Characteristic polynomial
The characteristic polynomial of a linear operator refers to the polynomial whose roots are the eigenvalues of the operator. It carries much information about the operator.
In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of linear recurrences.
Contents
[hide]Definition
Suppose is a
matrix (over a field
). Then the characteristic polynomial of
is defined as
, which is a
th degree polynomial in
. Here,
refers to the
identity matrix.
Written out, the characteristic polynomial is the determinant
![\[P_A(t) = \begin{vmatrix}t-a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & t-a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & t-a_{nn}\end{vmatrix}\]](http://latex.artofproblemsolving.com/7/b/7/7b733974014e5d4427145e4dc7636fb1236d6b4e.png)
Properties
An eigenvector is a non-zero vector that satisfies the relation
, for some scalar
. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number
is called the eigenvalue.
There are at most distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is
iff the column vectors of the matrix are linearly dependent. Observe that if
satisfies
, then the column vectors of the
matrix
are linearly dependent. Indeed, if we define
and let
denote the column vectors of
, this is equivalent to saying that there exists not all zero scalars
such that
Hence, there exists a non-zero vector such that
. Distributing and re-arranging,
, as desired. In the other direction, if
, then
. But then, the column vectors of
are linearly dependent, so it follows that
.
Note that if , then
. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the coefficient of the
term of
gives the negative of the trace of the matrix (which follows from Vieta's formulas).
By the Hamilton-Cayley Theorem, the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is . The minimal polynomial of
thus divides the characteristic polynomial
.
Linear recurrences
Let be a sequence of real numbers. Consider a monic homogenous linear recurrence of the form
![\[x_{n} = c_{n-1}x_{n-1} + c_{n-2}x_{n-2} + \cdots + c_{n-k}x_{n-k}, \quad (*)\]](http://latex.artofproblemsolving.com/b/d/7/bd75ededf41e3c408919adae6acf28e7d8400865.png)
where are real constants. The characteristic polynomial of this recurrence is defined as the polynomial
![\[P(x) = x^k - c_{k-1}x^{k-1} - c_{k-2}x^{k-2} - \cdots -c_1x - c_k.\]](http://latex.artofproblemsolving.com/8/7/8/878f8c52c9da432646133b88d86d0ffaf1def17a.png)
For example, let be the
th Fibonacci number defined by
, and
![\[F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.\]](http://latex.artofproblemsolving.com/9/9/8/99853f74f5b82d9785f30cdcaf97c3f1d4596853.png)
Then, its characteristic polynomial is .
The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial are distinct, then suppose the roots are
. Then, there exists real constants
such that
![\[x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\]](http://latex.artofproblemsolving.com/c/7/0/c70672f67d8a254ae54fbc6db99ac490d4c78edf.png)
If we evaluate different values of
(typically
), we can find a linear system in the
s that can be solved for each constant. Refer to the introductory problems below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields Binet's formula.
If there are roots with multiplicity greater than , suppose
. Then we would replace the term
with the expression
![\[(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.\]](http://latex.artofproblemsolving.com/a/5/2/a523f5db3ff00640e4c1fcaef9d4a51d157f69a9.png)
That is, there is now a polynomial in multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation
. It’s characteristic polynomial,
, has a
double root. Then, its closed form solution is of the type
.
Given a linear recurrence of the form , we often try to find a new sequence
such that
is a homogenous linear recurrence. Then, we can find a closed form for
, and then the answer is given by
.
Of the following proofs, the second and third are more approachable.
Proof 1 (Linear Algebra)
Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of linear algebra (upto the Spectral Theorem).
Let . Then, we can express our linear recurrence as the matrix
![\[A = \begin{pmatrix}a_1 & a_2 & a_3 & \cdots & a_{k-1} & a_{k} \\ 1 & 0 & 0 & \cdots &0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{pmatrix},\]](http://latex.artofproblemsolving.com/b/3/a/b3acadf49da3989f8c4abbf61fe299b0d38ce34e.png)
so that (try to verify this). The characteristic polynomial of
is precisely
. This is not difficult to show via Laplace's expansion and induction, and is left as an exercise to the reader.
