Question

In each case, find the coordinates of \(\mathbf{v}\) with respect to the basis \(B\) of the vector space \(V\). a. \(V=\mathbf{P}_{2}, \mathbf{v}=2 x^{2}+x-1, B=\left\\{x+1, x^{2}, 3\right\\}\) b. \(V=\mathbf{P}_{2}, \mathbf{v}=a x^{2}+b x+c, B=\left\\{x^{2}, x+1, x+2\right\\}\) \(\begin{aligned} \text { c. } & V=\mathbb{R}^{3}, \mathbf{v}=(1,-1,2), \\\ \quad B &=\\{(1,-1,0),(1,1,1),(0,1,1)\\} \\ \text { d. } V &=\mathbb{R}^{3}, \mathbf{v}=(a, b, c), \\ \quad B &=\\{(1,-1,2),(1,1,-1),(0,0,1)\\} \end{aligned}\) \(\begin{aligned} \text { e. } V &=\mathbf{M}_{22}, \mathbf{v}=\left[\begin{array}{rr}1 & 2 \\ -1 & 0\end{array}\right] \\ B &=\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right\\} \end{aligned}\)

Short answer

Coordinates are: a) (1/3, 2, -2/3), b) (a, b-a, c-b), c) (1,0,2), d) ((a+b)/2, (a-b)/2, c-(a+b)/2), e) (0, -1, 2, 1).

Step-by-step solution

Represent Vector with Basis in Part A

To determine the coordinates of \( \mathbf{v} = 2x^2 + x - 1 \) with respect to the basis \( B = \{ x+1, x^2, 3 \} \), express \( \mathbf{v} \) as a linear combination of the basis elements. Assume \( a(x+1) + b(x^2) + c(3) = 2x^2 + x - 1 \). Solving the system of equations gives us \( a = \frac{1}{3}, b = 2, c = -\frac{2}{3} \). Hence, the coordinate vector is \( \left( \frac{1}{3}, 2, -\frac{2}{3} \right) \).

Represent General Vector with Basis in Part B

Given \( \mathbf{v} = ax^2 + bx + c \) and basis \( B = \{ x^2, x+1, x+2 \} \), write \( \mathbf{v} \) as \( p x^2 + q(x+1) + r(x+2) = ax^2 + bx + c \). Expanding and equating coefficients, solve for \( p, q, r \) to get \( p = a, q = b-a, r = c-b \). Thus, the coordinate vector is \( (a, b-a, c-b) \).

Represent Point Vector with Basis in Part C

For \( \mathbf{v} = (1,-1,2) \) and basis \( B = \{ (1,-1,0), (1,1,1), (0,1,1) \} \), write \( \mathbf{v} = a(1,-1,0) + b(1,1,1) + c(0,1,1) \). Solve for \( a, b, c \) which leads to the equations \( a+b = 1 \), \(-a+b+c = -1 \), and \( b+c = 2 \). From solving this system, \( a = 1, b = 0, c = 2 \). The coordinate vector is \( (1,0,2) \).

Represent General Point Vector with Basis in Part D

Consider \( \mathbf{v} = (a,b,c) \) with basis \( B = \{ (1,-1,2), (1,1,-1), (0,0,1) \} \). Express \( \mathbf{v} = x(1,-1,2) + y(1,1,-1) + z(0,0,1) \). Solving gives \( x+y = a \), \(-x+y = b \), \( 2x-y+z = c \). Derived coordinates are \( x = \frac{a+b}{2} \), \( y = \frac{a-b}{2} \), \( z = c-x \). Thus, the coordinates are \( \left( \frac{a+b}{2}, \frac{a-b}{2}, c - \frac{a+b}{2} \right) \).

Represent Matrix with Basis in Part E

Express \( \mathbf{v} = \begin{bmatrix} 1 & 2 \ -1 & 0 \end{bmatrix} \) as a linear combination of \( B = \left\{ \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \right\} \). The equation is \( a\begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} + b\begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} + c\begin{bmatrix} 0 & 0 \ 1 & 1 \end{bmatrix} + d\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ -1 & 0 \end{bmatrix} \). Solving gives \( a = 0, b = -1, c = 2, d = 1 \). The coordinate vector is \( (0, -1, 2, 1) \).

Key concepts

Understanding Linear Algebra

Linear algebra is a branch of mathematics that deals with vector spaces and linear equations. It is fundamental in various scientific and engineering fields. One essential aspect of linear algebra is solving systems of linear equations, which involves finding values for variables that satisfy all equations simultaneously. Linear algebra provides tools for handling matrices, determinants, eigenvalues, and transformations.

In the context of our problem, we're interested in how vectors can be expressed in terms of a specific basis of a vector space. This involves concepts like vector addition, scalar multiplication, and the structure provided by matrices. Learning linear algebra helps us explore the relationships between dimensions in space and understand how different bases can transform the representation of vectors. It forms the backbone of mathematical modeling and data interpretation.

Basis Representation of Vectors

Basis representation refers to expressing any vector in a vector space as a unique combination of vectors from a basis of that space. A basis is a set of vectors that are linearly independent and span the vector space. In simpler terms, any vector in the space can be constructed as a linear combination of basis vectors. This simplifies vector operations and facilitates transformations like changing the coordinates of a system.

For example, in our exercise, different bases are given for specific problems. For each scenario, like the polynomial basis \(B = \{x+1, x^2, 3\}\) in part a or the matrix basis in part e, we find coefficients that combine these basis elements to generate the vector \(\mathbf{v}\). The resulting coefficients, or coordinates, uniquely identify the position of \(\mathbf{v}\) within the vector space concerning the given basis. This is crucial in computational practices, as it allows converting scenarios into manageable mathematical models.

Exploring Vector Spaces

A vector space is an algebraic structure formed by vectors, which can be added together and multiplied by scalars. Vector spaces are full of elegance, allowing us to generalize concepts like lines, planes, and more in any number of dimensions. The beauty of vector spaces lies in their ability to model numerous real-world phenomena mathematically.

Vector spaces must satisfy certain axioms:

• Vectors can be added, and this addition is commutative and associative.
• There exists a zero vector that, when added to any vector, yields the same vector back.
• Each vector has a negative counterpart that results in the zero vector when both are added.
• Vectors can be scaled (multiplied by a scalar) within the space and still yield another vector in the same space.

Understanding vector spaces enables us to dive into deep concepts such as linear transformations, eigenvalues, and more. Our exercise explores various forms of vector spaces, such as polynomials, real-number vectors, and matrices, demonstrating their wide-ranging applications and critical importance in maths and beyond.