A.8 Schur’s Theorem
Consider the following system of equations for x1,x2,
where m < n. Then the following theorem is a fundamental observation.
Theorem A.8.1 Let the system of equations be as just described in (1.34) where m < n. Then
there exists x≠0 such that the components satisfy each of the equations of (1.34). Here F is a field of
scalars. Think ℝ or ℂ for example.
Proof: The above system is of the form
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.8.2 A set of vectors in Fn, F = ℝ or ℂ,
is called an orthonormal set of
Theorem A.8.3 Let v1 be a unit vector
in Fn, n >
1. Then there exist vectors
is an orthonormal set of vectors.
Proof: The equation for x, v1Tx = 0 has a nonzero solution x by Theorem A.8.1. Pick such a solution
and divide by its magnitude to get v2 a unit vector such that v1T ⋅ v2 = 0. Now suppose v1,
have been chosen such that
is an orthonormal set of vectors. Then consider the
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 vk+1. ■
Definition A.8.4 If U is an n×n matrix whose columns form an orthonormal set of vectors, then U is
called an orthogonal matrix if it is real and a unitary matrix if it is complex. Note that from the way
we multiply matrices,
in case U is orthogonal. Thus U−1 = UT. If U is only unitary, then from the dot product in ℂn, we replace
the above with
Where the ∗ indicates to take the conjugate of the transpose.
Note the product of orthogonal or unitary matrices is orthogonal or unitary because
Two matrices A and B are similar if there is some invertible matrix S such that A = S−1BS. 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,
With this preparation, here is Schur’s theorem.
Theorem A.8.5 Let A be a real or complex n × n matrix. Then there exists a unitary matrix U such
where T is an upper triangular matrix having the eigenvalues of A on the main diagonal, listed according to
multiplicity as zeros of the characteristic equation. If A has all real entries and eigenvalues, then U can be
chosen 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
be an n × n
matrix. Then let v1
be a unit eigenvector for A
. There exists λ1
By Theorem A.8.3 there exists
, an orthonormal set in
. Let U0
be a matrix whose ith
column is vi
. Then from the above, it follows U0
is unitary. Then from the way you multiply matrices
is of the form
where A1 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
an upper triangular matrix. Consider
From the way we multiply matrices, this is a unitary matrix and
where T is upper triangular. Then let U = U0U1. 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