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Let's suppose that $h_1$, $h_2$, $b_1$ and $b_2$ are vectors of length $L\times 1$.

Where $h_1$ and $h_2$ are real and unknowns and $b_1$ and $b_2$ are known complexes

Is it possible to solve this expression?

$$h_1^T\times b_1=h_2^T\times b_2$$

I tried to see it as... $$(1\times L)\times (L \times 1)= (1\times L)\times (L \times 1)$$ then becomes $$(1\times L)= (1\times L)\times (L \times 1) \times (1\times L)$$ which indicates that $b_1$ and $b_2$ needs to be a matrix dimension. However linear algebra rule doesn't permit such move.

I want to see if I can find the relationship between $h_1$ and $h_2$. The most logical way is to move one $b$ to another side, but since they are vectors I have no idea how to.

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    Do you mean $1 \times L$, or do you mean $L \times 1$?2012-05-22
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    $1\times L$ vector. So yes indeed both RHS and LHS will end up with an single-element solution.2012-05-22
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    No, if $h$ and $b$ are $1 \times L$, $h^T$ is $L \times 1$, and $h^T b$ is $L \times L$.2012-05-22
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    Oh my bad! Sorry for the confusion. Yes you are right! It is $L\times 1$ indeed!2012-05-22

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