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.

Definition

Suppose $A$ is a $n \times n$ matrix (over a field $K$). Then the characteristic polynomial of $A$ is defined as $P_A(t) = \det(tI - A)$, which is a $n$th degree polynomial in $t$. Here, $I$ refers to the $n\times n$ identity matrix.

Written out, the characteristic polynomial is the determinant

\[P_A(t) = \begin{vmatrix}t-a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & t-a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & t-a_{nn}\end{vmatrix}\]

Properties

An eigenvector $\bold{v} \in K^n$ is a non-zero vector that satisfies the relation $A\bold{v} = \lambda\bold{v}$, for some scalar $\lambda \in K$. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number $\lambda$ is called the eigenvalue.

There are at most $n$ 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 $0$ iff the column vectors of the matrix are linearly dependent. Observe that if $\lambda$ satisfies $\det(\lambda I-A) = 0$, then the column vectors of the $n\times n$ matrix $\lambda I - A$ are linearly dependent. Indeed, if we define $T = \lambda I - A$ and let $\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n$ denote the column vectors of $T$, this is equivalent to saying that there exists not all zero scalars $v_i \in K$ such that

\[v_1 \cdot \bold{T}^1 + v_2 \cdot \bold{T}^2 + \cdots + v_n \cdot \bold{T}^n = \bold{O}.\]

Hence, there exists a non-zero vector $\bold{v} = (v_1, v_2, \ldots, v_n) \in K^n$ such that $(\lambda I - A) \bold{v} = \bold{O}$. Distributing and re-arranging, $A\bold{v} = \lambda\bold{v}$, as desired. In the other direction, if $A \bold{v} = \lambda \bold{v}$, then $\lambda I \bold{v} - A \bold{v} = \bold{O}$. But then, the column vectors of $\lambda I - A$ are linearly dependent, so it follows that $\det(\lambda I - A) = 0$.

Note that if $t = 0$, then $P_A(0) = \det (tI - A) = \det (-A) = (-1)^n \det (A)$. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the coefficient of the $t^{n-1}$ term of $P_A(t)$ 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 $P_A(A) = 0$. The minimal polynomial of $A$ thus divides the characteristic polynomial $p_A$.

Linear recurrences

Let $x_1, x_2, \ldots,$ be a sequence of real numbers. Consider a monic homogenous linear recurrence of the form

\[x_{n} = c_{n-1}x_{n-1} + c_{n-2}x_{n-2} + \cdots + c_{n-k}x_{n-k}, \quad (*)\]

where $c_1, \ldots, c_k$ are real constants. The characteristic polynomial of this recurrence is defined as the polynomial

\[P(x) = x^k - c_{k-1}x^{k-1} - c_{k-2}x^{k-2}  - \cdots -c_1x - c_k.\]

For example, let $F_n$ be the $n$th Fibonacci number defined by $F_1 = F_2 = 1$, and

\[F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.\]

Then, its characteristic polynomial is $x^2 - x - 1$.

The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial $P$ are distinct, then suppose the roots are $r_1,r_2, \cdots, r_k$. Then, there exists real constants $a_1, a_2, \cdots, a_k$ such that

\[x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\]

If we evaluate $k$ different values of $x_i$ (typically $x_0, x_1, \cdots, x_{k-1}$), we can find a linear system in the $a_i$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 $1$, suppose $r_1 = r_2 = \cdots = r_{k_1}$. Then we would replace the term $a_1 \cdot r^n + a_2 \cdot r_2^n + \cdots + a_{k_1} \cdot r_{k_1}^n$ with the expression

\[(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.\]

That is, there is now a polynomial in $n$ 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 $x_n = -2x_{n-1} -x_{n-2} \Longleftrightarrow x_n + 2x_{n-1} + x_{n-2} = 0$. It’s characteristic polynomial, $x^2 + 2x + 1$, has a $-1$ double root. Then, its closed form solution is of the type $x_n = (-1)^n(an + b)$.

Given a linear recurrence of the form $x_n - a_1x_{n-1} - \cdots - a_kx_{n-k} = f(n)$, we often try to find a new sequence $y_n = x_n - g(n)$ such that $y_n$ is a homogenous linear recurrence. Then, we can find a closed form for $y_n$, and then the answer is given by $x_n = y_n + g(n)$.

