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Chapter 11: Problem 29

Let \(W\) be the subspace of \(\mathbf{R}^{3}\) spanned by (1,1,0) and \((0,1,1) .\) Find a basis of the annihilator of \(W\).

Short Answer

Expert verified
The basis of the annihilator of the subspace \(W \subset \mathbf{R}^{3}\) spanned by \((1,1,0)\) and \((0,1,1)\) is \(\{ (-1,1,-1) \}\).

Step by step solution

01

Write down the given information

The subspace W of \(\mathbf{R}^{3}\) is spanned by (1,1,0) and (0,1,1).
02

Form a matrix

Create a matrix with the given vectors as its rows: \[W = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix}\]
03

Determine the row space

As we have only two non-zero rows, both spanning vectors are linearly independent. Therefore, the row space of the matrix is the subspace W itself.
04

Determine the annihilator

To determine the annihilator, we want to find all vectors in \(\mathbf{R}^{3}\) that will satisfy \(Wv = 0\), where v is a column vector of coefficients of the annihilator basis. In other words, we want to solve the following system: \[\begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}\] This system can be expressed as a set of linear equations: \[\begin{cases} x + y = 0\\ y + z = 0 \end{cases}\]
05

Solve the linear system

Notice that the second equation can be rewritten as \(z=-y\) Plug this into the first equation to get: \(x + y = 0 \Rightarrow x = -y\) Therefore, the general solution to this linear system is given by a vector v in the form: \[\begin{pmatrix}-y\\y\\-y\end{pmatrix} = y\begin{pmatrix}-1\\1\\-1\end{pmatrix}\]
06

Find the basis of the annihilator

The annihilator of W is generated by the vector: \[\begin{pmatrix}-1\\1\\-1\end{pmatrix}\] Thus, the basis of the annihilator of W is: \[\{ (-1,1,-1) \}\] So, we have found the basis of the annihilator of the given subspace.

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Most popular questions from this chapter

Show that if \(\phi \in V^{*}\) annihilates a subset \(S\) of \(V\), then \(\phi\) annihilates the linear span \(L(S)\) of \(S\). Hence, \(S^{0}=[\operatorname{span}(S)]^{0}\).

Let \(V\) be the vector space of polynomials of degree \(\leq 2\). Let \(a, b, c \in K\) be distinct scalars. Let \(\phi_{a}, \phi_{b}, \phi_{c}\) be the linear functionals defined by \(\phi_{a}(f(t))=f(a), \phi_{b}(f(t))=f(b), \phi_{c}(f(t))=f(c) .\) Show that \(\left\\{\phi_{a}, \phi_{b}, \phi_{c}\right\\}\) is linearly independent, and find the basis \(\left\\{f_{1}(t), f_{2}(t), f_{3}(t)\right\\}\) of \(V\) that is its dual.

Let \(\phi\) be the linear functional on \(\mathbf{R}^{2}\) defined by \(\phi(x, y)=3 x-2 y .\) For each of the following linear mappings \(T: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2},\) find \(\left(T^{t}(\phi)\right)(x, y, z)\) (a) \(T(x, y, z)=(x+y, y+z)\) (b) \(T(x, y, z)=(x+y+z, \quad 2 x-y)\)

Let \(W\) be a subspace of \(V\). For any linear functional \(\phi\) on \(W\), show that there is a linear functional \(\sigma\) on \(V\) such that \(\sigma(w)=\phi(w)\) for any \(w \in W ;\) that is, \(\phi\) is the restriction of \(\sigma\) to \(W\).

Show that (a) For any subset \(S\) of \(V, S \subseteq S^{00}\). (b) If \(S_{1} \subseteq S_{2}\), then \(S_{2}^{0} \subseteq S_{1}^{0}\).
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