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Projection onto Linearly Dependent Vectors

Now consider another example:

\begin{eqnarray*}
\underline{s}_0 &\isdef & [1,1], \\
\underline{s}_1 &\isdef & [-1,-1].
\end{eqnarray*}

The projections of $ x=[x_0,x_1]$ onto these vectors are

\begin{eqnarray*}
{\bf P}_{\underline{s}_0}(x) &=& \frac{x_0 + x_1}{2}\underline{s}_0, \\
{\bf P}_{\underline{s}_1}(x) &=& -\frac{x_0 + x_1}{2}\underline{s}_1.
\end{eqnarray*}

The sum of the projections is

\begin{eqnarray*}
{\bf P}_{\underline{s}_0}(x) + {\bf P}_{\underline{s}_1}(x) &=&
\frac{x_0 + x_1}{2}\underline{s}_0 - \frac{x_0 + x_1}{2}\underline{s}_1 \\
&\isdef & \frac{x_0 + x_1}{2}(1,1) - \frac{x_0 + x_1}{2} (-1,-1) \\
&=& \left(x_0+x_1,x_0+x_1\right) \neq x.
\end{eqnarray*}

Something went wrong, but what? It turns out that a set of $ N$ vectors can be used to reconstruct an arbitrary vector in $ \mathbb{C}^N$ from its projections only if they are linearly independent. In general, a set of vectors is linearly independent if none of them can be expressed as a linear combination of the others in the set. What this means intuitively is that they must ``point in different directions'' in $ N$ -space. In this example $ s_1 = - s_0$ so that they lie along the same line in $ 2$ -space. As a result, they are linearly dependent: one is a linear combination of the other ( $ s_1 = (-1)s_0$ ).


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8
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Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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