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Chapter 11: Problem 9

Find a basis of the annihilator \(W^{0}\) of the subspace \(W\) of \(\mathbf{R}^{4}\) spanned by \\[ v_{1}=(1,2,-3,4) \text { and } v_{2}=(0,1,4,-1) \\]

Short Answer

Expert verified
The basis for the annihilator \(W^{0}\) of the given subspace \(W\) is: \[\left\{ (11, -4, 1, 0), (-10, 1, 0, 1) \right\}\]

Step by step solution

01

Define the orthogonality condition

For a vector x to be in the annihilator W^0, it must be orthogonal to both v1 and v2, meaning that the dot product between x and v1, as well as the dot product between x and v2, must be equal to zero: \(x \cdot v1 = 0\) and \(x \cdot v2 = 0\) #Step 2: Set up a system of linear equations from the dot products#
02

Set up a system of linear equations

Let x = (x1, x2, x3, x4). Then, we have the following system of linear equations from the dot product formulas: \(x_1 + 2x_2 - 3x_3 + 4x_4 = 0\) \(x_2 + 4x_3 - x_4 = 0\) #Step 3: Write the augmented matrix and perform Gaussian elimination#
03

Augmented matrix and Gaussian elimination

We can write the augmented matrix for this system of linear equations and perform Gaussian elimination: \[ \left( \begin{array}{cccc|c} 1 & 2 & -3 & 4 & 0\\ 0 & 1 & 4 & -1 & 0 \end{array} \right) \] After performing Gaussian elimination, we have: \[ \left( \begin{array}{cccc|c} 1 & 0 & -11 & 10 & 0\\ 0 & 1 & 4 & -1 & 0 \end{array} \right) \] #Step 4: Solve the system and express the general solution#
04

Solve the system and express the general solution

Solving the system, we find: \(x_1 + 0x_2 - 11x_3 + 10x_4 = 0\) \(0x_1 + x_2 + 4x_3 - x_4 = 0\) Expressing the general solution: \(x_1 = 11x_3 - 10x_4\) \(x_2 = -4x_3 + x_4\) #Step 5: Write the basis for the annihilator W^0 #
05

Basis for the annihilator W^0

Let \(x_3 = s\) and \(x_4 = t\). Then, we can express x as a linear combination: \(x = s(11, -4, 1, 0) + t(-10, 1, 0, 1)\) So, the basis for the annihilator W^0 is: \(\left\{ (11, -4, 1, 0), (-10, 1, 0, 1) \right\}\)

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Most popular questions from this chapter

Let \(W\) be the subspace of \(\mathbf{R}^{4}\) spanned by \((1,2,-3,4), \quad(1,3,-2,6), \quad(1,4,-1,8) .\) Find a basis of the annihilator of \(W\).

Prove Theorem 11.7: Let \(T: V \rightarrow U\) be linear and let \(A\) be the matrix representation of \(T\) in the bases \(\left\\{v_{j}\right\\}\) of \(V\) and \(\left\\{u_{i}\right\\}\) of \(U\). Then the transpose matrix \(A^{T}\) is the matrix representation of \(T^{t}: U^{*} \rightarrow V^{*}\) in the bases dual to \(\left\\{u_{i}\right\\}\) and \(\left\\{v_{i}\right\\}\).

Prove Theorem 11.2: Let \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) be a basis of \(V\) and let \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) be the dual basis in \(V^{*} .\) For any \(u \in V\) and any \(\sigma \in V^{*},(\mathrm{i}) u=\sum_{i} \phi_{i}(u) v_{i}\) (ii) \(\sigma=\sum_{i} \phi\left(v_{i}\right) \phi_{i}\)

Let \(V\) be the vector space of polynomials over \(K\). For \(a \in K\), define \(\phi_{a}: V \rightarrow K\) by \(\phi_{a}(f(t))=f(a) .\) Show that \((\mathrm{a}) \phi_{a}\) is linear; \((\mathrm{b})\) if \(a \neq b,\) then \(\phi_{a} \neq \phi_{b}\).

Suppose \(V=U \oplus W .\) Prove that \(V^{0}=U^{0} \oplus W^{0}\).
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