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    Linear Algebra with Cubic Roots

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    Exercise. IV. This problem is a partial investigation of which n×n matrices over C have
    cube roots; that is, for which n × n matrices A over C there is an n × n B over C such
    that A = B3. Since C is algebraically closed, every n × n matrix over C is similar over C
    to a matrix in Jordan canonical form.

    A. Suppose that A is a nilpotent n × n matrix over C, and that there is an integer k,
    k ≥ n/3, such that Ak 6= 0. Show that A does not have a cube root. (Think about Exercise
    6.3.5 of Hoffman and Kunze, for this and for part C.) Thus not all n × n matrices over
    C have cube roots.

    B. Show that A (do not assume nilpotent) has a cube root iff the Jordan canonical
    form of A has a cube root.

    C. Find c1 ∈ Q such that if N is a nilpotent 2 × 2 matrix over C and M = I + c1N,
    then M3 = I + N. That is, M is a cube root of I + N. Find c2 ∈ Q such that if N is a
    nilpotent 3 × 3 matrix over C and M = I + c1N + c2N2, then M3 = I + N. Again, M is
    a cube root of I + N.

    It is not difficult to prove by induction that for any k and any nilpotent k × k matrix
    N over C, there is a similar formula for a cube root of I + N. (This can also be proved
    using the binomial series for (1 + t)1/3.) Assume that result for the remaining parts of
    this problem.

    D. Show that if c is a nonzero complex number and N is a nilpotent k × k matrix over C, then cI + N has a square root. (First look at I +1 cN.)
    E. Show that if A (do not assume nilpotent) is an invertible n × n matrix over C then
    A has a cube root.
    *6.3.5 As referred to in parts A and C states: Let V be an n-dimensional vector space
    and let T be a linear operator on V. Suppose that there exists some positive integer k so
    that T
    k = 0. Prove that T
    m = 0.

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    A. Proof:
    From the condition, there is an integer , such that . Now I claim that does not have a cubic root. If has a cubic root , then . Since is a nilpotent matrix, then is also a nilpotent matrix because implies that . Then we can find some with , such that . Then we have because . This is a contradiction. Therefore, does not have a cubit root.
    B. Proof:
    Suppose is the Jordan canonical form of . Then for some ...

    Solution Summary

    Linear algebra with cubic roots are examined. Partial investigation of cube roots are determined.

    $2.49

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