Mixing
If $A = A^*$ then $ lambda_{min} leqfrac{langle Av,v rangle}{langle v,v rangle}leqlambda_{max} $
$begingroup$
I'm leaning linear algebra and new to it. I have trouble with this problem and actually, I don't know what to do! any help or hint would be appreciated.
for the linear operator $Ain ell(V)$ wich $V$ is an Inner product space with finite dimentions. if $A = A^*$ then show that for every $v in V$ we have:
$$ lambda_{min} leqfrac{langle Av,vrangle}{langle v,vrangle}leqlambda_{max}, v neq 0$$ where the lambdas are eigenvalues.
linear-algebra eigenvalues-eigenvectors operator-theory inner-product-space
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
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add a comment |
$begingroup$
I'm leaning linear algebra and new to it. I have trouble with this problem and actually, I don't know what to do! any help or hint would be appreciated.
for the linear operator $Ain ell(V)$ wich $V$ is an Inner product space with finite dimentions. if $A = A^*$ then show that for every $v in V$ we have:
$$ lambda_{min} leqfrac{langle Av,vrangle}{langle v,vrangle}leqlambda_{max}, v neq 0$$ where the lambdas are eigenvalues.
linear-algebra eigenvalues-eigenvectors operator-theory inner-product-space
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
add a comment |
$begingroup$
I'm leaning linear algebra and new to it. I have trouble with this problem and actually, I don't know what to do! any help or hint would be appreciated.
for the linear operator $Ain ell(V)$ wich $V$ is an Inner product space with finite dimentions. if $A = A^*$ then show that for every $v in V$ we have:
$$ lambda_{min} leqfrac{langle Av,vrangle}{langle v,vrangle}leqlambda_{max}, v neq 0$$ where the lambdas are eigenvalues.
linear-algebra eigenvalues-eigenvectors operator-theory inner-product-space
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
I'm leaning linear algebra and new to it. I have trouble with this problem and actually, I don't know what to do! any help or hint would be appreciated.
for the linear operator $Ain ell(V)$ wich $V$ is an Inner product space with finite dimentions. if $A = A^*$ then show that for every $v in V$ we have:
$$ lambda_{min} leqfrac{langle Av,vrangle}{langle v,vrangle}leqlambda_{max}, v neq 0$$ where the lambdas are eigenvalues.
linear-algebra eigenvalues-eigenvectors operator-theory inner-product-space
linear-algebra eigenvalues-eigenvectors operator-theory inner-product-space
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
edited Jan 13 at 23:30
Peyman mohseni kiasari
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 13 at 22:56
Peyman mohseni kiasariPeyman mohseni kiasari
1089
1089
add a comment |
add a comment |
2 Answers
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oldest
votes
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Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $lambda_1,...,lambda_n$.
Let $[v]_mathcal{B}=hat{v}$. Then $hat{v}=sum a_i v_i$ for constants $a_i$, so $$left<Av,vright>=left<sum a_ilambda_i v_i,hat{v}right>leq lambda_{text{max}}left<sum a_iv_i,hat{v}right>=lambda_{text{max}}left<v,vright>.$$
We derive the result for $lambda_{text{min}}$ similarly.
answered Jan 13 at 23:16
MelodyMelody
80012
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add a comment |
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I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $lambda_1, ldots, lambda_n$. Then
$$langle Av, v rangle = v^* U D U^* v = sum_{i=1}^n lambda_i (U^* v)_i^2.$$
Use $lambda_{min} le lambda_i le lambda_{max}$ for all $i$ as well as the fact that $sum_{i=1}^n (U^* v)_i^2 = langle U^*v, U^* v rangle = langle v, vrangle$ to conclude.
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
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Isn't $A$ assumed a general linear operator and not a matrix?
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– Melody
Jan 13 at 23:17
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@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
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thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $lambda_1,...,lambda_n$.
Let $[v]_mathcal{B}=hat{v}$. Then $hat{v}=sum a_i v_i$ for constants $a_i$, so $$left<Av,vright>=left<sum a_ilambda_i v_i,hat{v}right>leq lambda_{text{max}}left<sum a_iv_i,hat{v}right>=lambda_{text{max}}left<v,vright>.$$
We derive the result for $lambda_{text{min}}$ similarly.
answered Jan 13 at 23:16
MelodyMelody
80012
$endgroup$
add a comment |
$begingroup$
Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $lambda_1,...,lambda_n$.
