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Chapter 10: Problem 65

Let \(C(f(t))\) denote the companion matrix to an arbitrary polynomial \(f(t)\). Show that \(f(t)\) is the characteristic polynomial of \(C(f(t))\)

Short Answer

Expert verified
The characteristic polynomial of the companion matrix \(C(f(t))\) is obtained by finding the determinant of the matrix \(C(f(t)) - tI_n\). By expanding the determinant using cofactor expansion and simplifying the expression, we show that the characteristic polynomial is equal to the given polynomial \(f(t)\).

Step by step solution

01

Definition of the companion matrix

Given a polynomial \[f(t) = t^n + a_{n-1}t^{n-1} + a_{n-2}t^{n-2} + \cdots + a_1t + a_0,\] its companion matrix \(C(f(t))\) is defined as follows: \[C(f(t)) = \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & 0 & \cdots & 0 & -a_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -a_{n-1} \end{pmatrix}.\]
02

Finding the characteristic polynomial

The characteristic polynomial of the companion matrix \(C(f(t))\) is given by the determinant of the matrix \(C(f(t)) - tI_n\), where \(I_n\) is the \(n \times n\) identity matrix. Let's subtract \(tI_n\) from \(C(f(t))\). \[C(f(t)) - tI_n = \begin{pmatrix} -t & 0 & 0 & \cdots & 0 & -a_0 \\ 1 & -t & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & -t & \cdots & 0 & -a_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -t & -a_{n-1} \end{pmatrix}.\] Now let's find the determinant of this matrix.
03

Evaluating the determinant using cofactor expansion

We will expand this determinant along the first row. The determinant of the matrix \(C(f(t)) - tI_n\) can be written as: \[\begin{vmatrix} -t & 0 & 0 & \cdots & 0 & -a_0 \\ 1 & -t & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & -t & \cdots & 0 & -a_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -t & -a_{n-1} \end{vmatrix} = (-1)^{n+1}(-t)^n \begin{vmatrix} \begin{array}{c|ccc|c} -t & 0 & \cdots & 0 & -a_1 \\ \hline 1 & -t & \cdots & 0 & -a_2 \\ \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -t & -a_{n-1} \end{array}\end{vmatrix} + (-1)^{2} a_0 \begin{vmatrix} \begin{array}{c|c} 1 & -t \\ \vdots & \\ 0 & -t \end{array}\end{vmatrix}.\] On expanding the determinant further, we get \[ = (-1)^{n+1}(-t)^n(t^{n-1} + a_{n-1}t^{n-2} + a_{n-2}t^{n-3} + \cdots + a_1) + (-1)^{2} a_0 (-t)^{n-1}.\]
04

Simplifying the expression

Simplify the expression above to obtain \[ = -t^n - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - \cdots - a_1t - a_0.\] This is equal to the given polynomial \(f(t)\). Therefore, we have shown that the characteristic polynomial of the companion matrix \(C(f(t))\) is equal to \(f(t)\).

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Key Concepts

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

Characteristic Polynomial
The characteristic polynomial is a key concept in linear algebra and it emerges from the study of eigenvalues and eigenvectors of a matrix. Specifically, it is a polynomial which provides insights into the matrix's eigenvalues, which are fundamental to many applications in both pure and applied mathematics. To find the characteristic polynomial of a matrix, we calculate the determinant of the matrix subtracted by a scalar multiple of the identity matrix. This process is symbolized by \( \det(A - \lambda I) \) where \lambda is a scalar.

As demonstrated in the exercise, the characteristic polynomial for the companion matrix \(C(f(t))\) is derived by subtracting \(tI_n\) from \(C(f(t))\) and then calculating the determinant of the resulting matrix. The characteristic equation is a cornerstone in determining the behavior of matrices and resonates deeply with the matrix's properties. It provides crucial information that leads to understanding the stability of systems, solving differential equations, and analyzing vibrations in mechanical structures, among other applications.
Matrix Determinant
The determinant of a matrix is a special scalar value that can be computed from its elements and provides profound information about the matrix. In many ways, the matrix determinant acts as a transformative bridge between different mathematical concepts and applications. For example, it indicates whether a matrix is invertible, reflects the volume scaling factor of a linear transformation, and it plays a vital role in solving systems of linear equations.

