$u , w \in S ^ { n - 1 }$ 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?
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: $$ [1 - u^Tw][1 - v^Tz] \geq 0 \implies\\ 1 - u^Tw - v^Tz + (u^Tw)(v^Tz) \geq 0 \implies\\ (u^Tw)(v^Tz) \geq (u^Tw) + (v^Tz) - 1 $$