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The spectral norm of real matrices with positive entries is increasing in its entries?
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Suppose that I restrict myself to $M_n times n(mathbbR_+)$ the set of real matrices with positive entries that are square and have size $n$, and I denote by $|cdot|_2$ the spectral norm of members of this set. I am wondering if the spectral norm is increasing in its entries, which it intuitively should.
Let me try to formalize this property:
$forall (A:[a_ij], B:[b_ij]) in M_n times n(mathbbR_+)^2$ such that $forall (i,j) in [n]^2, a_ij leq b_ij$ then $ |A|_2 leq |B|_2$
linear-algebra functional-analysis norm singularvalues
asked Jul 17 at 7:31
ippiki-ookami
350217
add a comment |Â
up vote
1
down vote
favorite
Suppose that I restrict myself to $M_n times n(mathbbR_+)$ the set of real matrices with positive entries that are square and have size $n$, and I denote by $|cdot|_2$ the spectral norm of members of this set. I am wondering if the spectral norm is increasing in its entries, which it intuitively should.
Let me try to formalize this property:
$forall (A:[a_ij], B:[b_ij]) in M_n times n(mathbbR_+)^2$ such that $forall (i,j) in [n]^2, a_ij leq b_ij$ then $ |A|_2 leq |B|_2$
linear-algebra functional-analysis norm singularvalues
asked Jul 17 at 7:31
ippiki-ookami
350217
1
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that I restrict myself to $M_n times n(mathbbR_+)$ the set of real matrices with positive entries that are square and have size $n$, and I denote by $|cdot|_2$ the spectral norm of members of this set. I am wondering if the spectral norm is increasing in its entries, which it intuitively should.
Let me try to formalize this property:
$forall (A:[a_ij], B:[b_ij]) in M_n times n(mathbbR_+)^2$ such that $forall (i,j) in [n]^2, a_ij leq b_ij$ then $ |A|_2 leq |B|_2$
linear-algebra functional-analysis norm singularvalues
asked Jul 17 at 7:31
ippiki-ookami
350217
Suppose that I restrict myself to $M_n times n(mathbbR_+)$ the set of real matrices with positive entries that are square and have size $n$, and I denote by $|cdot|_2$ the spectral norm of members of this set. I am wondering if the spectral norm is increasing in its entries, which it intuitively should.
Let me try to formalize this property:
$forall (A:[a_ij], B:[b_ij]) in M_n times n(mathbbR_+)^2$ such that $forall (i,j) in [n]^2, a_ij leq b_ij$ then $ |A|_2 leq |B|_2$
linear-algebra functional-analysis norm singularvalues
asked Jul 17 at 7:31
ippiki-ookami
350217
asked Jul 17 at 7:31
ippiki-ookami
350217
asked Jul 17 at 7:31
ippiki-ookami
350217
asked Jul 17 at 7:31
ippiki-ookami
350217
350217
1
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06
add a comment |Â
1
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06
1
1
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06
add a comment |Â
1 Answer
1
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1
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accepted
First note that $$||A||_2=sup__2=1||Ax||_2$$if we denote the entries of $A$ by $a_ij$ we conclude that$$||A||_2=sup__2=1sqrtsum_i=1^n(a_i1x_1+a_i2x_2+cdots +a_inx_n)^2$$since all the entries of $A$ are non-negative a supremum is achieved when $x_ige 0$ for all $i$ (or $x_ile 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $Isubseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_i1x_1+a_i2x_2+cdots +a_inx_n+sum_knotin Ia_ik$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\||B||_2=||By^*||_2$$therefore $$||By^*||_2ge ||Bx^*||_2=||Ax^*+Xx^*||_2ge||Ax^*||_2$$the last equality is true because of the following lemma
For $r_1,r_2in left(R^ge0right)^n$ we have $$||r_1+r_2||_2ge||r_1||_2$$where the equality is iff $r_2=0$.
proof: use the definition.
Therefore our proof is complete.
P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.
Conclusion: your theorem is right and the equality holds iff $$A=B$$
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First note that $$||A||_2=sup__2=1||Ax||_2$$if we denote the entries of $A$ by $a_ij$ we conclude that$$||A||_2=sup__2=1sqrtsum_i=1^n(a_i1x_1+a_i2x_2+cdots +a_inx_n)^2$$since all the entries of $A$ are non-negative a supremum is achieved when $x_ige 0$ for all $i$ (or $x_ile 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $Isubseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_i1x_1+a_i2x_2+cdots +a_inx_n+sum_knotin Ia_ik$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\||B||_2=||By^*||_2$$therefore $$||By^*||_2ge ||Bx^*||_2=||Ax^*+Xx^*||_2ge||Ax^*||_2$$the last equality is true because of the following lemma
For $r_1,r_2in left(R^ge0right)^n$ we have $$||r_1+r_2||_2ge||r_1||_2$$where the equality is iff $r_2=0$.
proof: use the definition.
