372 CHAPTER 13. MATRICES AND THE INNER PRODUCT

Now letting x ∈ Rn−1, x ̸= 0, the induction hypothesis implies

(xT ,0

)A

(x

0

)= xT An−1x= (An−1x,x)> 0.

The dimension of {z ∈ Rn : zn = 0} is n−1 and the dimension of span(u1,u2) = 2 and sothere must be some nonzero x ∈ Rn which is in both of these subspaces of Rn. However,the first computation would require that (Ax,x) < 0 while the second would require that(Ax,x)> 0. This contradiction shows that all the eigenvalues must be positive. This provesthe if part of the theorem.

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

xT Ax≥ δ ∥x∥2

Note that for x ∈ Rk,

(xT 0

)A

(x

0

)= xT Akx≥ δ ∥x∥2

From Lemma 13.14.2, this implies that all the eigenvalues of Ak are positive. Hence fromLemma 13.14.2, it follows that det(Ak)> 0, being the product of its eigenvalues. ■

Corollary 13.14.5 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 13.14.2, 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)k det(Ak) . ■

13.15 The Singular Value DecompositionIn this section, A will be an m×n matrix. To begin with, here is a simple lemma observedearlier.

Lemma 13.15.1 Let A be an m×n matrix. Then A∗A is self adjoint and all its eigenvaluesare nonnegative.

Proof: It is obvious that A∗A is self adjoint. Suppose A∗Ax = λx. Then λ |x|2 =(λx,x) = (A∗Ax,x) = (Ax,Ax)≥ 0. ■

Definition 13.15.2 Let A be an m×n matrix. The singular values of A are the square rootsof the positive eigenvalues of A∗A.

With this definition and lemma here is the main theorem on the singular value decom-position. In all that follows, I will write the following partitioned matrix(

σ 00 0

)