- #1

- 49

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Suppose I have this matrix A:

Why is the matrix exponential like so:

when A is not simple (eigenvalues not distinct)?

Thanks.

- Thread starter AirForceOne
- Start date

- #1

- 49

- 0

Suppose I have this matrix A:

Why is the matrix exponential like so:

when A is not simple (eigenvalues not distinct)?

Thanks.

- #2

- 120

- 0

f(A) = U

where f(D) is the diagonal matrix with elements f(d

The proof is not hard you just expand f in a series (due to analyticity) put in U

In your case, the matrix is already diagonal. For any analytic function the formal definition is anyway given in terms of the series expansion and in the case of exponential

exp(A) = 1 + A + A^2/2! + ....

where you can show that this series is convergent etc etc.

- #3

- 49

- 0

Okay, I understand now. So A must be diagonalizable in order to compute its exponential. For A to be diagonalizable, its eigenvectors must be linearly independent. I calculated A's eigenvalues to be σ (repeated). When I calculated its eigenvectors, I get [anything, anything]. Does this translate to being linearly independent?^{-1}DU with d_{i}eigenvalues)

f(A) = U^{-1}f(D) U

where f(D) is the diagonal matrix with elements f(d_{i}).

The proof is not hard you just expand f in a series (due to analyticity) put in U^{-1}D U and do some algebra.

In your case, the matrix is already diagonal. For any analytic function the formal definition is anyway given in terms of the series expansion and in the case of exponential

exp(A) = 1 + A + A^2/2! + ....

where you can show that this series is convergent etc etc.

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