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I was reading this post here and I got really confused at the part where the claim is that the laplace transform diagonalize the derivative operator (http://www.quora.com/What-is-the-purpose-of-Laplace-transforms-in-controls-theory)

In my mind the diagonalization process is where by a matrix $A = P^{-1}\Sigma P$, where $\Sigma$ is our diagonal matrix. How does this have to do with laplace transform is very unclear.

So two questions:

  1. What does it mean for an operator to be diagonalized? (can someone explain this as simply as possible)

  2. How does diagonalization change the basis? In the case of laplace transform the claim is that the laplace transform changes the basis of impulse functions to the eigenvector...can someone verify?

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Due to the identity

$\frac{d}{dt}e^{-st}=-se^{-st}$

analogous to $Ax=\lambda x$ for the Laplace transform you have the eigenvector $e^{-st}$ with ist Eigenvalue $-s$. In functional Analysis, functions are regarded as generalized vectors and Operators like Differentiation as generalized matrices.

For example, you have the differential equation: $\frac{d}{dt}x + x = (\frac{d}{dt}+1)x = 0$. For some functions $x(t)$ you will see that the Operator $\Delta:=\frac{d}{dt}+1$ Change the function (similar as a Matrix makes a vector completely different). But when you do the Laplace transform to this differential equation you will get $sL(x)+L(x)=(s+1)L(x)$ and you observe that the Laplace transform diagonalized the Operator $\Delta$, because the new "Basis" $L(x)$ is only multiplied by $s+1$ (similar to Eigenvalues in linear algebra). The Operator $\Delta$ was diagonalized. Hence, the Basis transform was performed via Laplace transform $x \rightarrow L(x)$.

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  • $\begingroup$ Thank you. Just one quick comment, you said taking the laplace transform gives us the new basis $L(x)$. Can you clarify as to what is the old basis prior taking the laplace transform? $\endgroup$
    – Fraïssé
    Feb 7, 2015 at 19:27
  • $\begingroup$ The old Basis is an arbitrary function $x(t)$. The Laplace transform can be expressed similar to a Matrix multiplication $L(x)= \int_{0}^{\infty}G(s,t)xdt$ with the Matrix elements $G(s,t)$. $\endgroup$
    – kryomaxim
    Feb 7, 2015 at 19:30

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