## Saturday, November 18, 2006

### The Matrix?! Topic: Calculus/Linear Algebra.

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.