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Assume we have the following linear transformation:

$$C(x) = \tilde{x} = \left( \begin{array}{c} \langle x, a_1 \rangle\\ \vdots \\ \langle x, a_k \rangle\\ \vdots \\ \langle x, a_n \rangle \end{array} \right)$$

What is the inverse function, $D$, of $C$? i.e. what $D$ achieves the following:

$$ D( C(x) ) =D \circ C (x) = x$$

Note that $a_i$ is an orthonormal basis and each $a_i \in \mathbb{C}^D$ and $\langle a, b \rangle = a^*b = a^H b$.


If $D$ can be expressed in terms of matrices, that would be great! Specifically, if it can be expressed in terms of $C$, the unitary matrix of the orthonormal vectors.

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Orthonormality makes this easy. $$ \sum_{k=1}^n \langle x,a_k\rangle a_k = x, \tag 1 $$ so the inverse function is $$ (s_1,\ldots,s_n)\mapsto \sum_{k=1}^n s_k a_k $$ for scalars $s_1,\ldots,s_n$.

To prove $(1)$, let $\displaystyle y=\sum_{k=1}^n\langle x,a_k\rangle a_k$ and first note that $$ \langle y,a_\ell\rangle = \left\langle \sum_{k=1}^n \langle x , a_k\rangle a_k, a_\ell \right\rangle = \sum_{k=1}^n \langle \langle x,a_k\rangle a_k,a_\ell \rangle = \sum_{k=1}^n \langle x,a_k\rangle \langle a_k,a_\ell \rangle. $$ Note that every term in the last sum is $0$ except the one in which $k=\ell$, and in that case we have $\langle x,a_\ell\rangle\langle a_\ell,a_\ell\rangle = \langle x,a_\ell\rangle$, by orthonormality.

Thus $\langle y,a_k\rangle=\langle x,a_k\rangle$ for $x=1,\ldots, n$. Applying this in the case $y=a_k$ you see that $(1)$ holds in that case. When $y$ is a linear combination of $a_1,\ldots,a_n$, you can then see that $(1)$ still holds, by linearity.

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  • $\begingroup$ So, D(x) = ? ... sorry I am a little lost and can't really see what D(x) is suppose to be. (1) is obviously true always if x can be expressed as a linear combination of $a_k$, can't see its relation to C(x). Sorry. I do understand your reasoning for $\langle y, a_l \rangle$ but I am having a hard time relating it to D(x). Your y, is just x... no? i.e. x = y. $\endgroup$ – Pinocchio Jan 19 '15 at 1:05
  • $\begingroup$ I think I understand, you are suggesting D to be $D(\tilde{x}) = C^{-1}(\tilde{x}) = \sum_{i=1}^{n} a_k \tilde{x}_{k} = \sum_{i=1}^{n} a_k \langle x, a_k \rangle = x$, which is clearly mapping $\tilde{x}$ to the $x$ that generated it? $\endgroup$ – Pinocchio Jan 19 '15 at 2:18
  • $\begingroup$ If $C(x)=s$, the $D(s)=x$, and $D(s) = D(s_1,\ldots,s_n) = \sum_{k=1}^n s_k a_k$. That's what $D(s)$ (not $D(x)$) is. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 19 '15 at 4:23

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