**Definition**: A square matrix $ A $ is diagonalizable if there exist matrices $ V, D $ such that $ V $ is invertible, $ D $ is diagonal (has only entries on the main diagonal), and $ A = VDV^{-1} $.

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**Problem**: Solve the differential equation $ x^{\prime}(t) = Ax(t) $ where $ A $ is a diagonalizable $ n \times n $ square matrix and $ x(t) $ is a vector valued function (which must have $ n $ components).

**Solution**: Well this is a lot like the regular differential equation $ y^{\prime} = ky $ which has solution $ y = y_0 e^{kt} $. So let's guess that

$ x(t) = e^{At}x(0) $

is the solution. But what exactly does $ e^{At} $ mean? Consider the following.

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**Definition**: If $ f $ is a function and $ D $ is a diagonal square matrix, then we say that $ f(D) $ is simply the matrix with $ f $ applied to all the elements on the diagonal.

If $ A $ is diagonalizable, then we have $ A = VDV^{-1} $ and we define $ f(A) = Vf(D)V^{-1} $.

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So does $ x(t) = e^{At}x(0) $ work as a solution to the differential equation? It turns out that, by using the definition of the derivative we can show that

$ \frac{d}{dt}[e^{Dt}] = De^{Dt} $

and

$ \frac{d}{dt}[e^{At}x(0)] = VDe^{Dt}V^{-1}x(0) $.

However, if we use the function $ g(x) = xe^{xt} $, we get

$ Ae^{At}x(0) = g(A)x(0) = Vg(D)V^{-1}x(0) = VDe^{Dt}V^{-1}x(0) $,

so $ x^{\prime}(t) = \frac{d}{dt}[e^{At}x(0)] = Ae^{At}x(0) = Ax(t) $,

as desired. QED.

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Comment: Applying functions to matrices is a super cool thing and being able to solve differential equations with matrix coefficients is even better. Linear algebra provides all sorts of new ways to work with matrices.

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Practice Problem: Show that all symmetric square matrices are diagonalizable.

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