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Chapter 7: Problem 3

For each linear operator \(T\) on \(V\), find the minimal polynomial of \(T\). (a) \(\mathbf{V}=\mathbf{R}^{2}\) and \(\mathrm{T}(a, b)=(a+b, a-b)\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and \(\mathrm{T}(g(x))=g^{\prime}(x)+2 g(x)\) (c) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and $\mathrm{T}(f(x))=-x f^{\prime \prime}(x)+f^{\prime}(x)+2 f(x)$ (d) \(\mathrm{V}=\mathrm{M}_{n \times n}(R)\) and \(\mathrm{T}(A)=A^{t}\). Hint: Note that \(\mathrm{T}^{2}=1\).

Short Answer

Expert verified
(a) The minimal polynomial for T is \( m_T(x) = x-1 \). (b) The minimal polynomial for T is \( m_T(x) = (x - 2)^2 \). (c) The minimal polynomial for T is \( m_T(x) = (x - 1)(x + 1) \). (d) The minimal polynomial for T is \( m_T(x) = (x - 1)(x + 1) \).

Step by step solution

01

Expression of T in matrix form

For this problem, we have the linear operator T on the vector space V=R^2, given by T(a, b) = (a+b, a-b). We can transform T into a matrix form by considering the following equation: \( T(v) = Av \), where A is a matrix, and (a, b) is a vector. This means that \( (a+b, a-b) = A(a,b)=\begin{pmatrix} \;a_1&b_1\\a_2&b_2 \end{pmatrix}\begin{pmatrix} a\\b \end{pmatrix}\). Thus, A=\(\begin{pmatrix} 1&1\\1&-1 \end{pmatrix}\), which represents the linear operator in the given form.
02

Finding the minimal polynomial

Now, we need to find the minimal polynomial \( m_T(x) \) for T. We start by trying a linear operator: \( m_T(x) = xI-(A) \), where I is the identity matrix. This will give us \( m_T(x) = \begin{pmatrix} x-1&-1\\-1&x+1 \end{pmatrix}\). We want this matrix to be equal to the zero matrix, so \( x-1 = x+1 = -1 = 0 \). The only solution for this is x = 1. Therefore, the minimal polynomial for T is \( m_T(x) = x-1 \). (b)
03

Expression of T in matrix form

For this problem, we have the linear operator T on the vector space V=P2(R), given by T(g(x))=g'(x)+2g(x). To find the matrix form, we will use the standard basis for P2(R), which is \( (1, x, x^2) \). We can then apply T to each of the basis vectors to find the columns of the A matrix. \( T(1) = T' + 2(1) = 0 + 2 = 2 \newline T(x) = (x') + 2(x) = 1 + 2x \newline T(x^2) = (x^2)' + 2(x^2) = 2x + 2x^2 \) Now forming matrix A using these new transformations as columns: \[ A = \begin{pmatrix} 2&0&0\\0&2&2\\0&0&2 \end{pmatrix} \]
04

Finding the minimal polynomial

To find the minimal polynomial \( m_T(x) \) for T, we will first go for a quadratic polynomial: \( m_T(x) = x^2 - (\lambda_1+\lambda_2)x + \lambda_1\lambda_2 \), where 位1 and 位2 are the eigenvalues of A. A has an eigenvalue 位 = 2 with algebraic multiplicity 3. Plugging the eigenvalue into the polynomial expression, we get \( m_T(x) = (x - 2)(x - 2)\). Therefore, the minimal polynomial for T is \( m_T(x) = (x - 2)^2 \). (c)
05

Expression of T in matrix form

Proceeding similarly as in part b, the matrix form for part c is: \[ A = \begin{pmatrix} 0&1&0\\2&0&2\\0&0&-1 \end{pmatrix} \]
06

Finding the minimal polynomial

The matrix A has eigenvalues 1, -1, and -1. Thus, the polynomial we will try is \( m_T(x) = (x - 1)(x + 1)\), which gives: \[ m_T(x) = \begin{pmatrix} 1&-1&0\\-2&1&-2\\0&0&2 \end{pmatrix} \] Since \(m_T(x) = (x - 1)(x + 1)\) annihilates A, the minimal polynomial for T is \( m_T(x) = (x - 1)(x + 1) \). (d)
07

Matrix form not required

In this part, we are given that \(T^2 = \mathrm{I}\), where I is the identity matrix. This means that T is its own inverse.
08

Finding the minimal polynomial

Since T is its own inverse, we can conclude that the minimal polynomial for T is \( m_T(x) = x^2 - 1 = (x - 1)(x + 1) \) because it annihilates T.

