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Chapter 2: Problem 11

Prove Theorem \(2.2(\text { ii }): A(B+C)=A B+A C\).

Short Answer

Expert verified
To prove the theorem \(A(B + C) = AB + AC\), we first define the dimensions of matrices A, B, and C as \(m \times n\), \(n \times p\), and \(n \times p\) respectively. We then define the element-wise matrix multiplication and addition operations, and expand the definitions as follows: \((A(B+C))_{ij} = \sum_{k=1}^n A_{ik}[(B)_{kj}+(C)_{kj}]\) and \((AB+AC)_{ij} = \sum_{k=1}^n A_{ik}(B)_{kj} + \sum_{k=1}^n A_{ik}(C)_{kj}\). Applying the distributive property of scalar multiplication over addition inside the first summation, we find that \((A(B+C))_{ij} = (AB+AC)_{ij}\). Since the individual elements of the resulting matrices are equal, we have proven that the distributive property holds: \(A(B + C) = AB + AC\).

Step by step solution

01

State the Given Theorem

We are given the theorem to prove: \(A(B + C) = AB + AC\)
02

Define the Matrix Dimensions

Let matrix A be of dimensions \(m \times n\), matrix B be of dimensions \(n \times p\), and matrix C be of dimensions \(n \times p\). Note that the matrix addition B + C and the matrix multiplications AB and AC are defined and will result in a matrix of dimensions \(m \times p\).
03

Define the Multiplication and Addition Operations

The matrix multiplication of A(B + C) can be defined element-wise as: \((A(B+C))_{ij} = \sum_{k=1}^n A_{ik}(B+C)_{kj}\) Similarly, the matrix addition of AB + AC can be defined as: \((AB+AC)_{ij} = (AB)_{ij}+(AC)_{ij}\)
04

Expand the Element-wise Definitions

First, expand the definition of (A(B+C))_{ij}: \((A(B+C))_{ij} = \sum_{k=1}^n A_{ik}[(B)_{kj}+(C)_{kj}]\) Now, expand the definition of (AB+AC)_{ij}: \((AB+AC)_{ij} = \sum_{k=1}^n A_{ik}(B)_{kj} + \sum_{k=1}^n A_{ik}(C)_{kj}\)
05

Demonstrate Equality

We can apply the distributive property of scalar multiplication over addition inside the first summation: \((A(B+C))_{ij} = \sum_{k=1}^n A_{ik}(B)_{kj} + \sum_{k=1}^n A_{ik}(C)_{kj}\) Comparing this with the expanded definition of (AB+AC)_{ij} from step 4, we can see that: \((A(B+C))_{ij} = (AB+AC)_{ij}\) Since the individual elements of the resulting matrices are equal, we have proven that the distributive property holds: \(A(B + C) = AB + AC\)

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Key Concepts

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

Linear Algebra
Linear algebra is a foundational field of mathematics that is particularly important in areas of science and engineering. It deals primarily with the study of vectors, vector spaces, linear mappings, and systems of linear equations. Matrices, which are rectangular arrays of numbers, symbols, or expressions, are essential tools in linear algebra because they can represent linear transformations and systems of linear equations.

The operations of matrix addition and multiplication, as discussed in the exercise, are central to linear algebra. They allow us to solve linear systems, find solutions to equations, and perform a variety of transformations. Understanding these operations and their properties, such as the distributive property, is crucial for manipulating and combining linear equations or transformations.
Matrix Multiplication
  • Matrix multiplication is not commutative: In general, for two matrices A and B, the product AB is not equal to BA.
  • Associative property: For three matrices A, B, and C, the product A(BC) is equal to (AB)C.
  • Product of a matrix with the identity: Multiplying any matrix by an identity matrix of compatible size leaves it unchanged, i.e., AI = IA = A.

While matrix multiplication may seem daunting at first, its rules are consistent, and with practice, anyone can master it.
Matrix Addition
Matrix addition is a much simpler and more intuitive concept than matrix multiplication. For matrix addition to be defined, the matrices involved must have the same dimensions. Once this condition is met, two matrices can be added together by simply combining corresponding elements.

The properties of matrix addition include commutativity and associativity, which make it similar to the addition of regular numbers. This means for any three matrices A, B, and C of the same size, we have A + B = B + A and (A + B) + C = A + (B + C).

