Mixing
Monotone matrix norms
[Ciarlet 2.2-10]
- Let $mathscr{S}_n$ be the set of symmetric matrices and $mathscr{S}_n^+$ the subset of non-negative definite symmetric matrices. A matrix norm $|cdot|$ to be monotone if
$$Ainmathscr{S}_n^+; wedge; B-Ainmathscr{S}_n^+ Rightarrow |A| leq |B|.$$
Show that the norms $|cdot|_2$ and $|cdot|_F$ (Frobenus norm) are monotone. - More generally, show that if a matrix norm $|cdot|$ is invariant under unitary transformations, that is, if $|A| = |AU| = |UA|$ for every unitary matrix $U$, then it is monotone.
- Let $|cdot|$ be a monotone norm and $mbox{cond}(cdot)$ the condition number function associated with it. Prove that
$$A,Binmathscr{S}_n^* Rightarrow mbox{cond}(A+B) leq maxleft{mbox{cond}(A),; mbox{cond}(B)right}$$
where $mathscr{S}_n^*$ denotes the subset of positive definite symmetric matrices.
I already have proved (1), and I proved that $lambda_i(A) leq lambda_i(B)$, $forall i=1,2,ldots,n$ and $forall A,B-Ainmathscr{S}_n^+$. But I have had problems in order to prove (2) and (3). For (2), i proved that
begin{eqnarray*}
|A| & = & |U^*AU| = |mbox{diag}(lambda_i(A))|,\[0.3cm]
|B| & = & |V^*BV| = |mbox{diag}(lambda_i(B))|.
end{eqnarray*}
but I don't know what I should do next. Please help me and thanks in advance.
linear-algebra norm
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
add a comment |
[Ciarlet 2.2-10]
- Let $mathscr{S}_n$ be the set of symmetric matrices and $mathscr{S}_n^+$ the subset of non-negative definite symmetric matrices. A matrix norm $|cdot|$ to be monotone if
$$Ainmathscr{S}_n^+; wedge; B-Ainmathscr{S}_n^+ Rightarrow |A| leq |B|.$$
Show that the norms $|cdot|_2$ and $|cdot|_F$ (Frobenus norm) are monotone. - More generally, show that if a matrix norm $|cdot|$ is invariant under unitary transformations, that is, if $|A| = |AU| = |UA|$ for every unitary matrix $U$, then it is monotone.
- Let $|cdot|$ be a monotone norm and $mbox{cond}(cdot)$ the condition number function associated with it. Prove that
$$A,Binmathscr{S}_n^* Rightarrow mbox{cond}(A+B) leq maxleft{mbox{cond}(A),; mbox{cond}(B)right}$$
where $mathscr{S}_n^*$ denotes the subset of positive definite symmetric matrices.
I already have proved (1), and I proved that $lambda_i(A) leq lambda_i(B)$, $forall i=1,2,ldots,n$ and $forall A,B-Ainmathscr{S}_n^+$. But I have had problems in order to prove (2) and (3). For (2), i proved that
begin{eqnarray*}
|A| & = & |U^*AU| = |mbox{diag}(lambda_i(A))|,\[0.3cm]
|B| & = & |V^*BV| = |mbox{diag}(lambda_i(B))|.
end{eqnarray*}
but I don't know what I should do next. Please help me and thanks in advance.
linear-algebra norm
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
add a comment |
[Ciarlet 2.2-10]
- Let $mathscr{S}_n$ be the set of symmetric matrices and $mathscr{S}_n^+$ the subset of non-negative definite symmetric matrices. A matrix norm $|cdot|$ to be monotone if
$$Ainmathscr{S}_n^+; wedge; B-Ainmathscr{S}_n^+ Rightarrow |A| leq |B|.$$
Show that the norms $|cdot|_2$ and $|cdot|_F$ (Frobenus norm) are monotone. - More generally, show that if a matrix norm $|cdot|$ is invariant under unitary transformations, that is, if $|A| = |AU| = |UA|$ for every unitary matrix $U$, then it is monotone.
- Let $|cdot|$ be a monotone norm and $mbox{cond}(cdot)$ the condition number function associated with it. Prove that
$$A,Binmathscr{S}_n^* Rightarrow mbox{cond}(A+B) leq maxleft{mbox{cond}(A),; mbox{cond}(B)right}$$
where $mathscr{S}_n^*$ denotes the subset of positive definite symmetric matrices.
