A First Course in Linear Algebra, 2017a

7.4. Orthogonality 413
by the assumption that A is positive definite.
♠
There is yet another way to recognize whether a matrix is positive definite which is described in terms
of these submatrices. We state the result, the proof of which can be found in more advanced texts.
Theorem 7.77: Positive Matrix and Determinant of A k
Let A be a symmetric matrix. Then A is positive definite if and only if det(A k ) is greater than 0 for
every submatrix A k , k = 1,···,n.
Proof. We prove this theorem by induction on n. It is clearly true if n = 1. Suppose then that it is true for
n−1wheren ≥ 2. Since det(A)=det(A n ) > 0, it follows that all the eigenvalues are nonzero. We need to
show that they are all positive. Suppose not. Then there is some even number of them which are negative,
even because the product of all the eigenvalues is known to be positive, equaling det(A). Picktwo,λ 1 and
λ 2 and let A⃗u i = λ i ⃗u i where ⃗u i ≠⃗0 fori = 1,2 and ⃗u 1 •⃗u 2 = 0. Now if ⃗y = α 1 ⃗u 1 + α 2 ⃗u 2 is an element of
span{⃗u 1 ,⃗u 2 }, then since these are eigenvalues and ⃗u 1 •⃗u 2 = 0, a short computation shows
(α 1 ⃗u 1 + α 2 ⃗u 2 ) T A(α 1 ⃗u 1 + α 2 ⃗u 2 )
= |α 1 | 2 λ 1 ‖⃗u 1 ‖ 2 + |α 2 | 2 λ 2 ‖⃗u 2 ‖ 2 < 0.
Now letting⃗x ∈ R n−1 , we can use the induction hypothesis to write
[ x
T
0 ] [ ] ⃗x
A =⃗x T A
0
n−1 ⃗x > 0.
Now the dimension of {⃗z ∈ R n : z n = 0} is n − 1 and the dimension of span{⃗u 1 ,⃗u 2 } = 2 and so there must
be some nonzero⃗x ∈ R n which is in both of these subspaces of R n . However, the first computation would
require that⃗x T A⃗x < 0 while the second would require that⃗x T A⃗x > 0. This contradiction shows that all the
eigenvalues must be positive. This proves the if part of the theorem. The converse can also be shown to
be correct, but it is the direction which was just shown which is of most interest.
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Corollary 7.78: Symmetric and Negative Definite Matrix
Let A be symmetric. Then A is negative definite if and only if
(−1) k det(A k ) > 0
for every k = 1,···,n.
Proof. This is immediate from the above theorem when we notice, that A is negative definite if and only
if −A is positive definite. Therefore, if det(−A k ) > 0forallk = 1,···,n, it follows that A is negative
definite. However, det(−A k )=(−1) k det(A k ).
♠