x|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
|M| = max_{|x| = 1} sqrt{(Mx)^*Mx} = sqrt{max_{|x| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$. It may help to set $y = U^*x$, noting that $|y| = |x|$.
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
The norm of an operator (as derived from the norm on the space) is given by
$$
|M| = max_{x neq 0} frac{|Mx|}{|x|} = max_{|x| = 1} |M x|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
|M| = max_{|x| = 1} sqrt{(Mx)^*Mx} = sqrt{max_{|x| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$. It may help to set $y = U^*x$, noting that $|y| = |x|$.
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
The norm of an operator (as derived from the norm on the space) is given by
$$
|M| = max_{x neq 0} frac{|Mx|}{|x|} = max_{|x| = 1} |M x|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
|M| = max_{|x| = 1} sqrt{(Mx)^*Mx} = sqrt{max_{|x| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$. It may help to set $y = U^*x$, noting that $|y| = |x|$.
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
The norm of an operator (as derived from the norm on the space) is given by
$$
|M| = max_{x neq 0} frac{|Mx|}{|x|} = max_{|x| = 1} |M x|
$$
If you're referring to the Euclidean norm (i.e. the 2-norm), then
$$
|M| = max_{|x| = 1} sqrt{(Mx)^*Mx} = sqrt{max_{|x| = 1} x^*M^*Mx}
$$
Hint: note that we can write $M = UDU^*$ for real diagonal matrix $D$ and unitary matrix $U$. It may help to set $y = U^*x$, noting that $|y| = |x|$.
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
edited Apr 8 '14 at 2:08
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
answered Apr 8 '14 at 1:53
OmnomnomnomOmnomnomnom
128k791183
128k791183
add a comment |
add a comment |
$begingroup$
This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,ldots,e_n$, (let $lambda_i$ be the evalues arranged so that $|lambda_1| geq cdots |lambda_n|$). For $v in V$ let $a^v_1,ldots,a_n^v$ be the coefficients satisfying
$v = sum_{i=1}^n a_i^v e_i$. Then
begin{align}
|M |^2
&= sup{ langle Mv,Mvrangle mid |v| = 1 } \
&= supleft{ leftlangle Msum_{i=1}^n a^v_i e_i,
M sum_{i=1}^n a^v_i e_irightrangle mid |v| = 1 right} \
&= supleft{ leftlangle sum_{i=1}^n a^v_i lambda_i e_i,
sum_{i=1}^n a^v_i lambda_i e_irightrangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2 |lambda_i|^2 langle e_i,
e_irangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2|lambda_i|^2 mid |v| = 1 right}
end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $|v|=1$. The max occurs when all mass is put at the largest component, $|lambda_1|^2$. Therefore
$$
|M|^2
= |lambda_1|^2.
$$
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
$endgroup$
add a comment |
$begingroup$
This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,ldots,e_n$, (let $lambda_i$ be the evalues arranged so that $|lambda_1| geq cdots |lambda_n|$). For $v in V$ let $a^v_1,ldots,a_n^v$ be the coefficients satisfying
$v = sum_{i=1}^n a_i^v e_i$. Then
begin{align}
|M |^2
&= sup{ langle Mv,Mvrangle mid |v| = 1 } \
&= supleft{ leftlangle Msum_{i=1}^n a^v_i e_i,
M sum_{i=1}^n a^v_i e_irightrangle mid |v| = 1 right} \
&= supleft{ leftlangle sum_{i=1}^n a^v_i lambda_i e_i,
sum_{i=1}^n a^v_i lambda_i e_irightrangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2 |lambda_i|^2 langle e_i,
e_irangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2|lambda_i|^2 mid |v| = 1 right}
end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $|v|=1$. The max occurs when all mass is put at the largest component, $|lambda_1|^2$. Therefore
$$
|M|^2
= |lambda_1|^2.
$$
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
$endgroup$
add a comment |
$begingroup$
This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,ldots,e_n$, (let $lambda_i$ be the evalues arranged so that $|lambda_1| geq cdots |lambda_n|$). For $v in V$ let $a^v_1,ldots,a_n^v$ be the coefficients satisfying
$v = sum_{i=1}^n a_i^v e_i$. Then
begin{align}
|M |^2
&= sup{ langle Mv,Mvrangle mid |v| = 1 } \
&= supleft{ leftlangle Msum_{i=1}^n a^v_i e_i,
M sum_{i=1}^n a^v_i e_irightrangle mid |v| = 1 right} \
&= supleft{ leftlangle sum_{i=1}^n a^v_i lambda_i e_i,
sum_{i=1}^n a^v_i lambda_i e_irightrangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2 |lambda_i|^2 langle e_i,
e_irangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2|lambda_i|^2 mid |v| = 1 right}
end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $|v|=1$. The max occurs when all mass is put at the largest component, $|lambda_1|^2$. Therefore
$$
|M|^2
= |lambda_1|^2.
$$
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
$endgroup$
This follows from the Min-max theorem, where the subspace of the Dirichlet form is the whole space.
More detail:
If $M$ is self adjoint, there is an orthonormal basis consisting of evectors of $M$, say $e_1,ldots,e_n$, (let $lambda_i$ be the evalues arranged so that $|lambda_1| geq cdots |lambda_n|$). For $v in V$ let $a^v_1,ldots,a_n^v$ be the coefficients satisfying
$v = sum_{i=1}^n a_i^v e_i$. Then
begin{align}
|M |^2
&= sup{ langle Mv,Mvrangle mid |v| = 1 } \
&= supleft{ leftlangle Msum_{i=1}^n a^v_i e_i,
M sum_{i=1}^n a^v_i e_irightrangle mid |v| = 1 right} \
&= supleft{ leftlangle sum_{i=1}^n a^v_i lambda_i e_i,
sum_{i=1}^n a^v_i lambda_i e_irightrangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2 |lambda_i|^2 langle e_i,
e_irangle mid |v| = 1 right} \
&= supleft{ sum_{i=1}^n |a_i^v|^2|lambda_i|^2 mid |v| = 1 right}
end{align}
(the cross terms were $0$ because of orthogonality). Now, you have a sum of squares, and the coefficients $|a_i^v|^2$ must sum to $1$ since $|v|=1$. The max occurs when all mass is put at the largest component, $|lambda_1|^2$. Therefore
$$
|M|^2
= |lambda_1|^2.
$$
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
edited Apr 8 '14 at 2:39
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
answered Apr 8 '14 at 2:00
user139388user139388
3,129715
3,129715
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