I already have proved (1), and I proved that $lambda_i(A) leq lambda_i(B)$, $forall i=1,2,ldots,n$ and $forall A,B-Ainmathscr{S}_n^+$. But I have had problems in order to prove (2) and (3). For (2), i proved that
begin{eqnarray*}
|A| & = & |U^*AU| = |mbox{diag}(lambda_i(A))|,\[0.3cm]
|B| & = & |V^*BV| = |mbox{diag}(lambda_i(B))|.
end{eqnarray*}
but I don't know what I should do next. Please help me and thanks in advance.
linear-algebra norm
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
[Ciarlet 2.2-10]
- Let $mathscr{S}_n$ be the set of symmetric matrices and $mathscr{S}_n^+$ the subset of non-negative definite symmetric matrices. A matrix norm $|cdot|$ to be monotone if
$$Ainmathscr{S}_n^+; wedge; B-Ainmathscr{S}_n^+ Rightarrow |A| leq |B|.$$
Show that the norms $|cdot|_2$ and $|cdot|_F$ (Frobenus norm) are monotone. - More generally, show that if a matrix norm $|cdot|$ is invariant under unitary transformations, that is, if $|A| = |AU| = |UA|$ for every unitary matrix $U$, then it is monotone.
- Let $|cdot|$ be a monotone norm and $mbox{cond}(cdot)$ the condition number function associated with it. Prove that
$$A,Binmathscr{S}_n^* Rightarrow mbox{cond}(A+B) leq maxleft{mbox{cond}(A),; mbox{cond}(B)right}$$
where $mathscr{S}_n^*$ denotes the subset of positive definite symmetric matrices.
I already have proved (1), and I proved that $lambda_i(A) leq lambda_i(B)$, $forall i=1,2,ldots,n$ and $forall A,B-Ainmathscr{S}_n^+$. But I have had problems in order to prove (2) and (3). For (2), i proved that
begin{eqnarray*}
|A| & = & |U^*AU| = |mbox{diag}(lambda_i(A))|,\[0.3cm]
|B| & = & |V^*BV| = |mbox{diag}(lambda_i(B))|.
end{eqnarray*}
but I don't know what I should do next. Please help me and thanks in advance.
linear-algebra norm
linear-algebra norm
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
edited Feb 26 '17 at 3:21
Martin Argerami
124k1176174
124k1176174
asked Apr 27 '13 at 17:56
FASCH
652926
asked Apr 27 '13 at 17:56
FASCH
652926
asked Apr 27 '13 at 17:56
FASCH
652926
652926
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
For part 2:
To simplify notation, define a function $g:mathbb R^ntomathbb R$ by $g(x_1,ldots,x_n)=|text{diag}(x_j)|$. Since $|cdot|$ is a unitarily invariant matrix norm, we have that
$g$ is a norm on $mathbb R^n$
$g(x_1,ldots,x_n)=g(|x_1|,ldots,|x_n|)$ (this comes from the unitary invariance)
$g(x_1,ldots,x_n)=g(x_{sigma(1)},ldots,x_{sigma(n)})$ for any permutation $sigma$.
Such a $g$ is called a gauge function.
Now, if $tin[0,1]$, then (writing $x=(x_1,ldots,x_n)$)
begin{align}
g(tx_1,x_2,ldots,x_n)&=gleft(frac{1+t}2,x+frac{1-t}2(-x_1,x_2,ldots,x_nright)\ \
&leqfrac{1+t}2,g(x)+frac{1-t}2g(-x_1,x_2,ldots,x_n)\ \
&=frac{1+t}2g(x)+frac{1-t}2g(x)=g(x).
end{align}
Applying the above inductively, we get
$$
g(t_1x_1,ldots,t_nx_n)leq g(x)
$$
whenever $t_1,ldots,t_nin [0,1]$.
Since $0leqlambda_j(A)leqlambda_j(B)$ for all $j$, we have $lambda_j(A)=t_jlambda_j(B)$ for appropriate $t_1,ldots,t_nin[0,1]$.
Then
begin{align}
|A|&=|text{diag}(lambda_j(A))|=|text{diag}(t_j,lambda_j(B))|\ \
&leq |text{diag}(lambda_j(B))|=|B|
end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $text{cond}(A)leqtext{cond}(B)$. Let $A'=A/|A|$, $B'=B/|B|$, and $t=frac{|A|}{|A|+|B|}$.