If the roots of are distinct, then there exists a basis (of
) consisting of eigenvectors of
(since eigenvectors of different eigenvalues are linearly independent). That is, applying a change of bases, we can write
![\[A = U^{-1} D U\]](http://latex.artofproblemsolving.com/f/3/f/f3f98a3b09d9abc86281be917f83d73face669a6.png)
for a matrix and a diagonal matrix
. Then,
, and in general,
. Thus,
. Here,
and
are fixed (note that to find the values of
, we may need to trace the recurrence backwards. We only take the
th index for simplicity). It follows that
is a linear combination of the diagonal elements of
, namely
.
Suppose now that that the roots of are not distinct. Then, we can write the matrix in the Jordan normal form. For simplicity, first consider just a single root
repeated
times. The corresponding Jordan form of the matrix is given by (that is, the matrix is similar to the following):
![\[\begin{pmatrix} r & 1 & 0 & \cdots & 0 \\ 0 & r & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & r \end{pmatrix}\]](http://latex.artofproblemsolving.com/7/3/f/73fe9e5736874f7ae37bd2f19e4790103a1e58df.png)
Exponentiating this matrix to the th power will yield binomial coefficients as follows
![\[\begin{pmatrix} r^n & {n \choose 1}r^{n-1} & {n \choose 2}r^{n-2} & \cdots & {n \choose k}r^{n-k} \\ 0 & r & {n \choose 1}r^{n-1} & \cdots & {n \choose {k-1}}r^{n-k+1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & r^n \end{pmatrix}.\]](http://latex.artofproblemsolving.com/9/f/2/9f25ce2d55904f7ed10e4a4466a7f81c006b560e.png)
We can treat the binomial coefficient as a polynomial in . Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows.
See also a <url>viewtopic.php?t=290351 graph-theoretic</url> approach.
Proof 2 (Induction)
There are a couple of lower-level ways to prove this. One is by induction, though the proof is not very revealing; we can explicitly check that a sequence , for real numbers
, satisfies the linear recurrence relation
. If the two sequences are the same for the first
values of the sequence, it follows by induction that the two sequences must be exactly the same.
In particular, for , we can check this by using the identity
![\[ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).\]](http://latex.artofproblemsolving.com/e/8/4/e84fc725b70484cd60a8bbbb0316d01460b4e24c.png)
It is also possible to reduce the recurrence to a telescoping sum. However, the details are slightly messy.
Proof 3 (Partial fractions)
Another method uses partial fractions and generating functions, though not much knowledge of each is required for this proof. Let be a linear recurrence. Consider the generating function given by
![\[G(x) = G_0 + G_1x + G_2x^2 + \cdots\]](http://latex.artofproblemsolving.com/f/5/6/f56b20688c3b3e0ac0e36dcc0de1a87aefb41543.png)
Then, writing the following expressing out and carefully comparing coefficients (try it),
![\[a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{G(x) + R(x)}x,\]](http://latex.artofproblemsolving.com/a/3/e/a3e8c3205c4c00b416ef72d03dfc264f5a4a140a.png)
where is a remainder polynomial with degree
. Re-arranging,
![\[G(x) = \frac{R(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots + a_{1}x - 1}\]](http://latex.artofproblemsolving.com/a/2/7/a277d3f4b21dcfa4407ebc4aef7242c109f19504.png)
This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are , assuming that the roots are distinct. Using partial fraction decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write
![\[G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx }\]](http://latex.artofproblemsolving.com/3/4/e/34edf7be7d9d403353a6f69889b425b94505f4ff.png)
for some constants . Using the geometric series formula, we have
. Thus,
![\[G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.\]](http://latex.artofproblemsolving.com/9/8/7/987ab3a9a4dde675e329f3f6310129d55bb2325a.png)
Comparing coefficients with our original definition of gives
, as desired.
The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form for
.
A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these blogposts for further ideas.
