Solving Quadratic and Higher Degree Polynomial Equations using Circulant Matrices, Assignments of Mathematics

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Seminar on Advanced Topics in Mathematics

Solving Polynomial Equations

5 December 2006

Dr. Tuen Wai Ng, HKU

What do we mean by solving an equation?

Example 1. Solve the equation x^2 = 1.

x^2 = 1 x^2 − 1 = 0 (x − 1)(x + 1) = 0 x = 1 or = − 1

• Need to check that in fact (1)^2 = 1 and (−1)^2 = 1.

Exercise. Solve the equation

√ x +

x − a = 2

where a is a positive real number.

Example 2. Solve the equation x^2 = 5.

x^2 = 5 x^2 − 5 = 0 (x −

5)(x +

x =

5 or −

• But what is

5? Well,

5 is the positive real number that square to 5.

• We have ”learned” that the positive solution to the equation x^2 = 5 is the positive real number that square to 5 !!!
• So there is a sense of circularity in what we have done here.
• Same thing happens when we say that i is a solution of x^2 = − 1.

What are “solved” when we solve these equations?

• The equations x^2 = 5 and x^2 = − 1 draw the attention to an inadequacy in a certain number system (it does not contain a solution to the equation).
• One is therefore driven to extend the number system by introducing, or ‘adjoining’, a solution.
• Sometimes, the extended system has the good algebraic properties of the original one, e.g. addition and multiplication can be defined in a natural way.
• These extended number systems (e.g. Q(

5. or Q(i)) have the added advantage that more equations can be solved.

What do we mean by solving a polynomial

equation?

Meaning II:

Suppose we can solve the equation xn^ = c, i.e. taking roots, try to express the the roots of a degree n polynomial using only the usual algebraic operations (addition, subtraction, multiplication, division) and application of taking roots.

• In this sense, one can solve any polynomials of degree 2 , 3 or 4 and this is in general impossible for polynomials of degree 5 or above.

• The Babylonians (about 2000 B.C.) knew how to solve specific quadratic equations.
• The solution formula for solving the quadratic equations was mentioned in the Bakshali Manuscript written in India between 200 BC and 400 AD.
• Based on the work of Scipione del Ferro and Nicolo Tartaglia, Cardano published the solution formula for solving the cubic equations in his book Ars Magna (1545).
• Lodovico Ferrari, a student of Cardano discovered the solution formula for the quartic equations in 1540 (published in Ars Magna later).
• The formulae for the cubic and quartic are complicated, and the methods to derive them seem ad hoc and not memorable.

The eigenvalues and eigenvectors of circulant matrices are very easy to compute using the nth roots of unity.

• For the 3 × 3 matrix C in (1), we need the cube roots of unity:

1 , ω = (−1 + i

3)/ 2 and ω^2 = ω.

• Direct computations show that the eigenvalues of C are a + b + c, a + bω + cω^2 , and a + bω + cω^2 , with corresponding eigenvectors (1, 1 , 1)T^ , (1, ω, ω^2 )T^ , and (1, ω, ω^2 )T^.
• This result can be generalized to higher dimensions (n ≥ 3).

To begin with, we define a distinguished circulant matrix W with first row (0, 1 , 0 ,... , 0). W is just the identity matrix with its top row moved to the bottom, e.g. for n = 4,

W =

 ,^ W^

 ,^ W^

Direct checking shows that

i) Note that W T^ = W −^1 (i.e. W is an orthogonal matrix).

ii) The characteristic polynomial for W is p(t) = det(tI − W ) = tn^ − 1 , and hence the eigenvalues of W are the nth roots of unity.

iii) For each nth root of unity λ, vλ = (1, λ, λ^2 ,... , λn−^1 ) is an associated eigenvector.

Therefore, q(W ) = a 0 I + a 1 W + a 2 W 2 + a 3 W 3 is equal to

C =

a 0 a 1 a 2 a 3 a 3 a 0 a 1 a 2 a 2 a 3 a 0 a 1 a 1 a 2 a 3 a 0

ii) For any nth root of unity λ, q(λ) is an eigenvalue of C = q(W ).

[ Indeed, if W v = λv, then W kv = λkv and hence q(W )v = q(λ)v.]

Example 3. Consider the circulant matrix

C =

Read the polynomial q from the first row of C:

q(t) = 1 + 2t + t^2 + 3t^3.

Summary

Start with any circulant matrix C, one can generate both the roots and coefficients of a polynomial p.

Here, the polynomial p is the characteristic polynomial of C; the coefficients can be obtained from the identity p(t) = det (tI − C); the roots, i.e., the eigenvalues of C, can be found by applying q to the nth roots of unity.

This perspective leads to a unified method for solving general quadratic, cubic, and quartic equations.

In fact, given a polynomial p, we try to find a corresponding circulant C having p as its characteristic polynomial. The first row of C then defines a different polynomial q, and the roots of p are obtained by applying q to the nth roots of unity.

Solving polynomial equations using circulant matrices.

Quadratics. Let’s consider a general quadratic polynomial,

p(t) = t^2 + αt + β.

We also consider a general 2 × 2 circulant

C =

[

a b b a

]

The characteristic polynomial of C is

det

[

t − a −b −b t − a

]

= t^2 − 2 at + a^2 − b^2.

The roots of the original quadratic are now found by applying q to the two square roots of unity:

q(1) =

−α 2

α^2 4 −^ β q(−1) =

−α 2

α^2 4 −^ β.

• Observe that defining b with the opposite sign produces the same roots of p, although the values of q(1) and q(−1) are exchanged.

Cubics. A parallel analysis works for cubic polynomials.

We first notice that by simple algebra, for p(x) = xn^ + αn− 1 xn−^1 + · · · + α 1 x + α 0 , the substitution y = x − αn− 1 /n eliminates the term of degree n − 1.

Therefore, we only need to consider cubic polynomials of the form

p(t) = t^3 + βt + γ.

For a general 3 × 3 circulant matrix

C =

a b c c a b b c a

we want to find a, b, and c so that p is the characteristic polynomial of C.