We have, in the new notation, that $$tag{0}|(A')^{-1}|leq|(B')^{-1}|.$$ And
$$tag{1}
|tA'+(1-t)B'|leq t|A'|+(1-t)|B'|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$tag{2}
(tA'+(1-t)B')^{-1}leq t(A')^{-1}+(1-t)(B')^{-1}.
$$
Thus, using $(2)$ and $(0)$,
begin{align}tag{3}
|(tA'+(1-t)B')^{-1}|&leq|t(A')^{-1}+(1-t)(B')^{-1}|\
&leq t|(A')^{-1}|+(1-t)|(B')^{-1}|\ &leq|(B')^{-1}|
end{align}
Now, combining $(1)$, and $(3)$,
begin{align}tag{4}
|tA'+(1-t)B'|,|(tA'+(1-t)B')^{-1}|&leq |(tA'+(1-t)B')^{-1}|leq|(B')^{-1}|.
end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=frac1{|A|+|B|},left(A+Bright), (B')^{-1}=|B|,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
|A+B|,|(A+B)^{-1}|leq |B|,|B^{-1}|,
$$
as desired.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
add a comment |
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1 Answer
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1 Answer
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For part 2:
To simplify notation, define a function $g:mathbb R^ntomathbb R$ by $g(x_1,ldots,x_n)=|text{diag}(x_j)|$. Since $|cdot|$ is a unitarily invariant matrix norm, we have that
$g$ is a norm on $mathbb R^n$
$g(x_1,ldots,x_n)=g(|x_1|,ldots,|x_n|)$ (this comes from the unitary invariance)
$g(x_1,ldots,x_n)=g(x_{sigma(1)},ldots,x_{sigma(n)})$ for any permutation $sigma$.
Such a $g$ is called a gauge function.
Now, if $tin[0,1]$, then (writing $x=(x_1,ldots,x_n)$)
begin{align}
g(tx_1,x_2,ldots,x_n)&=gleft(frac{1+t}2,x+frac{1-t}2(-x_1,x_2,ldots,x_nright)\ \
&leqfrac{1+t}2,g(x)+frac{1-t}2g(-x_1,x_2,ldots,x_n)\ \
&=frac{1+t}2g(x)+frac{1-t}2g(x)=g(x).
end{align}
Applying the above inductively, we get
$$
g(t_1x_1,ldots,t_nx_n)leq g(x)
$$
whenever $t_1,ldots,t_nin [0,1]$.
Since $0leqlambda_j(A)leqlambda_j(B)$ for all $j$, we have $lambda_j(A)=t_jlambda_j(B)$ for appropriate $t_1,ldots,t_nin[0,1]$.
Then
begin{align}
|A|&=|text{diag}(lambda_j(A))|=|text{diag}(t_j,lambda_j(B))|\ \
&leq |text{diag}(lambda_j(B))|=|B|
end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $text{cond}(A)leqtext{cond}(B)$. Let $A'=A/|A|$, $B'=B/|B|$, and $t=frac{|A|}{|A|+|B|}$.
We have, in the new notation, that $$tag{0}|(A')^{-1}|leq|(B')^{-1}|.$$ And
$$tag{1}
|tA'+(1-t)B'|leq t|A'|+(1-t)|B'|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$tag{2}
(tA'+(1-t)B')^{-1}leq t(A')^{-1}+(1-t)(B')^{-1}.
$$
Thus, using $(2)$ and $(0)$,
begin{align}tag{3}
|(tA'+(1-t)B')^{-1}|&leq|t(A')^{-1}+(1-t)(B')^{-1}|\
&leq t|(A')^{-1}|+(1-t)|(B')^{-1}|\ &leq|(B')^{-1}|
end{align}
Now, combining $(1)$, and $(3)$,
begin{align}tag{4}
|tA'+(1-t)B'|,|(tA'+(1-t)B')^{-1}|&leq |(tA'+(1-t)B')^{-1}|leq|(B')^{-1}|.
end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=frac1{|A|+|B|},left(A+Bright), (B')^{-1}=|B|,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
|A+B|,|(A+B)^{-1}|leq |B|,|B^{-1}|,
$$
as desired.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
add a comment |
For part 2:
To simplify notation, define a function $g:mathbb R^ntomathbb R$ by $g(x_1,ldots,x_n)=|text{diag}(x_j)|$. Since $|cdot|$ is a unitarily invariant matrix norm, we have that
$g$ is a norm on $mathbb R^n$
$g(x_1,ldots,x_n)=g(|x_1|,ldots,|x_n|)$ (this comes from the unitary invariance)
$g(x_1,ldots,x_n)=g(x_{sigma(1)},ldots,x_{sigma(n)})$ for any permutation $sigma$.
