Linear Algebra II (pdf, 500 kB)

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9.4. Definition. Let V be an inner product space.
(1) For x ∈ V , we set x = 〈x, x〉 ≥ 0. x is a unit vector if x = 1.
(2) We say that x, y ∈ V are orthogonal, x ⊥ y, if 〈x, y〉 = 0.
(3) A subset S ⊂ V is orthogonal if x ⊥ y for all x, y ∈ S such that x = y. S
is an orthonormal set if in addition, x = 1 for all x ∈ S.
(4) v1, . . . , vn ∈ V is an orthonormal basis or ONB of V if the vectors form a
basis that is an orthonormal set.
9.5. Proposition. Let V be an inner product space.
(1) For x ∈ V and a scalar λ, we have λx = |λ| x.
(2) (Cauchy-Schwarz inequality) For x, y ∈ V , we have |〈x, y〉| ≤ x y,
with equality if and only if x and y are linearly dependent.
(3) (Triangle inequality) For x, y ∈ V , we have x + y ≤ x + y.
Note that these properties imply that · is a norm on V in the sense of Section 7.
In particular,
d(x, y) = x − y
defines a metric on V ; we call d(x, y) the distance between x and y.If V = R n with
the standard inner product, then this is just the usual euclidean distance.
Proof.
(1) We have
λx = 〈λx, λx〉 =
λ ¯ λ〈x, x〉 = |λ| 2 〈x, x〉 = |λ| 〈x, x〉 = |λ| x .
(2) This is clear when y = 0, so assume y = 0. Consider
z = x −
then 〈z, y〉 = 0. We find that
0 ≤ 〈z, z〉 = 〈z, x〉 = 〈x, x〉 −
〈x, y〉
y ;
y2 〈x, y〉
y2 〈y, x〉 = x2 |〈x, y〉|2
−
y2 ,
which implies the inequality. If x = λy, we have equality by the first part
of the proposition. Conversely, if we have equality, we must have z = 0,
hence x = λy (with λ = 〈x, y〉/y 2 ).
(3) We have
x + y 2 = 〈x + y, x + y〉 = 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉
= x 2 + 2 Re〈x, y〉 + y 2 ≤ x 2 + 2|〈x, y〉| + y 2
≤ x 2 + 2x y + y 2 = (x + y) 2 ,
using the Cauchy-Schwarz inequality.
Next we show that given any basis of a finite-dimensional inner product space,
we can modify it in order to obtain an orthonormal basis. In particular, every
finite-dimensional inner product space has orthonormal bases.