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Chapter 10: Problem 40

Let \(V\) be a vector space of odd dimension (greater than 1 ) over the real field \(\mathbf{R}\). Show that any linear operator on \(V\) has an invariant subspace other than \(V\) or \(\\{0\\}\).

Short Answer

Expert verified
Given a vector space \(V\) of odd dimension over the real field \(\mathbf{R}\), any linear operator \(T\) on \(V\) has an invariant subspace other than \(V\) itself or the zero subspace. This can be demonstrated by finding the characteristic polynomial of \(T\), identifying a real eigenvalue \(\lambda_0\) using the fundamental theorem of algebra, and finding its corresponding eigenvector \(v\). The invariant subspace is then given by \(W=\{kv:k\in\mathbf{R}\}\), which contains only scalar multiples of the eigenvector \(v\) and is distinct from both \(V\) and the zero subspace.

Step by step solution

01

Understanding the Problem Statement

We are given a vector space \(V\) of odd dimension over the real field \(\mathbf{R}\). We need to show that any linear operator \(T\) on \(V\) has an invariant subspace other than \(V\) itself or the zero subspace. Invariant subspace: A subspace \(W\) of \(V\) is called invariant under the linear operator \(T\) if for any vector \(v\) belonging to \(W\), \(T(v)\) also belongs to \(W\).
02

Find the Characteristic Polynomial

To find an invariant subspace, we need to find an eigenvector belonging to the linear operator \(T\). In order to do that, we need to find the eigenvalues of the linear operator. Let's begin by finding the characteristic polynomial of the linear operator T, \(p_T(\lambda) = \det (T-\lambda I)\), where \(\lambda\) are the eigenvalues of \(T\) and \(I\) is the identity matrix in \(V\).
03

Realize the Fundamental theorem of algebra

The fundamental theorem of algebra tells us that every non-constant polynomial equation with real or complex coefficients has at least one complex root. Since the characteristic polynomial \(p_T(\lambda)\) is a polynomial of odd degree (the dimension of \(V\) is odd), it must also have an odd number of roots. Therefore, it must have at least one real root. Let \(\lambda_0\) be one of these real roots.
04

Find an Eigenvector

Now that we have a real eigenvalue \(\lambda_0\), let's find the eigenvector associated with this eigenvalue. To find the eigenvector, we need to solve the following equation: $$(T-\lambda_0 I)v = 0$$ Where \(v\) is the eigenvector we are looking for.
05

Find the Invariant Subspace

Once we have found the eigenvector \(v\) corresponding to the eigenvalue \(\lambda_0\), we can define the invariant subspace \(W = \{kv : k \in \mathbf{R}\}\). Notice that any vector \(w\) in \(W\) can be written as \(w=kv\) for some scalar \(k\). Thus: $$T(w)=T(kv)=kT(v)=k\lambda_0 v = \lambda_0 w$$ So, \(T(w)\) is also in the subspace \(W\), which means that \(W\) is an invariant subspace under the linear operator \(T\). Note that \(W\) is neither equal to \(V\) nor equal to the zero subspace, since it contains only the scalar multiples of the eigenvector \(v\) and it is not zero. By following these steps, we have shown that any linear operator on a vector space of odd dimension greater than 1 over the real field \(\mathbf{R}\) has an invariant subspace other than the vector space itself and the zero subspace.

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Most popular questions from this chapter

Let \(T: V \rightarrow V^{\prime}\) be linear with kernel \(W\) and image \(U .\) Show that the quotient space \(V / W\) is isomorphic to \(U\) under the mapping \(\theta: V / W \rightarrow U\) defined by \(\theta(v+W)=T(v) .\) Furthermore, show that \(T=i \circ \theta \circ \eta,\) where \(\eta: V \rightarrow V / W\) is the natural mapping of \(V\) into \(V / W\) (i.e., \(\eta(v)=v+W\) ), and \(i: U \hookrightarrow V^{\prime}\) is the inclusion mapping (i.e., \(i(u)=u\) ). (See diagram.)

Suppose \(T_{1}\) and \(T_{2}\) are nilpotent operators that commute (i.e., \(T_{1} T_{2}=T_{2} T_{1}\) ). Show that \(T_{1}+T_{2}\) and \(T_{1} T_{2}\) are also nilpotent.

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

Find all invariant subspaces of \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -2\end{array}\right]\) viewed as an operator on \(\mathbf{R}^{2}\) By Problem \(10.1, \mathbf{R}^{2}\) and \\{0\\} are invariant under \(A .\) Now if \(A\) has any other invariant subspace, it must be one-dimensional. However, the characteristic polynomial of \(A\) is \\[ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}+1 \\] Hence, \(A\) has no eigenvalues (in \(\mathbf{R}\) ) and so \(A\) has no eigenvectors. But the one-dimensional invariant subspaces correspond to the eigenvectors; thus, \(\mathbf{R}^{2}\) and \\{0\\} are the only subspaces invariant under \(A\).

Determine all possible Jordan canonical forms \(J\) for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{5}\) and whose minimal polynomial \(m(t)=(t-2)^{2}\) \(J\) must be a \(5 \times 5\) matrix, because \(\Delta(t)\) has degree \(5,\) and all diagonal elements must be \(2,\) because 2 is the only eigenvalue. Moreover, because the exponent of \(t-2\) in \(m(t)\) is \(2, J\) must have one Jordan block of order \(2,\) and the others must be of order 2 or \(1 .\) Thus, there are only two possibilities: \\[ J=\operatorname{diag}\left(\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{l} 2 \end{array}\right]\right) \quad \text { or } \quad J=\operatorname{diag}\left(\left[\begin{array}{ll} 2 & 1 \\ 2 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \end{array}\right]\right) \\]
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