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Chapter 10: Problem 32

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

Short Answer

Expert verified
The possible rational canonical forms of the linear operator T are: (a) M = 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) M = 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) M = diag $\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],[3],[3]\right)$

Step by step solution

01

Find the Characteristic Polynomial (螖(t))

First, we know that there are two possible characteristic polynomials: \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)^{2}(t-3)^{3}\) or \(\Delta_{2}(t)=\left(t^{2}-2 t+5\right)(t-3)^{5}\). We can't tell which 螖(t) belongs to M, so we have to consider both cases.
02

Finding possible rational canonical forms (M) of T for each 螖(t)

The rational canonical form (M) of an operator is a matrix consisting of companion matrices relative to the invariant factors; it appears as a block diagonal matrix. We need to find companion matrices for factors of the polynomials, which are \(C\left(t^{2}-2 t+5\right)\) and \(C\left((t-3)^{3}\right)\), and arrange the block diagonal matrices in a way that the sum of the orders adds up to 7 (the dimension of V). Considering both 螖1(t) and 螖2(t), we can get the following possible rational canonical forms: (a) M = 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) M = 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) M = diag $\left(\left[\begin{array}{rr}0 & -5 \\\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\\ 1 & 0 & -27 \\\ 0 & 1 & 9\end{array}\right],[3],[3]\right)$ So, M can be any one of these matrices (a), (b), and (c).

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics that deals with vector spaces and linear mappings between these spaces. It involves concepts like vectors, matrices, determinants, and systems of linear equations. The subject is foundational in a variety of fields including science, engineering, technology, and social sciences, as it provides a way to model and solve complex problems with linear systems.

In our exercise, we are working in a seven-dimensional vector space, which implies dealing with seven independent vectors making up the space. Linear algebra provides us tools to handle linear transformations in this kind of multi-dimensional space.
Linear Operator
A linear operator is a mapping from one vector space to itself, or to another vector space, that preserves the operations of vector addition and scalar multiplication. In formal terms, if we denote a linear operator by \(T\), it must satisfy \(T(u + v) = T(u) + T(v)\) and \(T(c\cdot v) = c\cdot T(v)\), where \(u\) and \(v\) are vectors, and \(c\) is a scalar.

In the given problem, \(T : V \rightarrow V\) is such a linear operator operating in a seven-dimensional space. Our goal is to understand the structure of this operator through its rational canonical form, which neatly summarizes the action of \(T\) on the space \(V\).
Minimal Polynomial
The minimal polynomial of a linear operator is the monic polynomial of smallest degree for which the operator is a root. Essentially, it captures the simplest algebraic relationship that the operator satisfies. For an operator \(T\), if \(m(t)\) is the minimal polynomial, then \(m(T) = 0\), where \(m(T)\) denotes applying the polynomial \(m\) to the operator \(T\).

The exercise specifies that the minimal polynomial of \(T\) is \(m(t) = (t^2 - 2t + 5)(t - 3)^3\). This indicates the operator \(T\) satisfies this relationship, and it guides the possible structure of the rational canonical forms since the factors of the minimal polynomial will form the blocks within the rational canonical form matrix.
Characteristic Polynomial
The characteristic polynomial of a linear operator or matrix is a polynomial which provides important information about the matrix. It is defined as \(\det(T - tI)\), where \(\det\) denotes the determinant, \(T\) is the matrix representing our linear operator, \(t\) is a variable, and \(I\) is the identity matrix of the same dimension as \(T\).

The characteristic polynomial tells us about the eigenvalues of the matrix, which are values of \(t\) for which there exist non-zero vectors \(x\) such that \(T(x) = t\cdot x\). In our problem, the characteristic polynomial helps us in figuring out the potential forms that our rational canonical form matrices could take.
Companion Matrix
The companion matrix is a special type of square matrix that serves as a canonical form for a given monic polynomial. If the polynomial is \(p(t) = t^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0\), the companion matrix of \(p(t)\) is:

\[\begin{array}{cccc}0 & 0 & \dots & -a_0 \1 & 0 & \dots & -a_1 \0 & 1 & \dots & -a_2 \vdots & \vdots & \ddots & \vdots \0 & 0 & \dots & -a_{n-1}\end{array}\]


For the linear operator \(T\) mentioned in our exercise, the companion matrices derived from the minimal polynomial will be the building blocks for the rational canonical forms of \(T\). By assembling companion matrices corresponding to the factors of the minimal polynomial in block-diagonal form, we can reveal the essential structural information about the linear operator.

