Linear Algebra II (pdf, 500 kB)

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Proof. First observe that if a matrix B represents a positive definite symmetric
bilinear form, then det B > 0: by Thm. 8.25, there is an invertible matrix P such
that P ⊤ BP is diagonal with entries 1, −1, or 0, and the bilinear form is positive
definite if and only if all diagonal entries are 1, i.e., P ⊤ BP = I. But this implies
1 = det(P ⊤ BP ) = det B(det P ) 2 , and since (det P ) 2 > 0, this implies det B > 0.
Now if A is positive definite, then all Aj are positive definite, since they represent
the restriction of the bilinear form to subspaces. So det Aj > 0 for all j.
Conversely, assume that det Aj > 0 for all j. We use induction on n. For n = 1
(or n = 0), the statement is clear. For n ≥ 2, we apply the induction hypothesis
to An−1 and obtain that An−1 is positive definite. Then there is an invertible
matrix P ∈ Mat(n − 1, R) such that
P ⊤ 0
0 1
P 0
A
0 1
=
with some vector b ∈ Rn−1 and α ∈ R. Setting
I −b
Q =
,
0 1
I b
b ⊤ α
=: B ,
we get
Q ⊤
I
BQ =
0
0
,
β
and so A is positive definite if and only if β > 0. But we have (note det Q = 1)
β = det(Q ⊤ BQ) = det B = det(P ⊤ ) det A det P = (det P ) 2 det A ,
so β > 0, since det A = det An > 0, and A is positive definite.
9. Inner Product Spaces
In many applications, we want to measure distances an angles in a real vector
space. For this, we need an additional structure, a so-called inner product.
9.1. Definition. Let V be a real vector space. An inner product on V is a positive
definite symmetric bilinear form on V . It is usually written in the form (x, y) ↦→
〈x, y〉 ∈ R. Recall the defining properties:
(1) 〈λx + λ ′ x ′ , y〉 = λ〈x, y〉 + λ ′ 〈x ′ , y〉;
(2) 〈y, x〉 = 〈x, y〉;
(3) 〈x, x〉 > 0 for x = 0.
A real vector space together with an inner product on it is called a real inner
product space.
Recall that an inner product on V induces an injective homomorphism V → V ∗ ,
given by sending x ∈ V to the linear form y ↦→ 〈x, y〉; this homomorphism is an
isomorphism when V is finite-dimensional.
Frequently, it is necessary to work with complex vector spaces. In order to have
a similar structure there, we cannot use a bilinear form: if we want to have 〈x, x〉
to be real and positive, then we would get
〈ix, ix〉 = i 2 〈x, x〉 = −〈x, x〉 ,
which would be negative. The solution to this problem is to consider Hermitian
forms instead of symmetric bilinear forms. The difference is that they are
conjugate-linear in the second argument.