Odd basic problem

Shadow · **Posted:** Wed, 25 Apr 2012 13:54:42 UTC

A friend of mine posed the question if I have a basis $v_1,\ldots, v_k$ of $V=\mathbb{R}^k$ and a dual basis $\phi_1,\ldots,\phi_k$ which has the usual pairing rule $\phi_i(v_j)=\delta_{ij}$ and ||v_i||=1

for every i, can we conclude that $||\phi_j||=||\phi_k||$ for every j,k

? Note the assumption is only that the original basis is normal, nothing about orthogonality is assumed, otherwise this is trivial. I personally feel like it should just come from the transpose isomorphism, but he feels there might be something more subtle. If there's a reference or a really solid proof that actually has the maximization via the analysis definition taking the sup over unit vectors, that would probably be most convincing. Thanks, and any help appreciated.

outermeasure · **Posted:** Wed, 25 Apr 2012 14:23:00 UTC

Shadow wrote:

A friend of mine posed the question if I have a basis $v_1,\ldots, v_k$ of $V=\mathbb{R}^k$ and a dual basis $\phi_1,\ldots,\phi_k$ which has the usual pairing rule $\phi_i(v_j)=\delta_{ij}$ and ||v_i||=1

for every i, can we conclude that $||\phi_j||=||\phi_k||$ for every j,k

? Note the assumption is only that the original basis is normal, nothing about orthogonality is assumed, otherwise this is trivial. I personally feel like it should just come from the transpose isomorphism, but he feels there might be something more subtle. If there's a reference or a really solid proof that actually has the maximization via the analysis definition taking the sup over unit vectors, that would probably be most convincing. Thanks, and any help appreciated.

No.

Take $\mathbb{R}^3$ with v_1=e_1

, $v_2=\cos(\theta) e_1+\sin(\theta) e_2$ , v_3=e_3

, where $\theta$ is tiny. Then obviously $\phi^3=e^3$ (lifting the indices to show it is the dual), but $\phi^1=e^1-\cot(\theta) e^2$ has norm $\lvert\csc(\theta)\rvert\gg1$ .

Shadow · **Posted:** Wed, 25 Apr 2012 14:33:30 UTC

outermeasure wrote:

Shadow wrote:

A friend of mine posed the question if I have a basis $v_1,\ldots, v_k$ of $V=\mathbb{R}^k$ and a dual basis $\phi_1,\ldots,\phi_k$ which has the usual pairing rule $\phi_i(v_j)=\delta_{ij}$ and ||v_i||=1

for every i, can we conclude that $||\phi_j||=||\phi_k||$ for every j,k

? Note the assumption is only that the original basis is normal, nothing about orthogonality is assumed, otherwise this is trivial. I personally feel like it should just come from the transpose isomorphism, but he feels there might be something more subtle. If there's a reference or a really solid proof that actually has the maximization via the analysis definition taking the sup over unit vectors, that would probably be most convincing. Thanks, and any help appreciated.

No.

Take $\mathbb{R}^3$ with v_1=e_1

, $v_2=\cos(\theta) e_1+\sin(\theta) e_2$ , v_3=e_3

, where $\theta$ is tiny. Then obviously $\phi^3=e^3$ (lifting the indices to show it is the dual), but $\phi^1=e^1-\cot(\theta) e^2$ has norm $\lvert\csc(\theta)\rvert\gg1$ .

Excellent. Thanks a lot. He found that it was true for $\mathbb{R}^2$ and it seems I was wrong to doubt that it must be true for all finite dimensional spaces.

outermeasure · **Posted:** Wed, 25 Apr 2012 14:51:07 UTC

Shadow wrote:

outermeasure wrote:

Shadow wrote:

A friend of mine posed the question if I have a basis $v_1,\ldots, v_k$ of $V=\mathbb{R}^k$ and a dual basis $\phi_1,\ldots,\phi_k$ which has the usual pairing rule $\phi_i(v_j)=\delta_{ij}$ and ||v_i||=1

for every i, can we conclude that $||\phi_j||=||\phi_k||$ for every j,k

? Note the assumption is only that the original basis is normal, nothing about orthogonality is assumed, otherwise this is trivial. I personally feel like it should just come from the transpose isomorphism, but he feels there might be something more subtle. If there's a reference or a really solid proof that actually has the maximization via the analysis definition taking the sup over unit vectors, that would probably be most convincing. Thanks, and any help appreciated.

No.

Take $\mathbb{R}^3$ with v_1=e_1

, $v_2=\cos(\theta) e_1+\sin(\theta) e_2$ , v_3=e_3

, where $\theta$ is tiny. Then obviously $\phi^3=e^3$ (lifting the indices to show it is the dual), but $\phi^1=e^1-\cot(\theta) e^2$ has norm $\lvert\csc(\theta)\rvert\gg1$ .

Excellent. Thanks a lot. He found that it was true for $\mathbb{R}^2$ and it seems I was wrong to doubt that it must be true for all finite dimensional spaces.

Well, for $\mathbb{R}^2$ it is true because you have an (orientation-reversing) isometry swapping v_1

and

, but it is not true in general that any permutation of v_i

is an isometry when in higher dimensions.

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Odd basic problem

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