1
$\begingroup$

$u , w \in S ^ { n - 1 }$ 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?

$\endgroup$
0
$\begingroup$

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: $$ [1 - u^Tw][1 - v^Tz] \geq 0 \implies\\ 1 - u^Tw - v^Tz + (u^Tw)(v^Tz) \geq 0 \implies\\ (u^Tw)(v^Tz) \geq (u^Tw) + (v^Tz) - 1 $$

$\endgroup$
  • $\begingroup$ Very clear thanks. $\endgroup$ – Dylon Jan 12 at 6:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.