/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Let \(\mathbf{D}\) denote the di... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 44

Let \(\mathbf{D}\) denote the differential operator on the vector space \(V\) of functions with basis \(S=\\{\sin \theta, \cos \theta\\}\) (a) Find the matrix \(A=[\mathbf{D}]_{s}\) (b) Use \(A\) to show that \(\mathbf{D}\) is a zero of \(f(t)=t^{2}+1\)

Short Answer

Expert verified
(a) The matrix representation \(A\) of the differential operator \(\mathbf{D}\) with respect to the basis \(S=\{ \sin \theta, \cos \theta \}\) is: \[A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\] (b) To show that \(\mathbf{D}\) is a zero of \(f(t) = t^2 + 1\), we compute \(f(A)\) and get: \[f(A) = A^2 + I = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}\] Since \(f(A)\) is the zero matrix, \(\mathbf{D}\) is a zero of \(f(t) = t^2 + 1\).

Step by step solution

01

Find the differential operator acting on the basis functions \(\sin \theta\) and \(\cos \theta\)

To find the matrix \(A\), we need to determine the action of the differential operator on the functions in the basis \(S\). To do so, we will differentiate \(\sin\theta\) and \(\cos\theta\) with respect to \(\theta\). Using basic calculus, we get: \[\mathbf{D}(\sin\theta) = \cos\theta\] \[\mathbf{D}(\cos\theta) = -\sin\theta\]
02

Express the results in terms of the basis vectors

Now that we have computed the differential operator acting on the basis functions, we need to express the results in terms of the basis vectors. We can do this using a linear combination of the basis functions: \[\mathbf{D}(\sin\theta) = 0 \cdot \sin\theta + 1 \cdot \cos\theta\] \[\mathbf{D}(\cos\theta) = (-1) \cdot \sin\theta + 0 \cdot \cos\theta\]
03

Find the matrix representation of the differential operator

We can now use the coefficients from Step 2 to form the matrix representation \(A\) of the differential operator \(D\) with respect to the basis \(S\): \[A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\]
04

Use the matrix \(A\) to show that \(\mathbf{D}\) is a zero of \(f(t) = t^2 + 1\)

To show that the matrix \(A\) is a zero of the polynomial function \(f(t) = t^2 + 1\), we need to evaluate \(f(A)\). When a matrix is a zero of a polynomial, it means that the polynomial evaluated at the matrix yields the zero matrix. We have: \[f(A) = A^2 + I\] where \(I\) is the identity matrix. Compute the square of matrix \(A\): \[A^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix}\] Now compute \(f(A)\): \[f(A) = A^2 + I = \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}\] As \(f(A)\) is equal to the zero matrix, 賺\(\mathbf{D}\) is a zero of \(f(t) = t^2 + 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Representation
Matrix representation is a powerful method to express linear transformations, like differential operators, in a structured form.
In this context, the differential operator \( \mathbf{D} \) acts on the functions \( \sin \theta \) and \( \cos \theta \) which form the basis of our vector space.
To find the matrix representation, we compute how \( \mathbf{D} \) transforms the basis functions and express these transformations through linear combinations of the basis.
  • The derivative of \( \sin \theta \) is \( \cos \theta \), expressed as \( 0 \cdot \sin \theta + 1 \cdot \cos \theta \).
  • Similarly, the derivative of \( \cos \theta \) is \( -\sin \theta \), expressed as \( -1 \cdot \sin \theta + 0 \cdot \cos \theta \).

Thus, the matrix \( A \) capturing the effect of \( \mathbf{D} \) on \( S = \{ \sin \theta, \cos \theta \} \) is:\[ A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \]
This matrix succinctly represents how each function in the basis is transformed by the differential operator.
Polynomial Functions
In mathematics, polynomial functions are expressions involving variables raised to powers and multiplied by coefficients.
They form the building blocks for many algebraic operations and can help solve complex problems involving matrices and transformations.
To demonstrate a relationship between matrices and polynomials, we can substitute a matrix into a polynomial function.For example, given a polynomial \( f(t) = t^2 + 1 \), substituting the matrix \( A \) results in \( f(A) = A^2 + I \), where \( I \) is the identity matrix.
  • We first compute \( A^2 \), which is effectively multiplying \( A \) by itself.
  • The result \( A^2 = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} \) corresponds to \( -I \).
  • Adding \( I \) results in the zero matrix \[ \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \].