Such a $g$ is called a gauge function.
Now, if $tin[0,1]$, then (writing $x=(x_1,ldots,x_n)$)
begin{align}
g(tx_1,x_2,ldots,x_n)&=gleft(frac{1+t}2,x+frac{1-t}2(-x_1,x_2,ldots,x_nright)\ \
&leqfrac{1+t}2,g(x)+frac{1-t}2g(-x_1,x_2,ldots,x_n)\ \
&=frac{1+t}2g(x)+frac{1-t}2g(x)=g(x).
end{align}
Applying the above inductively, we get
$$
g(t_1x_1,ldots,t_nx_n)leq g(x)
$$
whenever $t_1,ldots,t_nin [0,1]$.
Since $0leqlambda_j(A)leqlambda_j(B)$ for all $j$, we have $lambda_j(A)=t_jlambda_j(B)$ for appropriate $t_1,ldots,t_nin[0,1]$.
Then
begin{align}
|A|&=|text{diag}(lambda_j(A))|=|text{diag}(t_j,lambda_j(B))|\ \
&leq |text{diag}(lambda_j(B))|=|B|
end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $text{cond}(A)leqtext{cond}(B)$. Let $A'=A/|A|$, $B'=B/|B|$, and $t=frac{|A|}{|A|+|B|}$.
We have, in the new notation, that $$tag{0}|(A')^{-1}|leq|(B')^{-1}|.$$ And
$$tag{1}
|tA'+(1-t)B'|leq t|A'|+(1-t)|B'|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$tag{2}
(tA'+(1-t)B')^{-1}leq t(A')^{-1}+(1-t)(B')^{-1}.
$$
Thus, using $(2)$ and $(0)$,
begin{align}tag{3}
|(tA'+(1-t)B')^{-1}|&leq|t(A')^{-1}+(1-t)(B')^{-1}|\
&leq t|(A')^{-1}|+(1-t)|(B')^{-1}|\ &leq|(B')^{-1}|
end{align}
Now, combining $(1)$, and $(3)$,
begin{align}tag{4}
|tA'+(1-t)B'|,|(tA'+(1-t)B')^{-1}|&leq |(tA'+(1-t)B')^{-1}|leq|(B')^{-1}|.
end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=frac1{|A|+|B|},left(A+Bright), (B')^{-1}=|B|,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
|A+B|,|(A+B)^{-1}|leq |B|,|B^{-1}|,
$$
as desired.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
add a comment |
For part 2:
To simplify notation, define a function $g:mathbb R^ntomathbb R$ by $g(x_1,ldots,x_n)=|text{diag}(x_j)|$. Since $|cdot|$ is a unitarily invariant matrix norm, we have that
$g$ is a norm on $mathbb R^n$
$g(x_1,ldots,x_n)=g(|x_1|,ldots,|x_n|)$ (this comes from the unitary invariance)
$g(x_1,ldots,x_n)=g(x_{sigma(1)},ldots,x_{sigma(n)})$ for any permutation $sigma$.
Such a $g$ is called a gauge function.
Now, if $tin[0,1]$, then (writing $x=(x_1,ldots,x_n)$)
begin{align}
g(tx_1,x_2,ldots,x_n)&=gleft(frac{1+t}2,x+frac{1-t}2(-x_1,x_2,ldots,x_nright)\ \
&leqfrac{1+t}2,g(x)+frac{1-t}2g(-x_1,x_2,ldots,x_n)\ \
&=frac{1+t}2g(x)+frac{1-t}2g(x)=g(x).
end{align}
Applying the above inductively, we get
$$
g(t_1x_1,ldots,t_nx_n)leq g(x)
$$
whenever $t_1,ldots,t_nin [0,1]$.
Since $0leqlambda_j(A)leqlambda_j(B)$ for all $j$, we have $lambda_j(A)=t_jlambda_j(B)$ for appropriate $t_1,ldots,t_nin[0,1]$.
