MU Algebra Worksheet

1Read the following, and then answer the questions below.
Definition
Suppose that A is a matrix with complex number entries aij .
We define the complex conjugate of A, written A, to be the matrix whose entries
are aij (i.e. we conjugate each entry of A).
Furthermore, we define the the conjugate transpose of A, written A+ , to be the
matrix A+ = ( A)t , i.e. the transpose of A.
First, there are some natural algebraic facts that can be verified in a straightforward
fashion:
Theorem
Theorem 1
If A and B are two square matrices with complex entries and z is any complex
number, then the following are true:
1. ( A+ )+ = A
2. (kA)+ = kA+
3. ( A + B)+ = A+ + B+
4. ( AB)+ = B+ A+
We now define a special kind of complex matrix:
1
You may want to try the question from the Test again as a warm-up. A blank copy of the test is on the Test 1 Instructions page.
1
Def
A square matrix A with complex entries is called super-symmetric if A+ = A.
Before continuing, you should write down some examples of super-symmetric and
non-super-symmetric matrices to get a feeling for them. You may notice that in all
of your super-symmetric matrices, the entries on the diagonal are real numbers.
This is true in general:
Theorem
Theorem 2
If A is a super-symmetric matrix, then the entries on the main diagonal of A must
be real numbers.
Proof: In order for A = [ aij ] to be super-symmetric, we must have aii = aii . But for
any complex number z = a + bi, z = a − bi, so if z = z, then bi = −bi ⇒ b = 0, and
so z = a = z is real.
1.1
True or False: If A is an n × n matrix (for n ≥ 2), then the matrix A + A+ is
super-symmetric.
1.2
True or False: If A is an n × n matrix (for n ≥ 2), then the matrix A − A+ is
super-symmetric.
1.3
2
True or False: If A is a super-symmetric matrix and A is invertible, then A−1 is
super-symmetric.2
Definition
Consider the following definition and answer the questions that follow:
A matrix A, with complex number entries, is said to be complex-diagonalizable
if we can write it as A = PDP−1 for D a diagonal matrix and P an invertible
matrix (where D and P are allowed to have complex number entries.)
The same facts about P and D apply when A is complex-diagonalizable: namely,
that D will have the set of eigenvalues of A on it’s diagonal, and that the columns
of P will consist of the corresponding eigenvectors of A (possibly with complex
numbers in them!)
Note that a matrix A being diagonalizable in the usual sense (with only real number
eigenvalues) implies that A is complex-diagonalizable (since real numbers are just
complex numbers with no imaginary part (i.e z real means z = a + 0i).) But
we will see shortly that there are complex-diagonalizable matrices which are not
diagonalizable.
Before we continue though, we record some theorems about diagonalization which
apply equally to complex-diagonalization:
2
Note that inversion and invertibility of complex matrices works in exactly the same way as for matrices with only real numbers.
2
Theorem
Theorem 3
1. If A has n distinct (complex) eigenvalues, then A is complex-diagonalizable.
2. A is complex-diagonalizable if and only if A has n basic eigenvectors (where
complex numbers are allowed in the eigenvectors.)
3. A is complex-diagonalizable if and only if for each (complex) eigenvalue λ, if
λ has algebraic multiplicity k, then λ has geometric multiplicity k as well.a
a
Algebraic and geometric multiplicity were in the Week 5 In-Class activities, but we repeat them here: the algebraic multiplicity of
eigenvalue λ is the number of times the factor ( x − λ) appears in the characteristic polynomial, and the geometric multiplicity of λ is the
number of basic eigenvectors for λ.
#
0 −1
is not diagonalizable.3
We know that A =
1 0
”
2.1
Show that A is however, complex-diagonalizable.
2.2
Complex-diagonalize the following matrix:


−2 0 0


A =  0 2 −2
0 2 0
(That is, find it’s eigenvalues (including complex ones), and for each eigenvalue determine
(possibly complex) eigenvectors, in order to create D and P in the usual way. You do not
have to determine P−1 explicitly, just P.)
Write all complex numbers appearing in P and D in Cartesian form (a + bi) in your
final answer.
3
Example
A fact about polynomials is that they can always be factored into linear terms if
we allow complex numbers. That is, if p( x ) = x n + an−1 x n−1 + … + a1 x + a0 is a
polynomial, then there are complex numbers α1 , …, αn (possibly some or all are real
numbers, and there can be repeats) such that p( x ) factors into ( x − α1 ) · … · ( x − αn ).
(The numbers αi are the roots of p( x ).)
The polynomial p( x ) = x 5 − 7x 4 + 24x 3 − 88x 2 + 135x − 225 factors into
= ( x − 3i )( x + 3i )( x − 1 + 2i )( x + 1 − 2i )( x − 5).
You may notice in the above example that two pairs of complex roots appear: ±3i
and 1 ± 2i. A subtle follow-up fact about polynomials with complex roots is that
this always happens:
3
See Example 5 in the Section 3.3 pre-class reading. Notice the wording in the pre-class reading: because the characteristic polynomial of
this matrix had no real eigenvalues, we said that it had “no eigenvalues” and so was not diagonalizable. Again, in this assignment (and
in the future only when we specifically mention they are allowed), we are considering complex eigenvalues, and so the matrix will be
complex-diagonalizable.
3
Theorem
Theorem 4
If α ∈ C is a root of a polynomial p( x ),a then necessarily α (the complex
conjugate) is as well.
In other words, if p( x ) has ( x − α) as a factor, then it also has ( x − α) as a factor.
a
Technically this fact only applies to polynomials with real number coefficients, not things like p( x ) = x 2 + (1 − i ) x − i = ( x − i )( x + 1).
3.1
Suppose that A is a 3 × 3 matrix with only one real eigenvalue λ1 , and that λ1 is
not repeated (i.e. the algebraic multiplicity of λ1 is 1). Explain why A must be
complex-diagonalizable.
3.2
(BONUS 2 points) Give an example of a 4 × 4 matrix A which is not complexdiagonalizable and which has no real eigenvalues. Make sure to justify all aspects
of your answer.
4