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Most popular questions from this chapter

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

Let \(W=Z(v, T),\) and suppose the \(T\) -annihilator of \(v\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial of degree \(d .\) Show that \(f(T)^{s}(W)\) is a cyclic subspace generated by \(f(T)^{s}(v)\) and that it has dimension \(d(n-s)\) if \(n>s\) and dimension 0 if \(n \leq s\).

The subspaces \(W_{1}, \ldots, W_{r}\) are said to be independent if \(w_{1}+\cdots+w_{r}=0, w_{i} \in W_{i},\) implies that each \(w_{i}=0 .\) Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if the \(W_{i}\) are independent. [Here \(\operatorname{span}\left(W_{i}\right)\) denotes the linear span of the \(\left.W_{i} .\right]\)

Let \(V\) be a vector space and \(W\) a subspace of \(V .\) Show that the natural map \(\eta: V \rightarrow V / W\), defined by \(\eta(v)=v+W,\) is linear. For any \(u, v \in V\) and any \(k \in K,\) we have \\[ n(u+v)=u+v+W=u+W+v+W=\eta(u)+\eta(v) \\] and \\[ \eta(k v)=k v+W=k(v+W)=k \eta(v) \\] Accordingly, \(\eta\) is linear..

Prove Theorem 10.4: Suppose \(W_{1}, W_{2}, \ldots, W_{r}\) are subspaces of \(V\) with respective bases \\[ B_{1}=\left\\{w_{11}, w_{12}, \ldots, w_{1 n_{1}}\right\\}, \quad \ldots, \quad B_{r}=\left\\{w_{r 1}, w_{r 2}, \ldots, w_{r n_{r}}\right\\} \\] Then \(V\) is the direct sum of the \(W_{i}\) if and only if the union \(B=\bigcup_{i} B_{i}\) is a basis of \(V\) Suppose \(B\) is a basis of \(V\). Then, for any \(v \in V\) \\[ v=a_{11} w_{11}+\cdots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{m_{r}} w_{m_{r}}=w_{1}+w_{2}+\cdots+w_{r} \\] where \(w_{i}=a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i} .\) We next show that such a sum is unique. Suppose \\[ v=w_{1}^{\prime}+w_{2}^{\prime}+\cdots+w_{r}^{\prime}, \quad \text { where } \quad w_{i}^{\prime} \in W_{i} \\] Because \(\left\\{w_{i 1}, \ldots, w_{i n}\right\\}\) is a basis of \(W_{i}, w_{i}^{\prime}=b_{i 1} w_{i 1}+\cdots+b_{i n_{i}} w_{i n_{i}},\) and so \\[ v=b_{11} w_{11}+\dots+b_{1 n_{1}} w_{1 n_{1}}+\cdots+b_{r 1} w_{r 1}+\cdots+b_{m_{r}} w_{r m_{r}} \\] Because \(B\) is a basis of \(V, a_{i j}=b_{i j},\) for each \(i\) and each \(j .\) Hence, \(w_{i}=w_{i}^{\prime},\) and so the sum for \(v\) is unique. Accordingly, \(V\) is the direct sum of the \(W_{i}\) Conversely, suppose \(V\) is the direct sum of the \(W_{i}\). Then for any \(v \in V, v=w_{1}+\cdots+w_{r},\) where \(w_{i} \in W_{i} .\) Because \(\left\\{w_{i j_{i}}\right\\}\) is a basis of \(W_{i},\) each \(w_{i}\) is a linear combination of the \(w_{i j},\) and so \(v\) is a linear combination of the elements of \(B\). Thus, \(B\) spans \(V\). We now show that \(B\) is linearly independent. Suppose \\[ a_{11} w_{11}+\dots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{r n_{r}} w_{r n_{r}}=0 \\] Note that \(a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i} .\) We also have that \(0=0+0 \cdots 0 \in W_{i} .\) Because such a sum for 0 is unique, \\[ a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}}=0 \quad \text { for } i=1, \ldots, r \\] The independence of the bases \(\left\\{w_{i j}\right\\}\) implies that all the \(a\) 's are \(0 .\) Thus, \(B\) is linearly independent and is a basis of \(V\)
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