CODEWITHSUNDEEP
matrixdifferential-equationsexponential
user166271
user166271

Reputation: 1

Matrix Exponential

So, I'm basically wondering if for a matrix A of the type

(0 a)

(b 0)

there is an easy way to calculate e^At, since if it were

(0 b)

(-b 0)

One would just have the rotation matrix where the angular frequency of the terms would be b, with the sines and cosines. Because going through the hassle of getting the Jordan Block and Change of Basis seems to be too much work for such a simple one. Then again,I might be wrong

As an exemple, say A =

(0 9 (-1 0)

Where the eigenvalues are +-3i. How would I proceed?

Upvotes: 0

Views: 200

Answers (1)

dmuir
dmuir

Reputation: 4431

Let your matrix be M. Note that 

M*M  = a*b*I.

From this deduce that

pow(M, 2*n) = pow(a*b,n)*I
pow(M, 2*n+1) = pow( a*b,n)*M

Plug these into the power series for exp and rearrange to get

exp(t*M) = cosh( t*sqrt(a*b))*I + (1.0/sqrt(a*b))*sinh( t*sqrt(a*b))

As a check, diffentiate this expression wrt t and check that you get 

d/dt exp(t*M) = M*exp(t*M)

Note that if a*b<0 the expressions above are still valid, but can be rewritten, using 

cosh(i*x) = cos(x) and sinh(i*x) = i*sin(x)

in terms of cos and sin

Upvotes: 1

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