# Linear Algebra with Cubic Roots

Exercise. IV. This problem is a partial investigation of which nÃ—n matrices over C have

cube roots; that is, for which n Ã— n matrices A over C there is an n Ã— n B over C such

that A = B3. Since C is algebraically closed, every n Ã— n matrix over C is similar over C

to a matrix in Jordan canonical form.

A. Suppose that A is a nilpotent n Ã— n matrix over C, and that there is an integer k,

k â‰¥ 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 Ã— 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 âˆˆ Q such that if N is a nilpotent 2 Ã— 2 matrix over C and M = I + c1N,

then M3 = I + N. That is, M is a cube root of I + N. Find c2 âˆˆ Q such that if N is a

nilpotent 3 Ã— 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 Ã— 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 Ã— 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 Ã— 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

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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.