Signal/Vector Reconstruction from Projections

We now arrive finally at the main desired result for this section:

Theorem: The projections of any vector $x\in{\bf C}^N$ onto any orthogonal basis set for ${\bf C}^N$ can be summed to reconstruct $x$ exactly.

Proof: Let $\{\sv_0,\ldots,\sv_{N-1}\}$ denote any orthogonal basis set for ${\bf C}^N$. Then since $x$ is in the space spanned by these vectors, we have

$$x= \alpha_0\sv_0 + \alpha_1\sv_1 + \cdots + \alpha_{N-1}\sv_{N-1} \protect$$ (5.3)

for some (unique) scalars $\alpha_0,\ldots,\alpha_{N-1}$. The projection of $x$ onto $\sv_k$ is equal to

$${\bf P}_{\sv_k}(x) = \alpha_0{\bf P}_{\sv_k}(\sv_0) + \alpha_1{\bf P}_{\sv_k}(\sv_1) + \cdots + \alpha_{N-1}{\bf P}_{\sv_k}(\sv_{N-1})$$

(using the linearity of the projection operator which follows from linearity of the inner product in its first argument). Since the basis vectors are orthogonal, the projection of $\sv_l$ onto $\sv_k$ is zero for $l\neq k$:

$${\bf P}_{\sv_k}(\sv_l) \isdef \frac{\left<\sv_l,\sv_k\right>}{\l... ...ay}{ll} \underline{0}, & l\neq k \\ [5pt] \sv_k, & l=k. \\ \end{array}\right.$$

We therefore obtain

$${\bf P}_{\sv_k}(x) = 0 + \cdots + 0 + \alpha_k{\bf P}_{\sv_k}(\sv_k) + 0 + \cdots + 0 = \alpha_k\sv_k.$$

Therefore, the sum of projections onto the vectors $\sv_k$, $k=0,1,\ldots, N-1$, is just the linear combination of the $\sv_k$ which forms $x$:

$$\sum_{k=0}^{N-1} {\bf P}_{\sv_k}(x) = \sum_{k=0}^{N-1} \alpha_k \sv_k = x$$

by Eq.$\,$(5.3). $\Box$