306 CHAPTER 12. SELF ADJOINT OPERATORS

To show the converse, note that, as above, (Ax,x) = xTAx. Suppose that A is positivedefinite. Then this is equivalent to having

xTAx ≥ δ ∥x∥2

Note that for x ∈ Rk,

(xT 0

)A

(x

0

)= xTAkx ≥ δ ∥x∥2

From Lemma 12.4.2, this implies that all the eigenvalues of Ak are positive. Hence fromLemma 12.4.2, it follows that det (Ak) > 0, being the product of its eigenvalues. ■

Corollary 12.4.5 Let A be a self adjoint n× n matrix. Then A is negative definite if andonly if det (Ak) (−1)

k> 0 for every k = 1, · · · , n.

Proof: This is immediate from the above theorem by noting that, as in the proof ofLemma 12.4.2, A is negative definite if and only if −A is positive definite. Therefore,det (−Ak) > 0 for all k = 1, · · · , n, is equivalent to having A negative definite. However,

det (−Ak) = (−1)kdet (Ak) . ■

12.5 The Square Root

With the above theory, it is possible to take fractional powers of certain elements of L (X,X)where X is a finite dimensional inner product space. I will give two treatments of this, thefirst pertaining to the square root only and the second more generally pertaining to the kth

root of a self adjoint nonnegative matrix.

Theorem 12.5.1 Let A ∈ L (X,X) be self adjoint and nonnegative. Then there exists aunique self adjoint nonnegative B ∈ L (X,X) such that B2 = A and B commutes with everyelement of L (X,X) which commutes with A.

Proof: By Theorem 12.3.2, there exists an orthonormal basis of eigenvectors of A, say{vi}ni=1 such that Avi = λivi. Therefore, by Theorem 12.2.4, A =

∑i λivi ⊗ vi where each

λi ≥ 0.Now by Lemma 12.4.2, each λi ≥ 0. Therefore, it makes sense to define

B ≡∑i

λ1/2i vi ⊗ vi.

It is easy to verify that

(vi ⊗ vi) (vj ⊗ vj) =

{0 if i ̸= j

vi ⊗ vi if i = j.

Therefore, a short computation verifies that B2 =∑

i λivi ⊗ vi = A. If C commutes withA, then for some cij ,

C =∑ij

cijvi ⊗ vj

and so since they commute,∑i,j,k

cijvi ⊗ vjλkvk ⊗ vk =∑i,j,k

cijλkδjkvi ⊗ vk =∑i,k

cikλkvi ⊗ vk