In the provided solution, calculating the determinant of \(C(f(t)) - tI_n\) is essential in expressing the characteristic polynomial. The determinant is especially crucial when discussing eigenvalues, as it relates to the roots of the characteristic polynomial. The process of finding determinants can be complex for larger matrices, but several methods, such as cofactor expansion, can facilitate the calculation.
Cofactor Expansion
Cofactor expansion, also known as Laplace's expansion, is a technique used to calculate the determinant of a matrix. It involves breaking down the determinant into smaller parts, which makes it particularly useful for hand calculations when dealing with larger matrices. In simple terms, to find the determinant of a matrix, one selects any row or column and then sums the products of each element by its corresponding cofactor.

The solution uses cofactor expansion along the first row to calculate the determinant of the matrix \(C(f(t)) - tI_n\). This method might seem laborious for large matrices, but with the structured form of the companion matrix, it cleverly leads to a quick resolution of the determinant and hence the characteristic polynomial. Through the steps of this method, the intricate relationships within the matrix are unraveled, leading to the characteristic polynomial that holds key insights into the eigenvalues of the matrix.

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Most popular questions from this chapter

Suppose the characteristic polynomial of \(T: V \rightarrow V\) is \(\Delta(t)=f_{1}(t)^{m_{f}} f_{2}(t)^{\mathrm{n}_{2}} \ldots f_{r}(t)^{\mathrm{m}_{5}},\) where the \(f_{i}(t)\) are distinct monic imeducible polynomials, Let \(V=W_{1} \oplus \cdots \oplus W_{r}\) be the primary decomposition of \(V\) into \(T\) invariant subspaces. Show that \(f_{i}(t)^{n_{1}}\) is the characteristic polynomial of the restriction of \(T\) to \(W_{i}\).

Let \(V\) be a seven-dimensional vector space over \(\mathbf{R}\), and let \(T: V \rightarrow V\) be a linear operator with minimal polynomial \(m(t)=\left(t^{2}-2 t+5\right)(t-3)^{3} .\) Find all possible rational canonical forms \(M\) of \(T\) Because \(\operatorname{dim} V=7\), there are only two possible characteristic polynomials, \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)^{2}\) \((t-3)^{3}\) or \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)(t-3)^{5} .\) Moreover, the sum of the orders of the companion matrices must add up to 7. Also, one companion matrix must be \(C\left(t^{2}-2 t+5\right)\) and one must be \(C\left((t-3)^{3}\right)=\) \(C\left(t^{3}-9 t^{2}+27 t-27\right)\). Thus, \(M\) must be one of the following block diagonal matrices: (a) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],\left[\begin{array}{rr}0 & -9 \\ 1 & 6\end{array}\right]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],[3],[3]\right)\)(b) We have that $$ w_{k}=0+\cdots+0+w_{k}+0+\cdots+0 $$ is the unique sum corresponding to \(w_{k} \in W_{k}\); hence, \(E\left(w_{k}\right)=w_{k} .\) Then, for any \(v \in V\), $$ E^{2}(v)=E(E(v))=E\left(w_{k}\right)=w_{k}=E(v) $$ Thus, \(E^{2}=E\), as required.

Let \(E_{1}, \ldots, E_{r}\) be linear operators on \(V\) such that (i) \(E_{i}^{2}=E_{i}\) (i.e., the \(E_{i}\) are projections); (ii) \(E_{i} E_{j}=0, i \neq j\); (iii) \(I=E_{1}+\cdots+E_{r}\) Prove that \(V=\operatorname{Im} E_{1} \oplus \cdots \oplus \operatorname{Im} E_{r}\)

Suppose \(A\) is a supertriangular matrix (i.e., all entries on and below the main diagonal are 0 ). Show that \(A\) is nilpotent.

Prove Lemma 10.13: Let \(T: V \rightarrow V\) be a linear operator whose minimal polynomial is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial. Then \(V\) is the direct sum of \(T\) -cyclic subspaces \(Z_{i}=Z\left(v_{i}, T\right), i=1, \ldots, r,\) with corresponding \(T\) -annihilators \\[ f(t)^{n_{1}}, f(t)^{n_{2}}, \ldots, f(t)^{n_{r}}, \quad n=n_{1} \geq n_{2} \geq \cdots \geq n_{r} \\]
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