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I want to help one of my friends who studies engineering. He has a homework at the signal processing course. I think I realize what I have to do, but since I don't have their course, I do not fully understand what I have to do. It goes like this:

$$ H=\begin{pmatrix}1&-1&0&0&0&0&0&0 \\ 0&0&1&-1&0&0&0&0 \\ 0&0&0&0&1&-1&0&0\\ 0&0&0&0&0&0&1&-1 \\ 1&1&-1&-1&0&0&0&0\\ 0&0&0&0&1&1&-1&-1 \\ 1&1&1&1&-1&-1&-1&-1\\ 1&1&1&1&1&1&1&1 \end{pmatrix}$$

The exercise says to check that $H$ consists of independent line vectors (that's easy). Then it says to find the coordinates of $x=[1,1,...1]$ and $y=[1,-1,...,1,-1]$ in the Haar Basis defined by $H$.

Does this means to find the coordinates in the usual way, i.e. solve the equation $zH=x$?

(I do not need you to solve the system (I'll use Matlab for that), just say if I'm right or not. Thank you.)

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    Maybe this is more suited to http://dsp.stackexchange.com ?2012-01-11

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The Haar basis is the basis of the row vectors, not of the column vectors, so I presume you'd want to solve the equation $zH=x$ for row vectors instead. That also makes sense in that $x$ and $y$ are being specified as row vectors.

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    Thank you. I wrote it the other way around, but the idea is to solve those systems.2011-10-18
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    Isn't a normalization missing or is that omitted for practical/computational reasons? In any case, I think it would be worth pointing out that no matlab is needed, as up to multiplication with a diagonal matrix the inverse is $H^T$.2011-10-18
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    @t.b: Yeah, that is one of the first points of the exercise.2011-10-18
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    @Beni: I'm not sure what you mean by your first comment. Which are "those systems"? The one you wrote and the one I wrote? If so, why do you want to solve the one you wrote? I don't see how it has anything to do with finding coordinates in the Haar basis.2011-10-18
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    @joriki: Sorry about the confusion. In the problem there are two vectors for which I must find the coordinates. Those are the systems I want to solve, namely $zH=x$ and $zH=y$. I edited the question, to correct my mistake.2011-10-18
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    The multiplication $H\cdot H^T$ yields a diagonal matrix, but not the identity matrix.2011-10-18
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    @Beni: Yes, t.b. mentioned that. This is because the rows of $H$ are orthogonal but not orthonormal.2011-10-18
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    Ok, I didn't read carefully enough t.b.'s comment.2011-10-18