Inequality between Frobenius norm and L2 Norm
$begingroup$
$u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
Prove
$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$
$left| right| _ { F }$ is Frobenius norm defined for matrix.
$$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
$$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$
But when I tried to expand it, it seems hard to rearrange.
Any ideas?
probability matrices statistics inequality
edited Jan 11 at 19:43
Bernard
119k740113
asked Jan 11 at 19:25
DylonDylon
113
$endgroup$
add a comment |
$begingroup$
$u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
Prove
$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$
$left| right| _ { F }$ is Frobenius norm defined for matrix.
$$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
$$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$
But when I tried to expand it, it seems hard to rearrange.
Any ideas?
probability matrices statistics inequality
edited Jan 11 at 19:43
Bernard
119k740113
asked Jan 11 at 19:25
DylonDylon
113
$endgroup$
add a comment |
$begingroup$
$u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
Prove
$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$
$left| right| _ { F }$ is Frobenius norm defined for matrix.
$$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
$$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$
But when I tried to expand it, it seems hard to rearrange.
Any ideas?
probability matrices statistics inequality
edited Jan 11 at 19:43
Bernard
119k740113
asked Jan 11 at 19:25
DylonDylon
113
$endgroup$
$u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
Prove
$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$
$left| right| _ { F }$ is Frobenius norm defined for matrix.
$$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
$$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$
But when I tried to expand it, it seems hard to rearrange.
Any ideas?
probability matrices statistics inequality
probability matrices statistics inequality
edited Jan 11 at 19:43
Bernard
119k740113
asked Jan 11 at 19:25
DylonDylon
113
edited Jan 11 at 19:43
Bernard
119k740113
asked Jan 11 at 19:25
DylonDylon
113
edited Jan 11 at 19:43
Bernard
119k740113
edited Jan 11 at 19:43
Bernard
119k740113
edited Jan 11 at 19:43
Bernard
119k740113
119k740113
asked Jan 11 at 19:25
DylonDylon
113
asked Jan 11 at 19:25
DylonDylon
113
asked Jan 11 at 19:25
DylonDylon
113
113
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
We can expand each of the terms as follows. The Frobenius norm:
$$
|uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
= operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
= 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
= 2[1 - (u^Tw)(v^Tz)]
$$
The vector norm:
$$
|u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
$$
To compare these two, we make the following observation:
$$
[1 - u^Tw][1 - v^Tz] geq 0 implies\
1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
(u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
$$
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can expand each of the terms as follows. The Frobenius norm:
$$
|uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
= operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
= 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
= 2[1 - (u^Tw)(v^Tz)]
$$
The vector norm:
$$
|u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
$$
To compare these two, we make the following observation:
$$
[1 - u^Tw][1 - v^Tz] geq 0 implies\
1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
(u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
$$
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
add a comment |
$begingroup$
We can expand each of the terms as follows. The Frobenius norm:
$$
|uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
= operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
= 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
= 2[1 - (u^Tw)(v^Tz)]
$$
The vector norm:
$$
|u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
$$
To compare these two, we make the following observation:
$$
[1 - u^Tw][1 - v^Tz] geq 0 implies\
1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
(u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
$$
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
add a comment |
$begingroup$
We can expand each of the terms as follows. The Frobenius norm:
$$
|uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
= operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
= 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
= 2[1 - (u^Tw)(v^Tz)]
$$
The vector norm:
$$
|u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
$$
To compare these two, we make the following observation:
$$
[1 - u^Tw][1 - v^Tz] geq 0 implies\
1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
(u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
$$
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
We can expand each of the terms as follows. The Frobenius norm:
$$
|uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
= operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
= 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
= 2[1 - (u^Tw)(v^Tz)]
$$
The vector norm:
$$
|u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
$$
To compare these two, we make the following observation:
$$
[1 - u^Tw][1 - v^Tz] geq 0 implies\
1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
(u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
$$
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
answered Jan 11 at 19:46
OmnomnomnomOmnomnomnom
127k790178
127k790178
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
add a comment |
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
$begingroup$
Very clear thanks.
$endgroup$
– Dylon
Jan 12 at 6:23
add a comment |
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