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  1. MHB Apprentice

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    Apart from simplifying matrix powers, why do we want to diagonalize a matrix? Do they have any appealing application which can be used to motivate to study diagonal matrices.
    Thanks for any answers.

  2. Doctor Physicōrum
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    Simplifying matrix powers IS the main application for diagonalization. Why? Because of the very general ODE $\dot{\mathbf{x}}=A\mathbf{x}$ for constant $A$. If $A$ is diagonalizable, then the solution $\mathbf{x}=e^{At}\mathbf{x}_{0}$ makes sense only if you can exponentiate the $At$. To do that, you can form the Taylor series using matrices. Then, to compute that Taylor series, the computations are much more tractable with a diagonal matrix.
    Arma virumque canō, Trojae quī prīmus ab ōrīs Ītaliam fātō profugus Lāvīnaque vēnit lītora - multum ille et terrīs jactātus et altō vī superum, saevae memorem Jūnōnis ob īram, multa quoque et bellō passus, dum conderet urbem īnferretque deōs Latiō - genus unde Latīnum Albānīque patrēs atque altae moenia Rōmae. - Aeneid, by Publius Vergilius Maro.

  3. MHB Master
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    MHB Advanced Algebra Award (Jul-Dec 2013)

    multiplying (square) matrices is complicated, we have n2 inner products of rows and columns to consider, which is:

    n3 + n arithmetical operations in all (n products in each inner product, plus a summation, times n2).

    multiplying diagonal matrices is much simpler, the resulting product is ALSO diagonal, and requires only n operations:

    diag{a1,...,an}*diag{b1,...,bn} = diag{a1b1,...,anbn}

    even when n is small (like say n = 4), this is a tremendous savings of calculational effort (we only have 4 steps of arithmetic, rather than 68).

    it also making calculating the determinant MUCH more tractable: the determinant is invariant under a similarity transform. for an nxn matrix, normally calculating it requires computing n! n-fold products and then summing these, whereas computing the determinant of a diagonal matrix requires just computing ONE n-fold product.

    for example, computing a 5x5 determinant requires 121 arithmetical operations (even determining which 120 5-fold products to compute is tedious), whereas computing a 5x5 diagonal matrix's determinant can often be done in your head.

    morevoer, if A is diagonalizable, diagonalizing A illustrates a deep connection between the diagonalized matrix and the eigenvalues of A, and the diagonalizing matrix P and the eigenvectors of A (and since P is invertible, that the eigenvectors form an eigenbasis).

    the "catch" here is that not all matrices ARE diagonalizable. it turns out, however, that we can at least "semi-diagonalize" A into the sum:

    D + N, where D is diagonal, and N is nilpotent.

    this shows how important understanding nilpotent linear transformations is to "getting a good picture of bad matrices" (the diagonalizable ones being "good matrices").

    if a matrix function can be represented as a power series (such as in the exponential example Ackbach gives), then computing the matrix function becomes a LOT easier if our matrix is diagonalizable.

    unfortunately, the set of diagonalizable matrices isn't closed under matrix addition, which is a darn shame.

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