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Chapter 10: Problem 20

Determine all possible Jordan canonical forms for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{3}(t-5)^{2} .\) In each case, find the minimal polynomial \(m(t)\) Because \(t-2\) has exponent 3 in \(\Delta(t), 2\) must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad[5], \quad[5]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)\) (d) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5\end{array}\right], \quad[5]\right)\) (e) \(\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\\ 5\end{array}\right]\right)\) (f) \(\operatorname{diag}([2],[2],[2],[5],[5])\) The exponent in the minimal polynomial \(m(t)\) is equal to the size of the largest block. Thus, (a) \(m(t)=(t-2)^{3}(t-5)^{2}\) (b) \(m(t)=(t-2)^{3}(t-5)\) (c) \(\quad m(t)=(t-2)^{2}(t-5)^{2}\) (d) \(m(t)=(t-2)^{2}(t-5)\) (e) \(m(t)=(t-2)(t-5)^{2}\) \((\mathrm{f}) \quad m(t)=(t-2)(t-5)\)

Short Answer

Expert verified
The possible Jordan canonical forms are: (a) $\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\ & & 2\end{array}\right], \quad\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)$ with $m(t) = (t-2)^{3}(t-5)^{2}$ (b) $\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\ & & 2\end{array}\right], \quad[5], \quad[5]\right)$ with $m(t) = (t-2)^{3}(t-5)$ (c) $\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)$ with $m(t) = (t-2)^{2}(t-5)^{2}$ (d) $\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5\end{array}\right], \quad[5]\right)$ with $m(t) = (t-2)^{2}(t-5)$ (e) $\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)$ with $m(t) = (t-2)(t-5)^{2}$ (f) $\operatorname{diag}([2],[2],[2],[5],[5])$ with $m(t) = (t-2)(t-5)$

Step by step solution

01

Analyze the characteristic polynomial

Given the characteristic polynomial: \[\Delta(t) = (t-2)^{3}(t-5)^{2}\] The eigenvalues are 2 and 5, with respective multiplicities 3 and 2.
02

Construct possible Jordan canonical forms

Using the multiplicities of the eigenvalues, we can infer the sizes of the Jordan blocks. The possible Jordan canonical forms are listed below: (a) \[\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\ & & 2\end{array}\right], \quad\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\] (b) \[\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\ & & 2\end{array}\right], \quad[5], \quad[5]\right)\] (c) \[\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)\] (d) \[\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5\end{array}\right], \quad[5]\right)\] (e) \[\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)\] (f) \[\operatorname{diag}([2],[2],[2],[5],[5])\]
03

Find the minimal polynomial for each case

The exponent in the minimal polynomial m(t) is equal to the size of the largest block for each eigenvalue. Thus, (a) \(m(t) = (t-2)^{3}(t-5)^{2}\) (b) \(m(t) = (t-2)^{3}(t-5)\) (c) \(m(t) = (t-2)^{2}(t-5)^{2}\) (d) \(m(t) = (t-2)^{2}(t-5)\) (e) \(m(t) = (t-2)(t-5)^{2}\) (f) \(m(t) = (t-2)(t-5)\)

