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Chapter 10: Problem 36

Suppose \(W\) is invariant under \(T: V \rightarrow V\). Show that \(W\) is invariant under \(f(T)\) for any polynomial \(f(t)\)

Short Answer

Expert verified
To show that the subspace \(W\) is invariant under any polynomial \(f(T)\) for the linear transformation \(T: V \rightarrow V\), we consider an arbitrary vector \(w \in W\). We express \(f(T)\) as a linear combination of powers of \(T\), i.e., \(f(T)w = a_0w + a_1(Tw) + a_2(T^2w) + \dots + a_n(T^nw)\). Since \(W\) is invariant under \(T\), we know that \(Tw, T^2w, \dots, T^nw \in W\). As \(W\) is closed under scalar multiplication and vector addition, any linear combination of these vectors, like \(f(T)w\), must also be in \(W\). Therefore, \(W\) is invariant under \(f(T)\).

Step by step solution

01

Recall the definition of an invariant subspace

A subspace \(W\) of a vector space \(V\) is called invariant under the linear transformation \(T: V \rightarrow V\) if for any vector \(w \in W\), the transformed vector \(Tw \in W\) as well.
02

Define a polynomial function of the linear transformation

Given a polynomial \(f(t) = a_0 + a_1t + a_2t^2 + \dots + a_nt^n\), where \(a_i\) are scalar coefficients, the polynomial function \(f(T)\) of a linear transformation \(T\) is defined as: \[f(T) = a_0I + a_1T + a_2T^2 + \dots + a_nT^n\] where \(I\) is the identity transformation and \(T^k\) (\(k=2,3,\dots,n\)) represents applying the transformation \(T\) k times consecutively.
03

Show that \(W\) is invariant under \(f(T)\)

To show that \(W\) is invariant under \(f(T)\), we need to show that for any vector \(w \in W\), its image under \(f(T)\) is still in \(W\). Let's consider an arbitrary vector \(w \in W\). We have, \[ f(T)w = (a_0I + a_1T + a_2T^2 + \dots + a_nT^n)w \] Now, apply each term of the polynomial to \(w\): \[ f(T)w = a_0(Iw) + a_1(Tw) + a_2(T^2w) + \dots + a_n(T^nw) \] Since \(w\) is in \(W\), and \(W\) is invariant under \(T\), we know that \(Tw \in W\). Since \(W\) is a subspace and closed under scalar multiplication and addition, any power of \(T\) applied to \(w\) will also be in \(W\), i.e., \(T^kw \in W\) for \(k=1,2,\dots,n\). Additionally, \(Iw = w \in W\). Therefore, any linear combination of the resulting vectors (with scalar coefficients \(a_i\)) will still be in \(W\), i.e., \[ f(T)w = a_0w + a_1(Tw) + a_2(T^2w) + \dots + a_n(T^nw) \in W \] Since \(f(T)w \in W\) for any arbitrary vector \(w \in W\), this implies that \(W\) is invariant under \(f(T)\).

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Let \(T: V \rightarrow V\) be linear. Let \(W\) be a \(T\) -invariant subspace of \(V\) and \(\bar{T}\) the induced operator on \(V / W\). Prove (a) The T-annihilator of \(v \in V\) divides the minimal polynomial of \(T\) (b) The \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) divides the minimal polynomial of \(T\) (a) The \(T\) -annihilator of \(v \in V\) is the minimal polynomial of the restriction of \(T\) to \(Z(v, T) ;\) therefore, by Problem \(10.6,\) it divides the minimal polynomial of \(T\) (b) The \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) divides the minimal polynomial of \(\bar{T},\) which divides the minimal polynomial of \(T\) by Theorem 10.16 Remark: In the case where the minimum polynomial of \(T\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial, then the \(T\) -annihilator of \(v \in V\) and the \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) are of the form \(f(t)^{m},\) where \(m \leq n\).
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