In Tao's Fourier Transform preprint, https://www.math.ucla.edu/~tao/preprints/fourier.pdf, they show explicitly that the Laplacian can be viewed as a Fourier multiplier. In the same paragraph, they write
...This identity shows that the Fourier transform diagonalizes the Laplacian
Why is the word "diagonalizes" used in this context? I know it from a Linear Algebra context, in which we diagonalize some matrix, but in this context it seems out of place. Am I missing something?
Consider a Hermitian $n\times n$ matrix $A$. The spectral theorem guarantees the existence of a unitary matrix $U$ such that $A = U\Lambda U^\dagger$ or, equivalently, $\Lambda = U^\dagger AU$, where $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$. $A$ is related to a diagonal matrix $\Lambda$ via conjugation by a transformation $U$, so we say that $U$ diagonalizes $A$.
Now consider the Laplacian on some suitable space of functions $\mathcal{U}$ where both the Laplacian and Fourier transform are well-defined as maps from $\mathcal{U}$ to itself. The equation written in the OP says that for any function $u\in\mathcal{U}$, $$\mathcal{F}\Delta u = -4\pi|\xi|^2\mathcal{F}u.$$ Now define $v= \mathcal{F}^{-1}u$ and substitute to obtain $$\left(\mathcal{F}\Delta\mathcal{F}^{-1}\right)v = -4\pi|\xi|^2v.$$ If we now define an operator $\Lambda v := -4\pi|\xi|^2v$, we have $$\left(\mathcal{F}\Delta\mathcal{F}^{-1}\right)v = \Lambda v \quad \forall v\in\mathcal{U}.$$ $\Lambda$ is a diagonal operator because pointwise values if $w = \Lambda v$, when the value of $w$ at a specific point $\xi_0$ depends only on the value of $v$ at $\xi_0$. This is the same as in the finite-dimensional case where a diagonal matrix simple scales each component, except now we have a continuum of components. Now, we can say that we have represented $\Delta$ as a diagonal transformation $\Lambda$ via a similarity transformation $\mathcal{F}$, so we can say that $\mathcal{F}$ diagonalizes $\Delta$.