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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 fundamental concept in linear algebra, especially when dealing with matrix transformations. It is a polynomial that is derived from a square matrix, indicating crucial properties of the matrix. If we have a matrix \( A \), the characteristic polynomial \( \Delta(t) \) is given by:\[ \Delta(t) = \det(tI - A) \]where \( I \) is the identity matrix of the same size as \( A \) and \( \det \) denotes the determinant. The roots of the characteristic polynomial are known as the matrix's eigenvalues and provide insight into some of the matrix's intrinsic behavior. In the exercise, \( \Delta(t) = (t-2)^{3}(t-5)^{2} \) indicates that the eigenvalues are 2 and 5, with multiplicities of 3 and 2 respectively. These multiplicities determine the possible distributions of the Jordan blocks later.鈥�
  • The degree of the polynomial equals the size of the matrix \( A \).
  • It reveals the eigenvalues by setting the polynomial to zero and solving for \( t \).
  • The multiplicities of the eigenvalues in \( \Delta(t) \) reflect how many times each eigenvalue occurs as a root.
By understanding the characteristic polynomial, one can predict the basic structure of the Jordan canonical form and set the stage for determining the minimal polynomial.
Minimal Polynomial
The minimal polynomial is another essential polynomial associated with a matrix or a linear operator. It is the smallest degree monic polynomial that the matrix satisfies. This means if you take the linear operator \( T \) and substitute it into \( m(t) \), you should get the zero mapping:\[ m(T) = 0 \]In the exercise, the minimal polynomial for each configuration is identified by examining the largest size of the Jordan blocks formed by each eigenvalue. The degree of \( m(t) \) is dictated not only by the eigenvalues but also by the largest geometric multiplicity across Jordan blocks.
  • The minimal polynomial must annihilate the matrix; it cannot have any degree less than the largest Jordan block for any eigenvalue.
  • Unlike the characteristic polynomial, the minimal polynomial does not give all eigenvalues and their algebraic multiplicities directly.
  • It provides insights into the diagonalizability of the matrix: a matrix is diagonalizable if and only if its minimal polynomial contains only linear factors of degree 1.
In the problem, minimal polynomials range from \((t-2)(t-5)\) to \((t-2)^{3}(t-5)^{2}\), influenced by the configuration of Jordan blocks.
Eigenvalues
Eigenvalues are pivotal in understanding the behaviors of matrices and operators in linear transformations. They are values for which there exists a non-zero vector, known as an eigenvector, that when multiplied by the matrix \( A \) results in a scalar multiple of that vector. In simpler terms, these are scalars \( \lambda \) such that:\[ A \mathbf{v} = \lambda \mathbf{v} \]Where \( \mathbf{v} \) is the eigenvector corresponding to the eigenvalue \( \lambda \). Eigenvalues are determined as solutions to the characteristic equation formed by the characteristic polynomial.The exercise identifies 2 and 5 as eigenvalues with multiplicities 3 and 2. This detail about multiplicity helps determine the number of possible Jordan blocks.
  • They provide insight into the stability of systems and vibrational modes.
  • For a given eigenvalue, multiple eigenvectors can form a corresponding eigenspace.
  • They signify the stretch factor in an eigenvector's direction during a transformation.
The multiplicity of an eigenvalue is critical in determining Jordan forms, impacting both the characteristic and minimal polynomials.
Jordan Blocks
Jordan blocks are building blocks of a Jordan canonical form matrix, representing cycles of generalized eigenvectors corresponding to a single eigenvalue. Each Jordan block is of the form:\[\text{J} = \begin{bmatrix}\lambda & 1 & 0 & \dots & 0 \0 & \lambda & 1 & \dots & 0 \\vdots & \ddots & \ddots & \ddots & \vdots \0 & \dots & 0 & \lambda & 1 \0 & \dots & 0 & 0 & \lambda \\end{bmatrix}\]Each \( \lambda \) represents an eigenvalue, and the dimensions of the block correspond to the reach of that eigenvalue's influence over the vector space.
  • Jordan blocks visually and structurally prepare a matrix into a form that's easier to understand.
  • The largest size of a Jordan block for an eigenvalue dictates the highest power of that factor in the minimal polynomial.
  • Multiple Jordan blocks for the same eigenvalue are possible, reflecting the geometric multiplicity.
In the exercise, each proposed Jordan canonical form depicts different configurations of Jordan blocks based on the eigenvalues' multiplicities, thus impacting the shape of the minimal polynomial. Understanding Jordan blocks is crucial to grasping the intricacies of the Solvability in linear differential equation contexts and simplifying matrix functions.

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Most popular questions from this chapter

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) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],[3],[3]\right)\)

Suppose \(W\) is a subspace of a vector space \(V .\) Show that the operations in Theorem 10.15 are well defined; namely, show that if \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W,\) then (a) \(\quad(u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W\) and (b) \(\quad k u+W=k u^{\prime}+W \quad\) for any \(k \in K\) (a) Because \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W,\) both \(u-u^{\prime}\) and \(v-v^{\prime}\) belong to \(W .\) But then \((u+v)-\left(u^{\prime}+v^{\prime}\right)=\left(u-u^{\prime}\right)+\left(v-v^{\prime}\right) \in W .\) Hence, \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W\) (b) Also, because \(u-u^{\prime} \in W\) implies \(k\left(u-u^{\prime}\right) \in W,\) then \(k u-k u^{\prime}=k\left(u-u^{\prime}\right) \in W ;\) accordingly, \(k u+W=k u^{\prime}+W\).

