Page 164 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
152
matrix 7.40 has the required form 7.39, completing the proof in this
direction.
Conversely, now suppose that there is an orthonormal basis of V
with respect to which the matrix of S has the form required by the
theorem. Thus there is a direct sum decomposition
V = U 1 ⊕· · · U m ,
where each U j is a subspace of V of dimension 1 or 2. Furthermore,
any two vectors belonging to distinct U’s are orthogonal, and each S| U j
is an isometry mapping U j into U j .If v ∈ V, we can write
v = u 1 +· · ·+ u m ,

where each u j ∈ U j . Applying S to the equation above and then taking
norms gives
2 2
Sv = Su 1 + ··· + Su m
2 2
= Su 1 + ··· + Su m
2 2
= u 1 +· · ·+Hu m
2
= v .
Thus S is an isometry, as desired.

Polar and Singular-Value Decompositions

Recall our analogy between C and L(V). Under this analogy, a com-
plex number z corresponds to an operator T, and ¯ z corresponds to T .
∗
The real numbers correspond to the self-adjoint operators, and the non-
negative numbers correspond to the (badly named) positive operators.
Another distinguished subset of C is the unit circle, which consists of
the complex numbers z such that |z|= 1. The condition |z|= 1is
equivalent to the condition ¯ zz = 1. Under our analogy, this would cor-
respond to the condition T T = I, which is equivalent to T being an
∗
isometry (see 7.36). In other words, the unit circle in C corresponds to
the isometries.
Continuing with our analogy, note that each complex number z ex-
cept 0 can be written in the form

z z
z = |z|= ¯ zz,
|z| |z|