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Matrix polynomial

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A matrix polynomial is a linear combination of the powers of a square matrix.

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Powers

Remember that the powers of a square matrix A are obtained by multiplying A by itself several times.

In particular, given a positive integer $m$ and a $K imes K$ matrix A, we have[eq1]

A commonly adopted convention is that the 0-th power of A is the $K imes K$ identity matrix: [eq2]

Definition

We can now define matrix polynomials.

Definition Let A be a $K imes K$ matrix. Let $m$ be a non-negative integer. We say that $pleft( A ight) $ is a polynomial in A of degree $m$ if and only if[eq3]where I is the $K imes K$ identity matrix, [eq4] are scalars, and $a_{m} eq 0$.

Thus, $pleft( A ight) $ is a matrix having the same dimension as A, obtained as a linear combination of powers of A.

The scalars [eq5] are the so-called coefficients of the matrix polynomial.

In the above definition $m$ is assumed to be a non-negative integer. If [eq6] for any matrix A (i.e., the matrix polynomial is identically equal to the zero matrix), then we adopt the convention that the degree of p is $-infty $.

Example Let A be a square matrix. Then,[eq7]is a matrix polynomial in A of degree $2$.

Difference with respect to ordinary polynomials

Note that a matrix polynomial as defined above is not an ordinary polynomial. In fact, in an ordinary polynomial, the elements being raised to powers and their coefficients are required to come from the same field. Instead, in a matrix polynomial, the coefficients come from a field (the so-called field of scalars) while the matrices being raised to powers come from a different set, which is not even a field. For example, in a field the multiplication operation must be commutative, but matrix multiplication is not commutative. Actually, the set of $K imes K$ matrices, for fixed K, is a ring, an algebraic structure satisfying a weaker set of axioms than that satisfied by a field.

Nonetheless, if F is a field, such as the set of real numbers R or the set of complex numbers $U{2102} $, and $p:F ightarrow F$ is an ordinary polynomial[eq8]then we can use p to define, by extension, an associated matrix polynomial[eq9]provided that the entries of the matrix A belong to the field F.

It is common practice to switch back and forth between these two kinds of polynomials. For example, we often: 1) write a matrix polynomial; 2) derive its associated ordinary polynomial; 2) use the theory of ordinary polynomials to write the polynomial in a different form (e.g., we factorize it); 3) use the new form of the ordinary polynomial (e.g., its factorization) to rewrite the original matrix polynomial.

Example Define the matrix polynomial[eq10]Its associated ordinary polynomial is[eq11]which can be re-written as[eq12]Then, we have[eq13]

In theory, every time that we switch back and forth between the two kinds of polynomials, we should check whether the properties of ordinary polynomials that we are using hold also for matrix polynomials.

In practice, if we revise previous lectures on ordinary polynomials (e.g., polynomial division and greatest common divisors), we will realize that basically every definition, proof and proposition in those lectures is valid also for matrix polynomials. The reason is that, even if we lose some properties of fields when we deal with matrices, we do not need those properties to manipulate polynomials. Moreover, we can even use the commutative property of products, as explained in the next section.

Commutative property

We have already said that, in general, matrix multiplication is not commutative. However, the multiplication of two polynomials $pleft( A ight) $ and $qleft( A ight) $ in the same matrix A is commutative:[eq14]

This perhaps obvious fact (a proof of which can be found in a solved exercise at the end of this lecture) is extremely useful, and it is used to prove important results in linear algebra (e.g., the result below on null spaces).

Polynomials in linear operators

We can also define polynomials in linear operators, but, as we will shortly argue, we do not need to develop a separate theory because (as long as we deal with finite-dimensional vector spaces) a polynomial in a linear operator is always associated to an analogous matrix polynomial and we can study the properties of the latter.

Definition Let $S$ be a vector space and $f:S ightarrow S$ a linear operator. Let $m$ be a non-negative integer. We say that $pleft( f ight) $ is a polynomial in $f$ of degree $m$ if and only if[eq15]where [eq16] are scalars, $a_{m} eq 0$, I is the identity operator that associates each member of $S$ to itself, and $f^{j}$ denotes the operator obtained by composing $j$ times $f$:[eq17]

Let [eq18] be a basis for the vector space $S$. Remember that, provided $S$ is finite-dimensional, any linear operator $f:S ightarrow S$ is associated to a square matrix, called matrix of the linear operator with respect to $B$ and denoted by [eq19] such that, for any $sin S$,[eq20]where [eq21] and [eq22] respectively denote the coordinate vectors of $s$ and $fleft( s ight) $ with respect to $B$.

Moreover, the composition of operators can be performed by multiplying their respective matrices:[eq23]

Therefore, a polynomial $pleft( f ight) $ such as the one defined above is an operator whose matrix is a matrix polynomial in [eq24].[eq25]

Thus, as we have said at the beginning of this section, we do not need a separate theory for operator polynomials and we can deal with them by using polynomials in the respective matrices.

Null space of a matrix polynomial

A simple albeit often used result follows.

Proposition Let $S$ be the space of Kx1 vectors. Let A be a $K imes K$ matrix. Let $pleft( A ight) $ be a polynomial in A. Then, the null space of $pleft( A ight) $ is an invariant subspace of $S$ under the linear transformation defined by A.

Proof

Denote by [eq26] the null space of $pleft( A ight) $. Choose any [eq27]. We have [eq28]Thus, [eq29], which proves that [eq30] is invariant under A.

Eigenvalues of a matrix polynomial

Once we know the eigenvalues of A, we can easily compute the eigenvalues of $pleft( A ight) $.

Proposition Let A be a $K imes K$ matrix. Let $pleft( A ight) $ be a polynomial in A. Let $lambda $ be an eigenvalue of A. Then, [eq31] is an eigenvalue of $pleft( A ight) $.

Proof

We have previously demonstrated that, if $lambda $ is an eigenvalue of A associated to the eigenvector x, then $lambda ^{j}$ is an eigenvalue of $A^{j}$ associated to the same eigenvector x. Suppose that [eq32]Choose any eigenvalue $lambda $ of A and an associated eigenvector x. Then,[eq33]which proves the proposition.

Annihilating polynomial

A frequently used concept is that of an annihilating polynomial.

Definition Let A be a $K imes K$ matrix. Let $pleft( A ight) $ be a polynomial in A. We say that $pleft( A ight) $ is an annihilating polynomial if and only if[eq34]

Cayley-Hamilton theorem

In the lecture on the Cayley-Hamilton theorem, we will carefully explain one of the most important results in the theory of matrix polynomials, which states that the characteristic polynomial is an annihilating polynomial.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let A be a square matrix. Transform the ordinary polynomial[eq35]into a polynomial in A.

Solution

The matrix polynomial is[eq36]

Exercise 2

Let A be a square matrix. Transform the ordinary polynomial[eq37]into a polynomial in A.

Solution

The matrix polynomial is[eq38]

Exercise 3

Prove that the multiplication of two polynomials $pleft( A ight) $ and $qleft( A ight) $ in the same matrix A is commutative:[eq39]

Solution

Suppose that[eq40]and[eq41]Then[eq42]

How to cite

Please cite as:

Taboga, Marco (2021). "Matrix polynomial", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/matrix-polynomial.

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