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Problem 563

Let \(\mathrm{V}\) be the vector space of all polynomial functions from \(\mathrm{R}\) into \(\mathrm{R}\) with degree less than or equal to \(2 .\) Find a basis for \(\mathrm{V}\) by using the following procedures i) Find three linear functionals on \(\mathrm{V}\) ii) Use these functionals as a basis for \(\mathrm{V}^{*}\), the dual of \(\mathrm{V}\). iii) Use the functionals to find a basis for \(\mathrm{V}\).Annthilators, Transposes \& Adjoint

Short Answer

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To find a basis for the vector space V of all polynomial functions from R into R with degree less than or equal to 2, we first find three linear functionals \(L_1\), \(L_2\), and \(L_3\) that evaluate polynomials at specific points (0, 1, and -1, respectively). Then, we use these functionals to form a basis for the dual space V* as \(\mathcal{B}^* = \{L_1, L_2, L_3\}\). Finally, we find the annihilators for each functional and solve the resulting system of equations to obtain the basis for V: \(\mathcal{B} = \{1 + x, x^2\}\).
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Step 1: The given vector space V consists of all polynomial functions from R into R with degree less than or equal to 2. These polynomial functions can be written in the form \(p(x) = ax^2 + bx + c\), where a, b, and c are real numbers. #Step 2: Find three linear functionals on V#

To find three linear functionals on V, we need to find three functions that take a polynomial from V as input and output a scalar. Let's consider the evaluation functionals, which evaluate a polynomial at a particular point. Specifically, let's consider the functionals \(L_1\), \(L_2\), and \(L_3\) defined as follows: - \(L_1(p) = p(0)\) - \(L_2(p) = p(1)\) - \(L_3(p) = p(-1)\) These functionals take in a polynomial \(p(x) = ax^2 + bx + c\) and output the scalar value of the polynomial at the specific x-value chosen. These functionals are linear since the evaluation of any linear combination of polynomials is the same linear combination of their values. #Step 3: Use functionals as a basis for V*#

Step 2: Now we want to use the functionals \(L_1, L_2\), and \(L_3\) as a basis for the dual space V*. Recall that the dual of a vector space consists of all linear functionals on that space. The dimension of V equals 3 since all polynomials in V have degree less than or equal to 2. Since V* has the same dimension as V, we know that having three linearly independent functionals will form a basis for V*. Given that we have three linear functionals, we can now form a basis for V* by combining them into a set: \(\mathcal{B}^* = \{L_1, L_2, L_3\}\). #Step 4: Use functionals to find a basis for V#

Finally, we will use the functionals to find a basis for V. To do this, we will construct an annihilator for each functional and solve the resulting system of equations. An annihilator of a functional L is a polynomial p in V such that L(p) = 0. Let's find the annihilators for each of our functionals: - \(L_1\): \(L_1(p) = p(0) = c\), so the annihilator is a polynomial p with \(c = 0\). - \(L_2\): \(L_2(p) = p(1) = a + b + c\), so the annihilator is a polynomial p with \(a + b + c = 0\). - \(L_3\): \(L_3(p) = p(-1) = a - b + c\), so the annihilator is a polynomial p with \(a - b + c = 0\). Now, let's solve the system of equations formed by the annihilators: \[ \begin{cases} c = 0 \\ a + b + c = 0 \\ a - b + c = 0 \end{cases} \] Solving this system, we find that \(a = 1\), \(b = 1\), and \(c = 0\). Therefore, a basis for V consists of the following polynomials: \[ \mathcal{B} = \{1 + x, x^2\} \] In conclusion, we have found a basis for the vector space V of all polynomial functions of degree less than or equal to 2 using the procedures given: \(\mathcal{B} = \{1 + x, x^2\}\).

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Most popular questions from this chapter

Chapter 22

Let \(\mathrm{V}\) be the space of polynomials over the field of complex numbers, with the inner product $$ (\mathrm{f} / \mathrm{g})={ }^{1} \int_{0} \mathrm{f}(\mathrm{t}) \mathrm{g}_{-}(\mathrm{t}) \mathrm{dt} $$ Let \(\mathrm{D}\) be the differentiation operator on \(\mathrm{C}[\mathrm{x}]\). Show that \(\mathrm{D}\) has no adjoint.Multilinear Functionals

Chapter 22

Let \(\mathrm{V}\) be the space of polynomials over the field of complex numbers, with the inner product $$ (\mathrm{f} / \mathrm{g})=1 \int_{0} \mathrm{f}(\mathrm{t}) \mathrm{g}(\mathrm{t}) \mathrm{dt} $$ (where the bar indicates complex conjugation). Consider the operator 'multiplication by \(\mathrm{f}^{\prime}\), that is, the linear operator \(\mathrm{M}_{\mathrm{f}}\) defined by \(\mathrm{M}_{\mathrm{f}}(\mathrm{g})=\mathrm{fg}\). What is the adjoint of this operator?

Chapter 22

Let \(\mathrm{V}\) be an \(\mathrm{n}\) -dimensional vector space over the field \(\mathrm{F}\) and let \(\mathrm{T}\) be a linear operator on \(\mathrm{V} .\) Suppose \(\mathrm{B}=\left(\alpha_{1}, \ldots, \alpha_{\mathrm{n}}\right\\}\) and $B^{\prime}=\left\\{\alpha_{1}^{\prime}, \ldots, \alpha_{n}^{\prime}\right\\}\( are two ordered bases for \)\mathrm{V}$. Show how the transpose transformation \(\mathrm{T}^{\mathrm{t}}\) can be used to derive the formula for changing bases, that is, to find \(\mathrm{T}\) in the ordered basis B' given the matrix of \(\mathrm{T}\) in \(\mathrm{B}\).

Chapter 22

Let \(\mathrm{V}\) be \(\mathrm{K}^{\mathrm{n} \times \mathrm{n}}\) with the inner product $(\mathrm{A} / \mathrm{B})=\operatorname{tr}\left(\mathrm{B}^{*} \mathrm{~A}\right)\(. Let \)\mathrm{M}$ be a fixed \(n \times n\) matrix over \(K\) (a field). What is the adjoint of left multiplication by \(\mathrm{M}\) ?

Chapter 22

Let \(W\) be the subspace of \(R^{5}\) which is spanned by the Vectors $$ \begin{aligned} &\alpha_{1}=(2,-2,3,4,-1), \alpha_{2}=(-1,1,2,5,2) \\ &\alpha_{3}=(0,0,-1,-2,3) & \alpha_{4}=(1,-1,2,3,0) \end{aligned} $$ Describe \(\mathrm{W}^{\circ}\), the annihilator of \(\mathrm{W}\).

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