Let $[v]_mathcal{B}=hat{v}$. Then $hat{v}=sum a_i v_i$ for constants $a_i$, so $$left<Av,vright>=left<sum a_ilambda_i v_i,hat{v}right>leq lambda_{text{max}}left<sum a_iv_i,hat{v}right>=lambda_{text{max}}left<v,vright>.$$
We derive the result for $lambda_{text{min}}$ similarly.
answered Jan 13 at 23:16
MelodyMelody
80012
$endgroup$
add a comment |
$begingroup$
Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $lambda_1,...,lambda_n$.
Let $[v]_mathcal{B}=hat{v}$. Then $hat{v}=sum a_i v_i$ for constants $a_i$, so $$left<Av,vright>=left<sum a_ilambda_i v_i,hat{v}right>leq lambda_{text{max}}left<sum a_iv_i,hat{v}right>=lambda_{text{max}}left<v,vright>.$$
We derive the result for $lambda_{text{min}}$ similarly.
answered Jan 13 at 23:16
MelodyMelody
80012
$endgroup$
Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $lambda_1,...,lambda_n$.
Let $[v]_mathcal{B}=hat{v}$. Then $hat{v}=sum a_i v_i$ for constants $a_i$, so $$left<Av,vright>=left<sum a_ilambda_i v_i,hat{v}right>leq lambda_{text{max}}left<sum a_iv_i,hat{v}right>=lambda_{text{max}}left<v,vright>.$$
We derive the result for $lambda_{text{min}}$ similarly.
answered Jan 13 at 23:16
MelodyMelody
80012
edited Jan 13 at 23:29
answered Jan 13 at 23:16
MelodyMelody
80012
answered Jan 13 at 23:16
MelodyMelody
80012
answered Jan 13 at 23:16
MelodyMelody
80012
80012
add a comment |
add a comment |
$begingroup$
I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $lambda_1, ldots, lambda_n$. Then
$$langle Av, v rangle = v^* U D U^* v = sum_{i=1}^n lambda_i (U^* v)_i^2.$$
Use $lambda_{min} le lambda_i le lambda_{max}$ for all $i$ as well as the fact that $sum_{i=1}^n (U^* v)_i^2 = langle U^*v, U^* v rangle = langle v, vrangle$ to conclude.
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
add a comment |
$begingroup$
I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $lambda_1, ldots, lambda_n$. Then
$$langle Av, v rangle = v^* U D U^* v = sum_{i=1}^n lambda_i (U^* v)_i^2.$$
Use $lambda_{min} le lambda_i le lambda_{max}$ for all $i$ as well as the fact that $sum_{i=1}^n (U^* v)_i^2 = langle U^*v, U^* v rangle = langle v, vrangle$ to conclude.
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
add a comment |
$begingroup$
I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $lambda_1, ldots, lambda_n$. Then
$$langle Av, v rangle = v^* U D U^* v = sum_{i=1}^n lambda_i (U^* v)_i^2.$$
Use $lambda_{min} le lambda_i le lambda_{max}$ for all $i$ as well as the fact that $sum_{i=1}^n (U^* v)_i^2 = langle U^*v, U^* v rangle = langle v, vrangle$ to conclude.
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
$endgroup$
I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $lambda_1, ldots, lambda_n$. Then
$$langle Av, v rangle = v^* U D U^* v = sum_{i=1}^n lambda_i (U^* v)_i^2.$$
Use $lambda_{min} le lambda_i le lambda_{max}$ for all $i$ as well as the fact that $sum_{i=1}^n (U^* v)_i^2 = langle U^*v, U^* v rangle = langle v, vrangle$ to conclude.
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
answered Jan 13 at 23:08
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
add a comment |
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
Isn't $A$ assumed a general linear operator and not a matrix?
$endgroup$
– Melody
Jan 13 at 23:17
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
@Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa
$endgroup$
– angryavian
Jan 13 at 23:19
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
thanks, but why $langle Av, v rangle = v^* U D U^* v$?
$endgroup$
– Peyman mohseni kiasari
Jan 13 at 23:20
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty
$endgroup$
– Melody
Jan 13 at 23:21
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
$begingroup$
@Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices.
$endgroup$
– angryavian
Jan 13 at 23:22
add a comment |
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