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Most popular questions from this chapter

Find a basis for the subspace \(W\) of \(\mathbf{R}^{5}\) orthogonal to the vectors \(u_{1}=(1,1,3,4,1)\) and \(u_{2}=(1,2,1,2,1)\)

Prove Theorem 7.8: Suppose \(w_{1}, w_{2}, \ldots, w_{r}\) form an orthogonal set of nonzero vectors in \(V .\) Let \(v \in V\). Define \\[v^{\prime}=v-\left(c_{1} w_{1}+c_{2} w_{2}+\cdots+c_{r} w_{r}\right), \quad \text { where } \quad c_{i}=\frac{\left\langle v, w_{i}\right\rangle}{\left\langle w_{i}, w_{i}\right\rangle}\\] Then \(v^{\prime}\) is orthogonal to \(w_{1}, w_{2}, \dots, w_{r}\) For \(i=1,2, \ldots, r\) and using \(\left\langle w_{i}, w_{j}\right\rangle=0\) for \(i \neq j,\) we have \\[\begin{aligned} \left\langle v-c_{1} w_{1}-c_{2} x_{2}-\cdots-c_{r} w_{r}, w_{i}\right\rangle &=\left\langle v, w_{i}\right\rangle-c_{1}\left\langle w_{1}, w_{i}\right\rangle-\cdots-c_{i}\left\langle w_{i}, w_{i}\right\rangle-\cdots- c_{r}\left\langle w_{r}, w_{i}\right\rangle \\ &=\left\langle v, w_{i}\right\rangle-c_{1} \cdot 0-\cdots-c_{i}\left\langle w_{i}, w_{i}\right\rangle-\cdots-c_{r} \cdot 0 \\ &=\left\langle v, w_{i}\right\rangle-c_{i}\left\langle w_{i}, w_{i}\right\rangle=\left\langle v, w_{i}\right\rangle-\frac{\left\langle v, w_{i}\right\rangle}{\left\langle w_{i}, w_{i}\right\rangle}\left\langle w_{i}, w_{i}\right\rangle=0 \end{aligned}\\] The theorem is proved.

Label the following statements as true or false. (a) Eigenvectors of a linear operator \(\mathrm{T}\) are also generalized eigenvectors of \(\mathrm{T}\). (b) It is possible for a generalized eigenvector of a linear operator \(\mathrm{T}\) to correspond to a scalar that is not an eigenvalue of \(\mathrm{T}\). (c) Any linear operator on a finite-dimensional vector space has a Jordan canonical form. (d) A cycle of generalized eigenvectors is linearly independent. (e) There is exactly one cycle of generalized eigenvectors corresponding to each eigenvalue of a linear operator on a finite-dimensional vector space. (f) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\( be the distinct eigenvalues of \)\mathrm{T}$. If, for each \(i, \beta_{i}\) is a basis for \(\mathrm{K}_{\lambda_{i}}\), then $\beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}$ is a Jordan canonical basis for \(\mathrm{T}\). (g) For any Jordan block \(J\), the operator \(L_{J}\) has Jordan canonical form \(J\). (h) Let \(\mathrm{T}\) be a linear operator on an \(n\)-dimensional vector space whose characteristic polynomial splits. Then, for any eigenvalue \(\lambda\) of \(\mathrm{T}\), $\mathrm{K}_{\lambda}=\mathrm{N}\left((\mathrm{T}-\lambda \mathrm{I})^{n}\right)$.

Find the Fourier coefficient \(c\) and the projection of \(v=(1,-2,3,-4)\) along \(w=(1,2,1,2)\) in \(\mathbf{R}^{4}\) Compute \(\langle v, w\rangle=1-4+3-8=-8\) and \(\|w\|^{2}=1+4+1+4=10 .\) Then \\[c=-\frac{8}{10}=-\frac{4}{5} \quad \text { and } \quad \operatorname{proj}(v, w)=c w=\left(-\frac{4}{5},-\frac{8}{5},-\frac{4}{5},-\frac{8}{5}\right)\\]

Let \(\gamma_{1}, \gamma_{2}, \ldots, \gamma_{p}\) be cycles of generalized eigenvectors of a linear operator \(\mathrm{T}\) corresponding to an eigenvalue \(\lambda\). Prove that if the initial eigenvectors are distinct, then the cycles are disjoint.
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