Therefore our proof is complete.
P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.
Conclusion: your theorem is right and the equality holds iff $$A=B$$
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
add a comment |Â
up vote
1
down vote
accepted
First note that $$||A||_2=sup__2=1||Ax||_2$$if we denote the entries of $A$ by $a_ij$ we conclude that$$||A||_2=sup__2=1sqrtsum_i=1^n(a_i1x_1+a_i2x_2+cdots +a_inx_n)^2$$since all the entries of $A$ are non-negative a supremum is achieved when $x_ige 0$ for all $i$ (or $x_ile 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $Isubseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_i1x_1+a_i2x_2+cdots +a_inx_n+sum_knotin Ia_ik$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\||B||_2=||By^*||_2$$therefore $$||By^*||_2ge ||Bx^*||_2=||Ax^*+Xx^*||_2ge||Ax^*||_2$$the last equality is true because of the following lemma
For $r_1,r_2in left(R^ge0right)^n$ we have $$||r_1+r_2||_2ge||r_1||_2$$where the equality is iff $r_2=0$.
proof: use the definition.
Therefore our proof is complete.
P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.
Conclusion: your theorem is right and the equality holds iff $$A=B$$
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First note that $$||A||_2=sup__2=1||Ax||_2$$if we denote the entries of $A$ by $a_ij$ we conclude that$$||A||_2=sup__2=1sqrtsum_i=1^n(a_i1x_1+a_i2x_2+cdots +a_inx_n)^2$$since all the entries of $A$ are non-negative a supremum is achieved when $x_ige 0$ for all $i$ (or $x_ile 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $Isubseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_i1x_1+a_i2x_2+cdots +a_inx_n+sum_knotin Ia_ik$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\||B||_2=||By^*||_2$$therefore $$||By^*||_2ge ||Bx^*||_2=||Ax^*+Xx^*||_2ge||Ax^*||_2$$the last equality is true because of the following lemma
For $r_1,r_2in left(R^ge0right)^n$ we have $$||r_1+r_2||_2ge||r_1||_2$$where the equality is iff $r_2=0$.
proof: use the definition.
Therefore our proof is complete.
P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.
Conclusion: your theorem is right and the equality holds iff $$A=B$$
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
First note that $$||A||_2=sup__2=1||Ax||_2$$if we denote the entries of $A$ by $a_ij$ we conclude that$$||A||_2=sup__2=1sqrtsum_i=1^n(a_i1x_1+a_i2x_2+cdots +a_inx_n)^2$$since all the entries of $A$ are non-negative a supremum is achieved when $x_ige 0$ for all $i$ (or $x_ile 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $Isubseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_i1x_1+a_i2x_2+cdots +a_inx_n+sum_knotin Ia_ik$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\||B||_2=||By^*||_2$$therefore $$||By^*||_2ge ||Bx^*||_2=||Ax^*+Xx^*||_2ge||Ax^*||_2$$the last equality is true because of the following lemma
For $r_1,r_2in left(R^ge0right)^n$ we have $$||r_1+r_2||_2ge||r_1||_2$$where the equality is iff $r_2=0$.
proof: use the definition.
Therefore our proof is complete.
P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.
Conclusion: your theorem is right and the equality holds iff $$A=B$$
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
edited Aug 5 at 7:12
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
answered Jul 17 at 7:54
Mostafa Ayaz
8,6023630
8,6023630
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
add a comment |Â
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
You are saying that the spectral norm of a matrix is equal its maximum column 2-norm ? Take $A = bigl( beginsmallmatrix 2 & 1 \ 1 & 1 endsmallmatrix bigr)$. Its spectral norm is around 2.618 while its maximum column 2-norm is $sqrt5 approx 2.236$.
– ippiki-ookami
Jul 22 at 7:29
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Do you agree or disagree with my comment ?
– ippiki-ookami
Aug 5 at 6:13
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes surely! I'm sorry for such a very long delay in response and my first wrong proof but finally i've found the way of proof. Hope it helps you this time :)
– Mostafa Ayaz
Aug 5 at 7:14
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Yes I agree with your new proof. You incidentally show that the vector that achieves the spectral norm has entries all of the same sign, which I also find interesting in its own right.
– ippiki-ookami
Aug 5 at 10:55
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
Well. That was a nice (and of course tough!) question (+1) ......
– Mostafa Ayaz
Aug 5 at 14:10
add a comment |Â
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1
Spectral norm is a confusing term. We have the spectral radius and the operator norm. Which one are you interested in?
– Kavi Rama Murthy
Jul 17 at 8:03
The one induced by the $ell_2$ norm.
– ippiki-ookami
Jul 17 at 8:06