# Solving $h_1^T\times b_1=h_2^T\times b_2$ with vector dimension characterstics.

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 $L \times 1$? – Robert Israel May 22 '12 at 2:42
$1\times L$ vector. So yes indeed both RHS and LHS will end up with an single-element solution. – JuniorEngie May 22 '12 at 2:47
No, if $h$ and $b$ are $1 \times L$, $h^T$ is $L \times 1$, and $h^T b$ is $L \times L$. – Robert Israel May 22 '12 at 2:51
Oh my bad! Sorry for the confusion. Yes you are right! It is $L\times 1$ indeed! – JuniorEngie May 22 '12 at 2:53

If your vectors are column vectors, you have just one equation, since both sides are $1 \times 1$. So all you can do is solve for one entry of one of your vectors (assuming the entry of the other vector it's paired with is not $0$).
On the other hand, if your vectors are row vectors, both sides are $L \times L$ matrices. But those matrices have rank at most $1$; the rows are all multiples of $b_1$ or $b_2$ and the columns are all multiples of $h_1$ or $h_2$. If none of these vectors is $0$, then for some nonzero scalar $t$ we must have $h_1 = t h_2$ and $b_1 = t^{-1} b_2$.
The variables $h_1$ and $h_2$ are unknown, only $b$s are known. Thus the reason I want to see if an expression such as $h_1^T=h_2^T\times b_2 \times b_1^{-1 T}$ exists? – JuniorEngie May 22 '12 at 2:49