Differential equations
Given a monic linear homogenous differential equation of the form , then the characteristic polynomial of the equation is the polynomial
![\[P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.\]](http://latex.artofproblemsolving.com/6/3/5/63576203c0c69438e68aa3e49387bd8c5076e4c1.png)
Here, is short-hand for the differential operator.
If the roots of the polynomial are distinct, say , then the solutions of this differential equation are precisely the linear combinations
. Similarly, if there is a set of roots with multiplicity greater than
, say
, then we would replace the term
with the expression
.
In general, given a linear differential equation of the form , where
is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation
and a specific solution to
.
Proof
We can apply an induction technique similar to the section on linear recurrences above.
From linear algebra, we can use the following vector space decomposition theorem. Let be a linear operator for a vector spaces
over a field
. Suppose that there exists a polynomial
such that
, where
and
are non-zero polynomials such that
, and such that
. Then
. This allows us to reduce the differential equation into finding the solutions to the equation
, which has a basis of functions
.
Problems
Introductory
- Prove Binet's formula. Find a similar closed form equation for the Lucas sequence, defined with the starting terms
, and satisfying the recursion
.
- Let
denote the sequence defined by the recursion
, and
. Find a closed form expression for
.
- Given
,
, and the general relation
for
. Find a linear recurrence for
. (AHSME 1958, Problem 40)
Intermediate
- Let
denote the number of ternary sequences (consisting of
,
, and
s) of length
, such that they do not contain a substring of "10", "01", or "11". Find a closed form expression for
.
- Let
and
be sequences defined as follows:
![\[X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.\]](http://latex.artofproblemsolving.com/d/d/3/dd3269e2e349fcdeacff53693a9ae222f2189c0b.png)
- Let
be the largest integer that satisfies all of the following conditions:
- (i)
, for some positive integer
;
- (ii)
, for some positive integer
;
- (iii)
.
- (i)
- Find the remainder when
is divided by
. (2007 iTest, #47)
- Let
,
, and
be geometric sequences with different common ratios and let
for all integers
. If
,
,
,
,
, and
, find
. (Mock AIME 1 2006-2007, Problem 13)
- Find all possible values of
and
such that the sequence defined by:
![\[x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1\]](http://latex.artofproblemsolving.com/2/d/0/2d03f03aaae069ad7f649cd2991e4bdfb2c1f621.png)
- contains infinitely many natural numbers.
- Show that
can be written as a linear combination of the elementary symmetric polynomials
. In general, prove Newton's sums.
Olympiad
- Let
be a real number, and let
be a sequence such that
, and
for
. For which values of
does
for all positive integers
? (WOOT)
- Let
be the polynomial
, and
the polynomial
. Prove that we can find a polynomial
which is identically equal to
. For example,
. (1976 Putnam, Problem A2).
are distinct positive integers such that
holds for all positive integer
. Prove that there exists
such that
for
. (2010 Chinese MO, Problem 6)
Hints/Solutions
Introductory
- A proof of Binet's formula may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients.
- We will work out this problem in full detail. The recurrence relation is
, and its characteristic polynomial is given by
. The roots of this polynomial are
and
. Thus, a closed form solution is given by
. For
, we get
, and for
, we get
. Solving gives
. Thus, our answer is
.
- <url>viewtopic.php?t=282744 Discussion</url> (1958 AHSME, 40)
Intermediate
- <url>viewtopic.php?t=335138 Discussion</url>
- Answer (2007 iTest 47): The answer is
(before taking
).
- <url>viewtopic.php?t=287441 Discussion</url>
- Hint (Newton’s Sum): Let us work backwards. Suppose
are the roots of the characteristic polynomial of a linear recurrence. Then apply Vieta's formulas.
Olympiad
- Hint (WOOT): substitute the closed form solution. It is true for all
.
- Hint (1976 Putnam A2): do the even and odd cases separately.
- <url>viewtopic.php?search_id=804457492&t=327474 Discussion</url> (2010 Chinese MO, 6)