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Chapter 10: Problem 63

Prove that the characteristic polynomial of an operator \(T: V \rightarrow V\) is a product of its elementary divisors.

Short Answer

Expert verified
The characteristic polynomial of an operator \(T\), denoted \(p_T(\lambda)\), is given by \(\det(T - \lambda I)\), and the elementary divisors are monic polynomials associated with the eigenvalues of \(T\). We can express the characteristic polynomial as a product of the elementary divisors: \[p_T(\lambda) = \prod_{i=1}^{n} (\lambda - \mu_i)^{n_i},\] where \(n\) is the number of distinct eigenvalues of \(T\), and \(n_i\) is the multiplicity of the \(i\)-th eigenvalue \(\mu_i\). Since the elementary divisors divide the characteristic polynomial, their product will indeed equal the characteristic polynomial, proving that the characteristic polynomial of an operator is a product of its elementary divisors.

Step by step solution

01

Definitions

First, let's define the characteristic polynomial and elementary divisors of an operator. The characteristic polynomial of an operator \(T\), denoted \(p_T(\lambda)\), is the polynomial given by the determinant of the matrix \((T - \lambda I)\), where \(\lambda\) is a scalar and \(I\) is the identity matrix, i.e., \[p_T(\lambda) = \det(T - \lambda I).\] Elementary divisors of an operator \(T\) are the invariant factors of \(T\), which are the monic polynomials associated with \(T\) such that they have powers of the form \((\lambda - \mu_i)^{n_i}\), where \(\mu_i\) are eigenvalues of \(T\) and \(n_i\) are their corresponding multiplicities.
02

Write the characteristic polynomial in terms of the elementary divisors

Now let's express the characteristic polynomial in terms of the elementary divisors. We can write the characteristic polynomial as a product of the elementary divisors as follows: \[p_T(\lambda) = \prod_{i=1}^{n} (\lambda - \mu_i)^{n_i},\] where \(n\) is the number of distinct eigenvalues of \(T\), and \(n_i\) is the multiplicity of the \(i\)-th eigenvalue \(\mu_i\).
03

Prove the equality

To prove that the product of the elementary divisors is equal to the characteristic polynomial, first recall that the determinant of a matrix equals the product of its diagonal entries. Note that the elementary divisor of an eigenvalue \(\mu_i\) divides the characteristic polynomial. Since the characteristic polynomial can be expressed as a product of elementary divisors, which themselves divide the polynomial, the product of the elementary divisors will indeed equal the characteristic polynomial. Therefore, we have proven that the characteristic polynomial of an operator is a product of its elementary divisors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elementary Divisors
Elementary divisors are essentially factors that help in understanding the structure of linear operators through their characteristic polynomials. These divisors relate directly to the concept of invariant factors. Invariant factors are monic polynomials (meaning the leading coefficient is 1) derived from the linear operator.
  • Each elementary divisor corresponds to one of these invariant factors.
  • They typically take the form \[(\lambda - \mu_i)^{n_i},\]where \(\mu_i\) represents an eigenvalue of the operator, and \(n_i\) is the multiplicity or the times this eigenvalue repeats.
Understanding elementary divisors is crucial because they help in constructing the characteristic polynomial, as each divisor represents how many times an eigenvalue contributes to the overall polynomial.
Invariant Factors
Invariant factors are a powerful tool in understanding the algebraic structure of a linear operator. These factors are closely tied to elementary divisors and can be seen as the building blocks of a polynomial expression representing the operator.

  • Invariant factors are organized in increasing order of divisibility.
  • They capture the crucial insights about the operator's behavior.
  • The polynomials are ordered so that each one divides the next.
For a linear operator, the invariant factor decomposition aids in simplifying matrices to a form that clearly shows their eigenvalues and their multiplicities 鈥攁 concept that connects directly to the concept of elementary divisors.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental elements in linear algebra that relate directly to linear operators. They describe the action of an operator on a vector space and are key to understanding the structure and behavior of matrices.

  • An eigenvalue \(\mu\) of a linear operator \(T\) is a scalar such that there exists a non-zero vector \(v\) (an eigenvector) for which \(T(v) = \mu v\).
  • Eigenvectors provide direction, while the eigenvalues provide scaling factors on transformation by \(T\).
When the operator is represented in a matrix form, eigenvalues can be derived through the characteristic polynomial by solving \[\det(T - \lambda I) = 0,\]where \(I\) is the identity matrix. This polynomial reveals all the eigenvalues as its roots, giving insight into the operator's behavior.
Linear Operators
Linear operators are functions between vector spaces that respect the operations of vector addition and scalar multiplication. They are the foundation upon which many concepts in linear algebra are built.

  • A linear operator \(T: V \rightarrow V\) means \(T\) maps vectors in a space \(V\) back to \(V\).
  • The understanding of linear operators relies significantly on matrices.
  • Concepts like characteristic polynomials, eigenvalues, and eigenvectors are critical to unveiling the nature of these operators.
Matrices can represent linear operators, offering a finite-dimensional lens through which we can analyze their effects on vector spaces. The study of linear operators leads to insights into more complex mathematical constructions and is foundational in applied mathematics and theoretical physics.

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Most popular questions from this chapter

Let \(\left.T: V \text { be linear. Let } X=\operatorname{Ker} T^{i-2}, Y=\operatorname{Ker} T^{i-1}, Z=\operatorname{Ker} T^{i} . \text { Therefore (Problem } 10.14\right)\) \(X \subseteq Y \subseteq Z .\) Suppose \\[ \left\\{u_{1}, \ldots, u_{r}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{t}\right\\} \\] are bases of \(X, Y, Z,\) respectively.

Suppose \(V=U \oplus W\) and suppose \(T: V \rightarrow V\) is linear. Show that \(U\) and \(W\) are both \(T\) -invariant if and only if \(T E=E T,\) where \(E\) is the projection of \(V\) into \(U\).

Suppose dim \(V=n\). Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if \(\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right]=\operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\)

Prove Theorem 10.12: Let \(Z(v, T)\) be a \(T\) -cyclic subspace, \(T_{v}\) the restriction of \(T\) to \(Z(v, T),\) and \(m_{v}(t)=t^{k}+a_{k-1} t^{k-1}+\cdots+a_{0}\) the \(T\) -annihilator of \(v .\) Then, (i) The set \(\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is a basis of \(Z(v, T) ;\) hence, \(\operatorname{dim} Z(v, T)=k\) (ii) The minimal polynomial of \(T_{v}\) is \(m_{v}(t)\) (iii) The matrix of \(T_{v}\) in the above basis is the companion matrix \(C=C\left(m_{e}\right)\) of \(m_{v}(t)\) [which has \(1^{\prime}\) s below the diagonal, the negative of the coefficients \(a_{0}, a_{1}, \ldots, a_{k-1}\) of \(m_{v}(t)\) in the last column, and \(0^{\circ}\) s elsewhere.
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