Suppose \(T: V \rightarrow V\) is linear. Show that each of the following is invariant under \(T\) (a) \\{0\\} (b) \(V\) (c) kernel of \(T\) (d) image of \(T\) (a) We have \(T(0)=0 \in\\{0\\} ;\) hence, \\{0\\} is invariant under \(T\) (b) For every \(v \in V, T(v) \in V ;\) hence, \(V\) is invariant under \(T\) (c) Let \(u \in\) Ker \(T\). Then \(T(u)=0 \in\) Ker \(T\) because the kemel of \(T\) is a subspace of \(V\). Thus, Ker \(T\) is invariant under \(T\) (d) Because \(T(v) \in \operatorname{Im} T\) for every \(v \in V\), it is certainly true when \(v \in \operatorname{Im} T .\) Hence, the image of \(T\) is invariant under \(T\)

Find the rational canonical form for the four-square Jordan block with \(\lambda^{\prime}\) s on the diagonal.

Prove Theorem 10.1: Let \(T: V \rightarrow V\) be a linear operator whose characteristic polynomial factors into linear polynomials. Then \(V\) has a basis in which \(T\) is represented by a triangular matrix. The proof is by induction on the dimension of \(V\). If \(\operatorname{dim} V=1\), then every matrix representation of \(T\) is a \(1 \times 1\) matrix, which is triangular. Now suppose \(\operatorname{dim} V=n>1\) and that the theorem holds for spaces of dimension less than \(n\). Because the characteristic polynomial of \(T\) factors into linear polynomials, \(T\) has at least one eigenvalue and so at least one nonzero eigenvector \(v,\) say \(T(v)=a_{11} v .\) Let \(W\) be the one- dimensional subspace spanned by \(v\) Set \(\bar{V}=V / W .\) Then (Problem 10.26 ) \(\operatorname{dim} \bar{V}=\operatorname{dim} V-\operatorname{dim} W=n-1 .\) Note also that \(W\) is invariant under \(T .\) By Theorem \(10.16, T\) induces a linear operator \(T\) on \(V\) whose minimal polynomial divides the minimal polynomial of \(T .\) Because the characteristic polynomial of \(T\) is a product of linear polynomials, so is its minimal polynomial, and hence, so are the minimal and characteristic polynomials of \(\bar{T}\). Thus, \(\bar{V}\) and \(\bar{T}\) satisfy the hypothesis of the theorem. Hence, by induction, there exists a basis \(\left\\{\bar{v}_{2}, \ldots, \bar{v}_{n}\right\\}\) of \(\bar{V}\) such that \\[ \begin{array}{l} \bar{T}\left(\bar{v}_{2}\right)=a_{22} \bar{v}_{2} \\ \bar{T}\left(\bar{v}_{3}\right)=a_{32} \bar{v}_{2}+a_{33} \bar{v}_{3} \\ \bar{T}\left(\bar{v}_{n}\right)=a_{n 2} \bar{v}_{n}+a_{n 3} \bar{v}_{3}+\cdots+a_{n n} \bar{v}_{n} \end{array} \\] Now let \(v_{2}, \ldots, v_{n}\) be elements of \(V\) that belong to the cosets \(v_{2}, \ldots, v_{n},\) respectively. Then \(\left\\{v, v_{2}, \ldots, v_{n}\right\\}\) is a basis of \(V\) (Problem 10.26 ). Because \(\bar{T}\left(v_{2}\right)=a_{22} \bar{v}_{2},\) we have \\[ \bar{T}\left(\bar{v}_{2}\right)-a_{22} \bar{v}_{22}=0, \quad \text { and so } \quad T\left(v_{2}\right)-a_{22} v_{2} \in W \\] But \(W\) is spanned by \(v ;\) hence, \(T\left(v_{2}\right)-a_{22} v_{2}\) is a multiple of \(v,\) say, \\[ T\left(v_{2}\right)-a_{22} v_{2}=a_{21} v, \quad \text { and } \\] so \(\quad T\left(v_{2}\right)=a_{21} v+a_{22} v_{2}\) Similarly, for \(i=3, \ldots, n\) \\[ T\left(v_{i}\right)-a_{i 2} v_{2}-a_{i 3} v_{3}-\cdots-a_{i i} v_{i} \in W, \quad \text { and so } \quad T\left(v_{i}\right)=a_{i 1} v+a_{i 2} v_{2}+\cdots+a_{i i} v_{i} \\] Thus, \\[ T(v)=a_{11} v \\] \\[ \begin{array}{l} T\left(v_{2}\right)=a_{21} v+a_{22} v_{2} \\ T\left(v_{n}\right)=a_{n 1} v+a_{n 2} v_{2}+\cdots+a_{n n} v_{n} \end{array} \\] and hence the matrix of \(T\) in this basis is triangular.
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