How to find the order of a matrix?

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To determine the order of a matrix, one must raise it to powers until it becomes the identity matrix. In the case of the Heisenberg group over a field, the specific 3x3 upper triangular matrix with 1's on the diagonal does not yield the identity matrix upon repeated multiplication. Therefore, it suggests that this matrix has infinite order, as it never equals the identity for any positive integer power. The discussion highlights the confusion around finding the order of such matrices and the implications of infinite order. Clarification on the properties of the matrix and its behavior under exponentiation is needed.
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For example the Heisenberg group over Field.

H(F) is a 3x3 upper right triangle where the entries on the main diagonals are all 1's.

So by definition that I need to use this matrix and raise to the power where it becomes the identity matrix, then the number of that power would the order of H(F). But if I take the upper right triangle with all 1's in the main diagonal, it would never becomes identity matrix though no matter how many times I multiple them.

I couldn't find any examples from the material that I have. Am I doing something wrong here? Any suggestions?
 
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If you have g^m\ne1 for every positive integer m, you say that g has infinite order.
 
Thread 'How to define a vector field?'

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Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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