Page 189 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
178
Continue in this fashion for j = 4,...,m − 1, at each step solving for
j
a j so that the coefficient of N on the right side of the equation above
equals 0. Actually we do not care about the explicit formula for the
a j ’s. We need only know that some choice of the a j ’s gives a square
root of I + N.
The previous lemma is valid on real and complex vector spaces.
However, the next result holds only on complex vector spaces.
On real vector spaces 8.32 Theorem: Suppose V is a complex vector space. If T ∈L(V)
there exist invertible is invertible, then T has a square root.
operators that have no
square roots. For Proof: Suppose T ∈L(V) is invertible. Let λ 1 ,...,λ m be the dis-
example, the operator tinct eigenvalues of T, and let U 1 ,...,U m be the corresponding sub-
of multiplication by −1 spaces of generalized eigenvectors. For each j, there exists a nilpotent
on R has no square
operator N j ∈L(U j ) such that T| U j = λ j I + N j (see 8.23(c)). Because T
root because no real
is invertible, none of the λ j ’s equals 0, so we can write
number has its square
equal to −1. N j
= λ j I +
T| U j
λ j
for each j. Clearly N j /λ j is nilpotent, and so I + N j /λ j has a square
root (by 8.30). Multiplying a square root of the complex number λ j by
.
a square root of I + N j /λ j , we obtain a square root S j of T| U j
A typical vector v ∈ V can be written uniquely in the form

v = u 1 +· · ·+ u m ,

where each u j ∈ U j (see 8.23). Using this decomposition, define an
operator S ∈L(V) by

Sv = S 1 u 1 +· · ·+ S m u m .

You should verify that this operator S is a square root of T, completing
the proof.

By imitating the techniques in this section, you should be able to
prove that if V is a complex vector space and T ∈L(V) is invertible,
th
then T has a k -root for every positive integer k.