Consider the following system of equations for x_{1},x_{2},
⋅⋅⋅
,x_{n}
n
∑ a x = 0,i = 1,2,⋅⋅⋅,m (1.34)
j=1 ij j
(1.34)
where m < n. Then the following theorem is a fundamental observation.
Theorem A.8.1Let the system of equations be as just described in (1.34) where m < n. Thenletting
T n
x ≡ (x1,x2,⋅⋅⋅,xn) ∈ F ,
there existsx≠0such that the components satisfy each of the equations of (1.34). Here F is a field ofscalars. Think ℝ or ℂ for example.
Proof: The above system is of the form
Ax = 0
where A is an m × n matrix with m < n. Therefore, from Theorem A.2.10, there exists x≠0 such that
Ax = 0.■
Definition A.8.2A set of vectors in F^{n}, F = ℝ or ℂ,
{x1,⋅⋅⋅,xk}
is called an orthonormal setofvectors if
{
xTx = δ ≡ 1 if i = j
i j ij 0 if i ⁄= j
Theorem A.8.3Let v_{1}be a unit vector
(|v1| = 1)
in F^{n}, n > 1. Then there exist vectors
{v2,⋅⋅⋅,vn}
such that
{v1,⋅⋅⋅,vn}
is an orthonormal set of vectors.
Proof: The equation for x,v_{1}^{T}x = 0 has a nonzero solution x by Theorem A.8.1. Pick such a solution
and divide by its magnitude to get v_{2} a unit vector such that v_{1}^{T}⋅ v_{2} = 0. Now suppose v_{1},
⋅⋅⋅
,v_{k}
have been chosen such that
{v1,⋅⋅⋅,vk}
is an orthonormal set of vectors. Then consider the
equations
--
vjTx = 0 j = 1,2,⋅⋅⋅,k
This amounts to the situation of Theorem A.8.1 in which there are more variables than equations.
Therefore, by this theorem, there exists a nonzero x solving all these equations. Divide by its magnitude
and this gives v_{k+1}. ■
Definition A.8.4If U is an n×n matrix whose columns form an orthonormal set of vectors, then U iscalled an orthogonal matrix if it is real and a unitary matrix if it is complex.Note that from the waywe multiply matrices,
UT U = UU T = I
in case U is orthogonal. Thus U^{−1} = U^{T}. If U is only unitary, then from the dot product in ℂ^{n}, we replacethe above with
U∗U = U U∗ = I.
Where the ∗ indicates to take the conjugate of the transpose.
Note the product of orthogonal or unitary matrices is orthogonal or unitary because
(U1U2)T (U1U2 ) = UT2 U T1 U1U2 = I
(U U )∗ (U U ) = U∗U ∗U U = I.
1 2 1 2 2 1 1 2
Two matrices A and B are similar if there is some invertible matrix S such that A = S^{−1}BS. Note that
similar matrices have the same characteristic equation because by Theorem A.5.13 which says the
determinant of a product is the product of the determinants,
det(λI − A) = det(λI − S−1BS ) = det (S −1(λI − B )S)
= det(S−1)det(λI − B )det(S) = det(S−1S )det(λI − B) = det(λI − B)
With this preparation, here is Schur’s theorem.
Theorem A.8.5Let A be a real or complex n × n matrix. Then there exists a unitary matrix U suchthat
U ∗AU = T, (1.35)
(1.35)
where T is an upper triangular matrix having the eigenvalues of A on the main diagonal, listed according tomultiplicity as zeros of the characteristic equation. If A has all real entries and eigenvalues, then U can bechosen to be orthogonal.
Proof: The theorem is clearly true if A is a 1 × 1 matrix. Just let U = 1 the 1 × 1 matrix which has 1
down the main diagonal and zeros elsewhere. Suppose it is true for
(n − 1)
×
(n− 1)
matrices
and let A be an n × n matrix. Then let v_{1} be a unit eigenvector for A. There exists λ_{1} such
that
, an orthonormal set in F^{n}. Let U_{0} be a matrix whose i^{th}
column is v_{i}. Then from the above, it follows U_{0} is unitary. Then from the way you multiply matrices
U_{0}^{∗}AU_{0} is of the form
( )
λ1 ∗ ⋅⋅⋅ ∗
|| 0 ||
|| . ||
( .. A1 )
0
where A_{1} is an n− 1 ×n− 1 matrix. The above matrix is similar to A so it has the same eigenvalues and
indeed the same characteristic equation. Now by induction there exists an
(n − 1)
×
(n− 1)
unitary matrix
U^
_{1} such that
^U∗A1 ^U1 = Tn− 1,
1
an upper triangular matrix. Consider
( )
1 0
U1 ≡ ^
0 U1
From the way we multiply matrices, this is a unitary matrix and
where T is upper triangular. Then let U = U_{0}U_{1}. Then U^{∗}AU = T. If A is real having real eigenvalues, all
of the above can be accomplished using the real dot product and using real eigenvectors. Thus U can be
orthogonal. ■