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Consider the inverse Fourier Transform and the Fourier Transform:

$$f(x) = \int_{-\infty}^\infty F(k)e^{2\pi i k x}dk \\ F(k) = \int_{-\infty}^\infty f(x)e^{-2\pi i k x}dx$$

The Fourier transform is linear, since if $f(x)$ and $g(x)$ have Fourier transforms $F(k)$ and $G(k)$, then

$$\int_{-\infty}^\infty[af(x)+bg(x)]e^{-2 \pi ikx}dx = a \int_{-\infty}^\infty f(x) e^{-2 \pi ikx}dx+b \int_{-\infty}^\infty g(x) e^{-2 \pi ikx}dx = aF(k)+bG(k)$$

Is the inverse Fourier transform a “linear transform”?

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    $\begingroup$ Ofcourse, by the same logic as you used for the forward transform. $\endgroup$ – Rajesh Dachiraju Jan 20 '15 at 7:53
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Claim: Let $T$ be an invertible linear transformation. Then $T^{-1}$ is a linear transformation

Proof: $$av+w=T^{-1}T(av+w)=T^{-1}(aT(v)+bT(w))$$ Now write $v'=T(v)$ and $w'=T(w)$. We get $$aT^{-1}(v')+bT^{-1}(w')=T^{-1}(av'+bw')$$

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