Hence, the matrix \( A \) satisfies being a zero of the polynomial \( f(t) = t^2 + 1 \). This shows how matrices can be evaluated and interpreted through polynomials.
Vector Space Basis
A vector space basis is a set of vectors that are linearly independent and span the entire space.
This means any vector in the space can be written as a combination of basis vectors.
In our exercise, the basis \( S = \{ \sin \theta, \cos \theta \} \) forms a framework to express transformations via the differential operator.
  • The choice of these trigonometric functions as a basis is convenient due to their cyclical derivatives, allowing clear transformation through \( \mathbf{D} \).
  • These functions span the functional space of interest, giving a full picture of possible outcomes using this basis.
  • Linearity ensures each basis function can be transformed independently, and results can be expressed as linear combinations of the basis.

Understanding vector space bases allows deeper insight into transformations and operator actions, providing a complete toolset for analyzing mathematical problems involving linear transformations.

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Most popular questions from this chapter

For each of the following linear transformations (operators) \(L\) on \(\mathbf{R}^{2},\) find the matrix \(A\) that represents \(L\) (relative to the usual basis of \(\mathbf{R}^{2}\) ): (a) \(L\) is defined by \(L(1,0)=(2,4)\) and \(L(0,1)=(5,8)\) (b) \(L\) is the rotation in \(\mathbf{R}^{2}\) counterclockwise by \(90^{\circ}\) (c) \(L\) is the reflection in \(\mathbf{R}^{2}\) about the line \(y=-x\)

For each linear transformation \(L\) on \(\mathbf{R}^{2}\), find the matrix \(A\) representing \(L\) (relative to the usual basis of \(\mathbf{R}^{2}\) ): (a) \(L\) is the rotation in \(\mathbf{R}^{2}\) counterclockwise by \(45^{\circ}\). (b) \(L\) is the reflection in \(\mathbf{R}^{2}\) about the line \(y=x\). (c) \(L\) is defined by \(L(1,0)=(3,5)\) and \(L(0,1)=(7,-2)\). (d) \(L\) is defined by \(L(1,1)=(3,7)\) and \(L(1,2)=(5,-4)\).