Then
begin{align}
|A|&=|text{diag}(lambda_j(A))|=|text{diag}(t_j,lambda_j(B))|\ \
&leq |text{diag}(lambda_j(B))|=|B|
end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $text{cond}(A)leqtext{cond}(B)$. Let $A'=A/|A|$, $B'=B/|B|$, and $t=frac{|A|}{|A|+|B|}$.
We have, in the new notation, that $$tag{0}|(A')^{-1}|leq|(B')^{-1}|.$$ And
$$tag{1}
|tA'+(1-t)B'|leq t|A'|+(1-t)|B'|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$tag{2}
(tA'+(1-t)B')^{-1}leq t(A')^{-1}+(1-t)(B')^{-1}.
$$
Thus, using $(2)$ and $(0)$,
begin{align}tag{3}
|(tA'+(1-t)B')^{-1}|&leq|t(A')^{-1}+(1-t)(B')^{-1}|\
&leq t|(A')^{-1}|+(1-t)|(B')^{-1}|\ &leq|(B')^{-1}|
end{align}
Now, combining $(1)$, and $(3)$,
begin{align}tag{4}
|tA'+(1-t)B'|,|(tA'+(1-t)B')^{-1}|&leq |(tA'+(1-t)B')^{-1}|leq|(B')^{-1}|.
end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=frac1{|A|+|B|},left(A+Bright), (B')^{-1}=|B|,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
|A+B|,|(A+B)^{-1}|leq |B|,|B^{-1}|,
$$
as desired.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
For part 2:
To simplify notation, define a function $g:mathbb R^ntomathbb R$ by $g(x_1,ldots,x_n)=|text{diag}(x_j)|$. Since $|cdot|$ is a unitarily invariant matrix norm, we have that
$g$ is a norm on $mathbb R^n$
$g(x_1,ldots,x_n)=g(|x_1|,ldots,|x_n|)$ (this comes from the unitary invariance)
$g(x_1,ldots,x_n)=g(x_{sigma(1)},ldots,x_{sigma(n)})$ for any permutation $sigma$.
Such a $g$ is called a gauge function.
Now, if $tin[0,1]$, then (writing $x=(x_1,ldots,x_n)$)
begin{align}
g(tx_1,x_2,ldots,x_n)&=gleft(frac{1+t}2,x+frac{1-t}2(-x_1,x_2,ldots,x_nright)\ \
&leqfrac{1+t}2,g(x)+frac{1-t}2g(-x_1,x_2,ldots,x_n)\ \
&=frac{1+t}2g(x)+frac{1-t}2g(x)=g(x).
end{align}
Applying the above inductively, we get
$$
g(t_1x_1,ldots,t_nx_n)leq g(x)
$$
whenever $t_1,ldots,t_nin [0,1]$.
Since $0leqlambda_j(A)leqlambda_j(B)$ for all $j$, we have $lambda_j(A)=t_jlambda_j(B)$ for appropriate $t_1,ldots,t_nin[0,1]$.
Then
begin{align}
|A|&=|text{diag}(lambda_j(A))|=|text{diag}(t_j,lambda_j(B))|\ \
&leq |text{diag}(lambda_j(B))|=|B|
end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $text{cond}(A)leqtext{cond}(B)$. Let $A'=A/|A|$, $B'=B/|B|$, and $t=frac{|A|}{|A|+|B|}$.
We have, in the new notation, that $$tag{0}|(A')^{-1}|leq|(B')^{-1}|.$$ And
$$tag{1}
|tA'+(1-t)B'|leq t|A'|+(1-t)|B'|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$tag{2}
(tA'+(1-t)B')^{-1}leq t(A')^{-1}+(1-t)(B')^{-1}.
$$
Thus, using $(2)$ and $(0)$,
begin{align}tag{3}
|(tA'+(1-t)B')^{-1}|&leq|t(A')^{-1}+(1-t)(B')^{-1}|\
&leq t|(A')^{-1}|+(1-t)|(B')^{-1}|\ &leq|(B')^{-1}|
end{align}
Now, combining $(1)$, and $(3)$,
begin{align}tag{4}
|tA'+(1-t)B'|,|(tA'+(1-t)B')^{-1}|&leq |(tA'+(1-t)B')^{-1}|leq|(B')^{-1}|.
end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=frac1{|A|+|B|},left(A+Bright), (B')^{-1}=|B|,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
|A+B|,|(A+B)^{-1}|leq |B|,|B^{-1}|,
$$
as desired.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
answered Feb 26 '17 at 3:21
Martin Argerami
124k1176174
124k1176174
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