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 $\bold{y}_{n-1} = \begin{pmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_{n-k}  \end{pmatrix}$. Then, we can express our linear recurrence as the matrix

\[A = \begin{pmatrix}a_1 & a_2 & a_3 & \cdots & a_{k-1} & a_{k} \\ 1 & 0 & 0 & \cdots &0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{pmatrix},\]

so that $\bold{y}_n = A\bold{y}_{n-1}$ (try to verify this). The characteristic polynomial of $A$ is precisely $P_A(t) = \det (tI - A) = a_1 \cdot t^{k-1} + a_2 \cdot t^{k-2} \cdots + a_k = 0$. 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 $P_A$ are distinct, then there exists a basis (of $\mathbb{R}^{k-1}$) consisting of eigenvectors of $A$ (since eigenvectors of different eigenvalues are linearly independent). That is, applying a change of bases, we can write

\[A = U^{-1} D U\]

for a matrix $U$ and a diagonal matrix $D$. Then, $A^2 = U^{-1} D U U^{-1} D U = U^{-1} D^2 U$, and in general, $A^n = U^{-1} D^n U$. Thus, $\bold{y}_{n} = A^{n}\bold{y}_0 = U^{-1}D^{n+k}U\bold{y}_0$. Here, $U^{-1}, U,$ and $\bold{y}_0$ are fixed (note that to find the values of $\bold{y}_0$, we may need to trace the recurrence backwards. We only take the $0$th index for simplicity). It follows that $y_{n}$ is a linear combination of the diagonal elements of $D$, namely $r_1^n, r_2^n, \ldots, r_k^n$.

Suppose now that that the roots of $P_A$ are not distinct. Then, we can write the matrix in the Jordan normal form. For simplicity, first consider just a single root $r$ repeated $k$ times. The corresponding Jordan form of the matrix is given by (that is, the matrix is similar to the following):

\[\begin{pmatrix} r & 1 & 0 & \cdots & 0 \\ 0 & r & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & r \end{pmatrix}\].

Exponentiating this matrix to the $n$th power will yield binomial coefficients as follows

\[\begin{pmatrix} r^n & {n \choose 1}r^{n-1} & {n \choose 2}r^{n-2} & \cdots & {n \choose k}r^{n-k} \\ 0 & r & {n \choose 1}r^{n-1} & \cdots & {n \choose {k-1}}r^{n-k+1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & r^n \end{pmatrix}.\]

We can treat the binomial coefficient as a polynomial in $n$. 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 $\{a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\}$, for real numbers $a_1, a_2, \ldots, a_k$, satisfies the linear recurrence relation $(*)$. If the two sequences are the same for the first $k$ values of the sequence, it follows by induction that the two sequences must be exactly the same.

In particular, for $k=2$, we can check this by using the identity

\[ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).\]


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 $G_n = a_1G_{n-1} + \cdots + a_kG_{n-k}$ be a linear recurrence. Consider the generating function given by

\[G(x) = G_0 + G_1x + G_2x^2 + \cdots\]

Then, writing the following expressing out and carefully comparing coefficients (try it),

\[a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{G(x) + R(x)}x,\]

where $R(x)$ is a remainder polynomial with degree $\le k-1$. Re-arranging,

\[G(x) =  \frac{R(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots  + a_{1}x - 1}\]

This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are $1/r_1, 1/r_2, \ldots, 1/r_k$, 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

\[G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx }\]

for some constants $c_1, \ldots, c_k$. Using the geometric series formula, we have $\frac{c}{1-y} = c + cy + cy^2 + \ldots$. Thus,

\[G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.\]

Comparing coefficients with our original definition of $G(x)$ gives $G_n = c_1r_1^n + c_2r_2^n + \cdots + c_kr_k^n$, 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 $\frac{c_i}{(1-r_ix)^\alpha}$ for $\alpha > 1$.

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 $D^ny +c_{n-1}D^{n-1}y + \cdots + c_1Dy + c_0y = 0$, then the characteristic polynomial of the equation is the polynomial

\[P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.\]

Here, $D = \frac{d}{dx}$ is short-hand for the differential operator.

If the roots of the polynomial are distinct, say $r_1, r_2, \cdots, r_n$, then the solutions of this differential equation are precisely the linear combinations $y(x) = a_1e^{r_1x} + a_2e^{r_2x} + \cdots + a_ne^{r_nx}$. Similarly, if there is a set of roots with multiplicity greater than $1$, say $r_1, r_2, \cdots, r_k$, then we would replace the term $a_1e^{r_1x} + \cdots + a_ke^{r_kx}$ with the expression $(a_1x^{k-1}  + a_2x^{k-2} + \cdots + a_{k-1}x + a_k)e^{r_1x}$.

