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Linear Algebra with Cubic Roots

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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Solution Preview

Please see the attachment.

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.