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Chapter 2: Problem 17

Let \(T\) be a linear operator on \(V\), and let \(W\) be a subspace of \(V\). Prove that \(\mathrm{W}\) is T-invariant (as defined in the exercises of Section 2.1) if and only if \(\mathrm{W}^{0}\) is \(\mathrm{T}^{t}\)-invariant.

Short Answer

Expert verified
In conclusion, we proved that a subspace W is T-invariant if and only if its annihilator \(W^0\) is \(T^t\)-invariant by showing that: 1. If W is T-invariant, then for every vector \(w^0\) in \(W^0\), we have \(T^t(w^0) \in W^0\). 2. If \(W^0\) is \(T^t\)-invariant, then for every vector w in W, we have \(Tw \in W\). These results hold for arbitrary elements in their respective subspaces, confirming the equivalence between W being T-invariant and \(W^0\) being \(T^t\)-invariant.

Step by step solution

01

If W is T-invariant, then \(W^0\) is \(T^t\)-invariant.

Assume that W is T-invariant. This means that for every vector w in W, we have \(Tw \in W\). We need to show that \(W^0\) is \(T^t\)-invariant, which means for any vector \(w^0\) in \(W^0\), we have \(T^t(w^0) \in W^0\). Let \(w^0\) be an arbitrary element in \(W^0\). By definition of the annihilator, for every vector w in W, we have: \[\langle w^0, w \rangle = 0\] Now, we want to show that \(\langle T^t(w^0), w \rangle = 0\) for all w in W. This is the same as showing \(\langle w^0, Tw \rangle = 0\), since \(T^t(w^0)\) is the vector such that, for all w in W, we have \(\langle T^t(w^0), w \rangle = \langle w^0, Tw \rangle\). By the definition of W being T-invariant and our assumption, \(Tw \in W\). So, \(\langle w^0, Tw \rangle = 0\) because \(w^0\) annihilates every vector in W. Hence, \(T^t(w^0) \in W^0 \) for all \(w^0\) in \(W^0\). Therefore, if W is T-invariant, then \(W^0\) is \(T^t\)-invariant.
02

If \(W^0\) is \(T^t\)-invariant, then W is T-invariant.

Assume that \(W^0\) is \(T^t\)-invariant. This means that for every vector \(w^0\) in \(W^0\), we have \(T^t(w^0) \in W^0\). We need to show that W is T-invariant, which means for any vector w in W, we have \(Tw \in W\). Let w be an arbitrary element in W, and let \(w^0\) be an arbitrary element in \(W^0\). Since \(W^0\) is \(T^t\)-invariant, we have \(T^t(w^0) \in W^0 \). Now, consider the inner product \(\langle T^t(w^0), w \rangle\). We know this equals \(\langle w^0, Tw \rangle\). Also, since \(T^t(w^0) \in W^0\), we have \(\langle T^t(w^0), w \rangle = 0\). So, \(\langle w^0, Tw \rangle = 0\) for all \(w^0\) in \(W^0\). This implies that the vector Tw is orthogonal to every vector in \(W^0\). Since \(W^0\) is the orthogonal complement of W, this means that Tw must be in W. Hence, W is T-invariant. Therefore, if \(W^0\) is \(T^t\)-invariant, then W is T-invariant. In conclusion, we have proven that W is T-invariant if and only if \(W^0\) is \(T^t\)-invariant.

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Most popular questions from this chapter

Let \(\mathrm{V}=\mathrm{P}_{n}(F)\), and let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars in \(F\). (a) For \(0 \leq i \leq n\), define \(\mathrm{f}_{i} \in \mathrm{V}^{*}\) by \(\mathrm{f}_{i}(p(x))=p\left(c_{i}\right)\). Prove that \(\left\\{\mathrm{f}_{0}, \mathrm{f}_{1}, \ldots, \mathrm{f}_{n}\right\\}\) is a basis for \(\mathrm{V}^{*}\). Hint: Apply any linear combination of this set that equals the zero transformation to \(p(x)=\) \(\left(x-c_{1}\right)\left(x-c_{2}\right) \cdots\left(x-c_{n}\right)\), and deduce that the first coefficient is zero. (b) Use the corollary to Theorem \(2.26\) and (a) to show that there exist unique polynomials \(p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\) such that \(p_{i}\left(c_{j}\right)=\delta_{i j}\) for \(0 \leq i \leq n\). These polynomials are the Lagrange polynomials defined in Section 1.6. (c) For any scalars \(a_{0}, a_{1}, \ldots, a_{n}\) (not necessarily distinct), deduce that there exists a unique polynomial \(q(x)\) of degree at most \(n\) such that \(q\left(c_{i}\right)=a_{i}\) for \(0 \leq i \leq n .\) In fact, $$ q(x)=\sum_{i=0}^{n} a_{i} p_{i}(x) $$ (d) Deduce the Lagrange interpolation formula: $$ p(x)=\sum_{i=0}^{n} p\left(c_{i}\right) p_{i}(x) $$ for any \(p(x) \in \mathrm{V}\). (e) Prove that $$ \int_{a}^{b} p(t) d t=\sum_{i=0}^{n} p\left(c_{i}\right) d_{i} $$ where $$ d_{i}=\int_{a}^{b} p_{i}(t) d t $$ Suppose now that $$ c_{i}=a+\frac{i(b-a)}{n} \text { for } i=0,1, \ldots, n \text {. } $$ For \(n=1\), the preceding result yields the trapezoidal rule for evaluating the definite integral of a polynomial. For \(n=2\), this result yields Simpson's rule for evaluating the definite integral of a polynomial.

Refer to \(A=\left[\begin{array}{rr}2 & -5 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}4 & -2 \\ 1 & -6\end{array}\right], C=\left[\begin{array}{rr}6 & -4 \\ 3 & -2\end{array}\right].\) Find a nonzero column vector \(u\) such that \(C u=4 u.\)

Let \(B=\left[\begin{array}{lll}1 & 8 & 5 \\ 0 & 9 & 5 \\ 0 & 0 & 4\end{array}\right] .\) Find a triangular matrix \(A\) with positive diagonal entries such that \(A^{2}=B\)

Let \(V\) and \(W\) be nonzero vector spaces over the same field, and let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be a linear transformation. (a) Prove that \(T\) is onto if and only if \(T^{t}\) is one-to-one. (b) Prove that \(T^{t}\) is onto if and only if \(T\) is one-to-one.

Refer to the following matrices: $$A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 4 & 0 & -3 \\ -1 & -2 & 3 \end{array}\right], \quad C=\left[\begin{array}{rrrr} 2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3 \end{array}\right], \quad D=\left[\begin{array}{rr} 2 \\ -1 \\ 3 \end{array}\right].$$ Find (a) \(A^{T},\) (b) \(A^{T} B,\) (c) \(A^{T} C.\)
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