In general, given a linear differential equation of the form $Ly = f$, where $L = c_nD^n + c_{n-1}D^{n-1} + \cdots + c_0$ is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation $Ly = 0$ and a specific solution to $Ly = f$.

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 $A: V \to V$ be a linear operator for a vector spaces $V$ over a field $K$. Suppose that there exists a polynomial $f(x) \in K[x]$ such that $f = gh$, where $g$ and $h$ are non-zero polynomials such that $\text{gcd}\,(g,h) = 1$, and such that $f(A) = O$. Then $V = \ker g(A) \oplus \ker h(A)$. This allows us to reduce the differential equation into finding the solutions to the equation $(D - \lambda I)^my = 0$, which has a basis of functions $e^{\lambda t}, te^{\lambda t}, \ldots, t^{m-1}e^{\lambda t}$.

Problems

Introductory

  • Prove Binet's formula. Find a similar closed form equation for the Lucas sequence, defined with the starting terms $L_1 = 2, L_2 = 1$, and satisfying the recursion $L_n = L_{n-1} + L_{n-2}$.
  • Let $\{x_n\}$ denote the sequence defined by the recursion $x_0 = 3, x_1 = 1$, and $x_n = 2x_{n-1} + 3x_{n-2}$. Find a closed form expression for $x_n$.
  • Given $a_0 = 1$, $a_1 = 3$, and the general relation $a_n^2 - a_{n - 1}a_{n + 1} = ( - 1)^n$ for $n \ge 1$. Find a linear recurrence for $a_n$. (AHSME 1958, Problem 40)

Intermediate

  • Let $S_n$ denote the number of ternary sequences (consisting of $0$,$1$, and $2$s) of length $n$, such that they do not contain a substring of "10", "01", or "11". Find a closed form expression for $S_n$.
  • Let $\{X_n\}$ and $\{Y_n\}$ be sequences defined as follows:
\[X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.\]
Let $k$ be the largest integer that satisfies all of the following conditions:
(i) $|X_i - k| \le 2007$, for some positive integer $i$;
(ii) $|Y_j - k| \le 2007$, for some positive integer $j$;
(iii) $k < 10^{2007}$.
Find the remainder when $k$ is divided by $2007$. (2007 iTest, #47)
  • Find all possible values of $x_0$ and $x_1$ such that the sequence defined by:
\[x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1\]
contains infinitely many natural numbers.

Olympiad

  • Let $r$ be a real number, and let $x_n$ be a sequence such that $x_0 = 0, x_1 = 1$, and $x_{n+2} = rx_{n+1} - x_n$ for $n \ge 0$. For which values of $r$ does $x_1 + x_3 + \cdots + x_{2m-1} = x_m^2$ for all positive integers $m$? (WOOT)
  • Let $a(x,y)$ be the polynomial $x^2y + xy^2$, and $b(x,y)$ the polynomial $x^2 + xy + y^2$. Prove that we can find a polynomial $p_n(a, b)$ which is identically equal to $(x + y)^n + (-1)^n (x^n + y^n)$. For example, $p_4(a, b) = 2b^2$. (1976 Putnam, Problem A2).
  • $a_1,a_2,a_3,b_1,b_2,b_3$ are distinct positive integers such that $(n + 1)a_1^n + na_2^n + (n - 1)a_3^n|(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n$ holds for all positive integer $n$. Prove that there exists $k\in N$ such that $b_i = ka_i$ for $i = 1,2,3$. (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 $x_n -2x_{n-1} -3x_{n-2}$, and its characteristic polynomial is given by $x^2-2x-3$. The roots of this polynomial are $-1$ and $3$. Thus, a closed form solution is given by $x_n = a \cdot (-1)^n + b \cdot 3^n$. For $n = 0$, we get $x_0 = a + b = 3$, and for $n = 1$, we get $x_1 = -a + 3b = 1$. Solving gives $a = 2, b = 1$. Thus, our answer is $x_n = 2 \cdot (-1)^n + 3^n$.
  • <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 $k = 7468$ (before taking $\mod{2007}$).
  • <url>viewtopic.php?t=287441 Discussion</url>
  • Hint (Newton’s Sum): Let us work backwards. Suppose $a,b,c$ 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 $r$.
  • 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)
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