Matrix Exponential

Calculus Level 4

Given the matrix,

A = [ 1 2 − 1 4 ] A=\begin{bmatrix}1&2\\-1&4\end{bmatrix}

What is the matrix exponential, e A t e^{At} ?

[ e 2 t − e 3 t − 2 e 2 t + 2 e 3 t e 2 t − e 3 t − e 2 t − 2 e 3 t ] \begin{bmatrix}e^{2t}-e^{3t}&-2e^{2t} +2e^{3t}\\e^{2t}-e^{3t}&-e^{2t}-2e^{3t}\end{bmatrix} [ 2 e 2 t + e 3 t − 2 e 2 t + 2 e 3 t e 2 t − e 3 t − e 2 t − 2 e 3 t ] \begin{bmatrix}2e^{2t}+e^{3t}&-2e^{2t} +2e^{3t}\\e^{2t}-e^{3t}&-e^{2t}-2e^{3t}\end{bmatrix} [ 2 e t − e 3 t − 2 e 2 t + 2 e 3 t e 2 t − e 3 t − e 2 t + 2 e 5 t ] \begin{bmatrix}2e^{t}-e^{3t}&-2e^{2t} +2e^{3t}\\e^{2t}-e^{3t}&-e^{2t}+2e^{5t}\end{bmatrix} [ 2 e 2 t − e 3 t − 2 e 2 t + 2 e 3 t e 2 t − e 3 t − e 2 t + 2 e 3 t ] \begin{bmatrix}2e^{2t}-e^{3t}&-2e^{2t} +2e^{3t}\\e^{2t}-e^{3t}&-e^{2t}+2e^{3t}\end{bmatrix}

Samuel Hansen by Samuel Hansen , 29, USA

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1 solution

Samuel Hansen
Apr 10, 2017

There are several ways to solve this problem (such as with Laplace transforms) but perhaps the quickest way by hand is to diagonalize the matrix and then use the transformation identity,

e A t = D e Λ t D − 1 e^{At}= D e^{\Lambda t} D^{-1}

Where Λ \Lambda is the matrix of eigenvalues. First we find the characteristic equation,

ϕ = λ 2 − 5 λ + 6 \phi = \lambda^{2} -5 \lambda +6

And then the associated eigenvalues and eigenvectors. For λ \lambda =2 I choose the eigenvector v 1 = [ 2 1 ] v_{1} = \begin{bmatrix}2 \\ 1 \end{bmatrix} and for λ \lambda =3 I choose the eigenvector v 1 = [ 1 1 ] v_{1}=\begin{bmatrix}1\\1 \end{bmatrix} .

This gives,

D = [ 2 1 1 1 ] D= \begin{bmatrix}2 &1 \\ 1 & 1 \end{bmatrix}

And,

e Λ t = [ e 2 t 0 0 e 3 t ] e^{\Lambda t}= \begin{bmatrix}e^{2t} & 0 \\ 0 & e^{3t} \end{bmatrix} .

Applying the formula yields the desired result. Note that this method only works in the case that the matrix is diagonalizable. If we had a non diagonalizable matrix we would have to use generalized eigenvectors.

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