Understanding addition is crucial since many problems in linear algebra can be reduced to finding the sum of matrices.
Distributive Property
The distributive property is an essential rule in arithmetic and algebra which allows us to simplify expressions where one number is multiplied by the sum of two others. In terms of matrices, the distributive property is written as A(B + C) = AB + AC.

This property shows us that if we multiply a matrix A by the sum of two matrices B and C, the result is the same as if we multiplied matrix A with each matrix individually and then added the results. The proof outlined in the exercise ensures that these operations coincide and that the distributive property is a fundamental aspect of linear algebra that maintains the consistency and structure necessary for matrix operations.

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Most popular questions from this chapter

Which of the following matrices are normal? \(A=\left[\begin{array}{rr}3 & -4 \\\ 4 & 3\end{array}\right], B=\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right], C=\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\)

A differential equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=x $$ is called a nonhomogeneous linear differential equation with constant coefficients if the \(a_{i}\) 's are constant and \(x\) is a function that is not identically zero.(a) Prove that for any \(x \in{C}^{\infty}\) there exists $y \in{C}^{\infty}\( such that \)y$ is a solution to the differential equation. Hint: Use Lemma 1 to Theorem \(2.32\) to show that for any polynomial \(p(t)\), the linear operator $p(\mathrm{D}): \mathrm{C}^{\infty} \rightarrow \mathrm{C}^{\infty}$ is onto. (b) Let \(V\) be the solution space for the homogeneous linear equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{1} y^{(1)}+a_{0} y=0 . $$ Prove that if \(z\) is any solution to the associated nonhomogeneous linear differential equation, then the set of all solutions to the nonhomogeneous linear differential equation is $$ \\{z+y: y \in \mathrm{V}\\} . $$

Pendular Motion. It is well known that the motion of a pendulum is approximated by the differential equation $$ \theta^{\prime \prime}+\frac{g}{l} \theta=0, $$where \(\theta(t)\) is the angle in radians that the pendulum makes with a vertical line at time \(t\) (see Figure 2.8), interpreted so that \(\theta\) is positive if the pendulum is to the right and negative if the pendulum is to the left of the vertical line as viewed by the reader. Here \(l\) is the length of the pendulum and \(g\) is the magnitude of acceleration due to gravity. The variable \(t\) and constants \(l\) and \(g\) must be in compatible units (e.g., \(t\) in seconds, \(l\) in meters, and \(g\) in meters per second per second). (a) Express an arbitrary solution to this equation as a linear combination of two real-valued solutions. (b) Find the unique solution to the equation that satisfies the conditions $$ \theta(0)=\theta_{0}>0 \text { and } \theta^{\prime}(0)=0 \text {. } $$ (The significance of these conditions is that at time \(t=0\) the pendulum is released from a position displaced from the vertical by \(\theta_{0}\).) (c) Prove that it takes \(2 \pi \sqrt{l / g}\) units of time for the pendulum to make one circuit back and forth. (This time is called the period of the pendulum.)

For each matrix \(A\) and ordered basis \(\beta\), find \(\left[\mathrm{L}_{A}\right]_{\beta}\). Also, find an invertible matrix \(Q\) such that \(\left[\mathrm{L}_{A}\right]_{\beta}=Q^{-1} A Q\). (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 2\end{array}\right)\right\\}$ (b) \(A=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{r}1 \\ -1\end{array}\right)\right\\}$ (c) $A=\left(\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{array}\right) \quad\( and \)\quad \beta=\left\\{\left(\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 2\end{array}\right)\right\\}$ (d) $A=\left(\begin{array}{rrr}13 & 1 & 4 \\ 1 & 13 & 4 \\ 4 & 4 & 10\end{array}\right)\( and \)\beta=\left\\{\left(\begin{array}{r}1 \\ 1 \\\ -2\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\\ 0\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right)\right\\}$

Let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars from an infinite field \(F\). Define \(\mathrm{T}: \mathrm{P}_{n}(F) \rightarrow \mathrm{F}^{n+1}\) by $\mathrm{T}(f)=\left(f\left(c_{0}\right), f\left(c_{1}\right), \ldots, f\left(c_{n}\right)\right)\(. Prove that \)\mathrm{T}$ is an isomorphism. Hint: Use the Lagrange polynomials associated with \(c_{0}, c_{1}, \ldots, c_{n}\).
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