Consider the following bases of \(\mathbf{R}^{2}\) : $$ S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,-2),(3,-4)\\} \quad \text { and } \quad S^{\prime}=\left\\{v_{1}, v_{2}\right\\}=\\{(1,3),(3,8)\\} $$ (a) Find the coordinates of \(v=(a, b)\) relative to the basis \(S\). (b) Find the change-of-basis matrix \(P\) from \(S\) to \(S^{\prime}\). (c) Find the coordinates of \(v=(a, b)\) relative to the basis \(S^{\prime}\). (d) Find the change-of-basis matrix \(Q\) from \(S^{\prime}\) back to \(S\). (e) Verify \(Q=P^{-1}\). (f) Show that, for any vector \(v=(a, b)\) in \(\mathbf{R}^{2}, P^{-1}[v]_{S}=[v]_{S^{\prime}}\). (See Theorem 6.6.) (a) Let \(v=x u_{1}+y u_{2}\) for unknowns \(x\) and \(y\); that is, $$ \left[\begin{array}{l} a \\ b \end{array}\right]=x\left[\begin{array}{r} 1 \\ -2 \end{array}\right]+y\left[\begin{array}{r} 3 \\ -4 \end{array}\right] \quad \text { or } \begin{aligned} x+3 y=a \\ -2 x-4 y=b \end{aligned} \quad \text { or } \quad \begin{aligned} x+3 y &=a \\ 2 y &=2 a+b \end{aligned} $$ Solve for \(x\) and \(y\) in terms of \(a\) and \(b\) to get \(x=-2 a-\frac{3}{2} b\) and \(y=a+\frac{1}{2} b\). Thus, $$ (a, b)=\left(-2 a-\frac{3}{2}\right) u_{1}+\left(a+\frac{1}{2} b\right) u_{2} \quad \text { or } \quad[(a, b)]_{S}=\left[-2 a-\frac{3}{2} b, a+\frac{1}{2} b\right]^{T} $$ (b) Use part (a) to write each of the basis vectors \(v_{1}\) and \(v_{2}\) of \(S^{\prime}\) as a linear combination of the basis vectors \(u_{1}\) and \(u_{2}\) of \(S\); that is, $$ \begin{aligned} &v_{1}=(1,3)=\left(-2-\frac{9}{2}\right) u_{1}+\left(1+\frac{3}{2}\right) u_{2}=-\frac{13}{2} u_{1}+\frac{5}{2} u_{2} \\ &v_{2}=(3,8)=(-6-12) u_{1}+(3+4) u_{2}=-18 u_{1}+7 u_{2} \end{aligned} $$ Then \(P\) is the matrix whose columns are the coordinates of \(v_{1}\) and \(v_{2}\) relative to the basis \(S ;\) that is, $$ P=\left[\begin{array}{rr} -\frac{13}{2} & -18 \\ \frac{5}{2} & 7 \end{array}\right] $$ (c) Let \(v=x v_{1}+y v_{2}\) for unknown scalars \(x\) and \(y\) : $$ \left[\begin{array}{l} a \\ b \end{array}\right]=x\left[\begin{array}{l} 1 \\ 3 \end{array}\right]+y\left[\begin{array}{l} 3 \\ 8 \end{array}\right] \quad \text { or } \quad \begin{array}{r} x+3 y=a \\ 3 x+8 y=b \end{array} \quad \text { or } \quad \begin{array}{r} x+3 y=a \\ -y=b-3 a \end{array} $$ Solve for \(x\) and \(y\) to get \(x=-8 a+3 b\) and \(y=3 a-b\). Thus, $$ (a, b)=(-8 a+3 b) v_{1}+(3 a-b) v_{2} \quad \text { or } \quad[(a, b)]_{S^{\prime}}=[-8 a+3 b, \quad 3 a-b]^{T} $$ (d) Use part ( \(c)\) to express each of the basis vectors \(u_{1}\) and \(u_{2}\) of \(S\) as a linear combination of the basis vectors \(v_{1}\) and \(v_{2}\) of \(S^{\prime}\) $$ \begin{aligned} &u_{1}=(1,-2)=(-8-6) v_{1}+(3+2) v_{2}=-14 v_{1}+5 v_{2} \\ &u_{2}=(3,-4)=(-24-12) v_{1}+(9+4) v_{2}=-36 v_{1}+13 v_{2} \end{aligned} $$ Write the coordinates of \(u_{1}\) and \(u_{2}\) relative to \(S^{\prime}\) as columns to obtain \(Q=\left[\begin{array}{rr}-14 & -36 \\ 5 & 13\end{array}\right]\). (e) \(Q P=\left[\begin{array}{rr}-14 & -36 \\ 5 & 13\end{array}\right]\left[\begin{array}{rr}-\frac{13}{2} & -18 \\ \frac{5}{2} & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I\) (f) Use parts \((\mathrm{a}),(\mathrm{c})\), and (d) to obtain $$ P^{-1}[v]_{S}=Q[v]_{S}=\left[\begin{array}{rr} -14 & -36 \\ 5 & 13 \end{array}\right]\left[\begin{array}{r} -2 a-\frac{3}{2} b \\ a+\frac{1}{2} b \end{array}\right]=\left[\begin{array}{c} -8 a+3 b \\ 3 a-b \end{array}\right]=[v]_{S} $$

Show that all matrices similar to an invertible matrix are invertible. More generally, show that similar matrices have the same rank.

Two linear operators \(F\) and \(G\) on \(V\) are said to be similar if there exists an invertible linear operator \(T\) on \(V\), such that \(G=T^{-1} \circ F \circ T .\) Prove (a) \(F\) and \(G\) are similar if and only if, for any basis \(S\) of \(V,[F]_{S}\) and \([G]_{S}\) are similar matrices. (b) If \(F\) is diagonalizable (similar to a diagonal matrix), then any similar matrix \(G\) is also diagonalizable.
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