Definition 11.3.3The linear map, A^{∗}is called the adjoint of A. In the case whenA : X → X and A = A^{∗}, A is called a self adjoint map. Such a map is also called Hermitian.
Theorem 11.3.4Let M be an m×n matrix. Then M^{∗} =
(M-)
^{T}in words, the transposeof the conjugate of M is equal to the adjoint.
Proof:Using the definition of the inner product in ℂ^{n},
Since x,y are arbitrary vectors, it follows that M_{ji} =
(M ∗)
_{ij} and so, taking conjugates of both
sides,
----
M ∗ij = Mji ■
The next theorem is interesting. You have a p dimensional subspace of F^{n} where F = ℝ or ℂ.
Of course this might be “slanted”. However, there is a linear transformation Q which preserves
distances which maps this subspace to F^{p}.
Theorem 11.3.5Suppose V is a subspace of F^{n}having dimension p ≤ n. Then there exists aQ ∈ℒ
(Fn, Fn)
such that
QV ⊆ span (e1,⋅⋅⋅,ep)
and
|Qx|
=
|x |
for allx.Also
Q∗Q = QQ ∗ = I.
Proof:By Lemma 11.2.1 there exists an orthonormal basis for V,
{vi}
_{i=1}^{p}. By using the
Gram Schmidt process this may be extended to an orthonormal basis of the whole space
F^{n},
{v1,⋅⋅⋅,vp,vp+1,⋅⋅⋅,vn}.
Now define Q ∈ℒ
(Fn,Fn )
by Q
(vi)
≡ e_{i} and extend linearly. If ∑_{i=1}^{n}x_{i}v_{i} is an arbitrary
element of F^{n},
|| ( n ) ||2 ||n ||2 n || n ||2
||Q ∑ xv || = ||∑ x e || = ∑ |x |2 = ||∑ xv || .
| i=1 i i | |i=1 ii| i=1 i |i=1 i i|
It remains to verify that Q^{∗}Q = QQ^{∗} = I. To do so, let x,y∈ F^{n}. Then let ω be a complex
number such that