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Exponential of an upper triangular matrix is upper triangular? Or not? watch

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    Why is it the case?


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    where are the purists when you want them ?
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    surely you can use the definition of matrix multiplication to show that the product of two upper triangular matrices is upper triangular. Thus by induction any power, thus by definition the exponential.
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    Come to think of it, can you find an expression for the numbers along the leading diagonal of such a exponential?
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    (Original post by TeeEm)
    where are the purists when you want them ?
    Getting our beauty sleep!
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    (Original post by EricPiphany)
    Come to think of it, can you find an expression for the numbers along the leading diagonal of such a exponential?
    Yes, if you square an (upper/lower) triangular matrix, the elements on the leading diagonal are the squares of the elements on the leading diagonal of the original matrix. And so on...So the elements on the leading diagonal of the exponential of a triangular matrix are the exponentials of the elements on the leading diagonal of the original matrix.

    It may be worth pointing out an examiners' favourite here: If A is an upper triangular matrix such that A = I + B, with B strictly upper triangular, then I and B commute and \exp(A) = \exp(I+B) = \exp(I) \exp(B). You then notice that \exp(B) is especially easy to compute...
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    (Original post by Gregorius)
    Yes, if you square an (upper/lower) triangular matrix, the elements on the leading diagonal are the squares of the elements on the leading diagonal of the original matrix. And so on...So the elements on the leading diagonal of the exponential of a triangular matrix are the exponentials of the elements on the leading diagonal of the original matrix.

    It may be worth pointing out an examiners' favourite here: If A is an upper triangular matrix such that A = I + B, with B strictly upper triangular, then I and B commute and \exp(A) = \exp(I+B) = \exp(I) \exp(B). You then notice that \exp(B) is especially easy to compute...
    Nicey. Is exp(B) especially easy to compute because for n by n, only n - 2 matrix multiplications are needed to compute exp(B), the sum is finite?
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    (Original post by EricPiphany)
    Nicey. Is exp(B) especially easy to compute because for n by n, only n - 2 matrix multiplications are needed to compute exp(B), the sum is finite?
    It's easy because the matrix B is nilpotent - that is, there is an n such that B^n = 0, so the series for \exp(B) is finite.

    When you multiply a strictly upper triangular matrix by itself, the non-zero entries shift one up and to the right, further from the leading diagonal.
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    (Original post by Gregorius)
    It's easy because the matrix B is nilpotent - that is, there is an n such that B^n = 0, so the series for \exp(B) is finite.

    When you multiply a strictly upper triangular matrix by itself, the non-zero entries shift one up and to the right, further from the leading diagonal.
    Yep, and I suppose the same trick can be used if we let A = mI + B.
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    (Original post by EricPiphany)
    Yep, and I suppose the same trick can be used if we let A = mI + B.
    Yup.
 
 
 
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