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Chapter 2: Problem 47

Prove Theorem \(2.2(\) iii) and (iv): (iii) \((B+C) A=B A+C A\) (iv) \(k(A B)=(k A) B=A(k B)\)

Short Answer

Expert verified
In order to prove Theorem 2.2 (iii) \((B+C)A = BA + CA\), we showed that all elements of the resulting matrices are equal by algebraic manipulations and the definition of matrix multiplication. We expanded the addition inside the summation to separate the sums and use the distributive property of real numbers. This led to the conclusion that \((B+C)A = BA + CA\). For Theorem 2.2 (iv) \(k(AB) = (kA)B = A(kB)\), we analyzed all the elements of the three resulting matrices, and algebraically manipulated the expressions using the associative property of real numbers. By comparing the expressions, we found that they have the same form, which concludes that \(k(AB) = (kA)B = A(kB)\).

Step by step solution

01

Proving (iii) \((B+C)A = BA + CA\)

In order to prove the distributive matrix property, we need to focus on two operations: matrix addition and matrix multiplication. First, let's consider the addition of matrices B and C, and then, the product of their sum with matrix A. Let A, B, and C be matrices of appropriate dimensions, meaning that matrix addition and multiplication are possible. To prove this equality, we will show that all the elements of the matrix \((B+C)A\) are equal to corresponding elements of the matrix \(BA + CA\). For any element in the i-th row and j-th column of a matrix, we can denote it as \((B+C)_{ij}A_{jk} = BA_{ij} + CA_{ij}\). For the multiplication, we express the elements as follows: \(((B+C)A)_{ik} = \sum_{j=1}^n (B+C)_{ij}A_{jk}\) Now, expand the addition inside the sum: \(\sum_{j=1}^n (B_{ij} + C_{ij})A_{jk}\) Using the distributive property of real numbers, we can separate the sum: \(\sum_{j=1}^n (B_{ij}A_{jk} + C_{ij}A_{jk})\) Now, separate the summation: \(\sum_{j=1}^n B_{ij}A_{jk} + \sum_{j=1}^n C_{ij}A_{jk}\) This can be rewritten as: \( (BA)_{ik} + (CA)_{ik}\) As this holds for every element, using the definition of matrix addition, we conclude: \((B+C)A = BA + CA\)
02

Proving (iv) \(k(AB) = (kA)B = A(kB)\)

In order to prove the associative property for scalar multiplication and matrix multiplication, we must verify that all the elements of the three resulting matrices are equal. Let A and B be matrices of appropriate dimensions and k be a scalar. We will denote any element in the i-th row and j-th column of the resulting matrices as follows: \((k(AB))_{ij} = ((kA)B)_{ij} = (A(kB))_{ij}\) We will start with the product of the scalar k with the matrix AB. For the multiplication, we express the elements as follows: \((k(AB))_{ij} = k\sum_{p=1}^n A_{ip}B_{pj}\) Now, we move to the product of the matrix kA with the matrix B: \(((kA)B)_{ij} = \sum_{p=1}^n (kA)_{ip}B_{pj} = \sum_{p=1}^n (kA_{ip})B_{pj}\) Using the associative property of real numbers, we can rewrite the summation as: \(\sum_{p=1}^n k(A_{ip}B_{pj})\) Finally, let's consider the product of the matrix A with the matrix kB: \((A(kB))_{ij} = \sum_{p=1}^n A_{ip}(kB)_{pj} = \sum_{p=1}^n A_{ip}(kB_{pj})\) Again, we apply the associative property of real numbers: \(\sum_{p=1}^n (A_{ip}k)B_{pj}= \sum_{p=1}^n k(A_{ip}B_{pj})\) Comparing the three expressions, we can see that all have the same form: \(k\sum_{p=1}^n A_{ip}B_{pj} = \sum_{p=1}^n k(A_{ip}B_{pj}) = \sum_{p=1}^n k(A_{ip}B_{pj})\) Thus, we can conclude that: \(k(AB) = (kA)B = A(kB)\)

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Key Concepts

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

Matrix Addition
Matrix addition is a fundamental operation in matrix algebra where two matrices are added together to form a new matrix. The process is quite straightforward yet requires attention to detail.
  • For addition to be possible, both matrices must have the same dimensions, meaning they have the same number of rows and columns.
  • The elements of the resulting matrix are obtained by adding the corresponding elements of the given matrices.
For example, if matrix \( B \) and matrix \( C \) each have 2 rows and 3 columns, they can be added together as follows:\[B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \ b_{21} & b_{22} & b_{23} \end{bmatrix}, \quad C = \begin{bmatrix} c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \end{bmatrix}\]Then their sum matrix \( (B + C) \) is:\[B + C = \begin{bmatrix} b_{11}+c_{11} & b_{12}+c_{12} & b_{13}+c_{13} \ b_{21}+c_{21} & b_{22}+c_{22} & b_{23}+c_{23} \end{bmatrix}\]This operation is element-wise and fairly straightforward once you ensure the matrices are compatible in terms of dimensions. Breaking addition into these simple steps helps maintain organization and accuracy, especially when dealing with large matrices.
Matrix Multiplication
Matrix multiplication might seem complex at first, but with careful practice, it becomes manageable. Unlike matrix addition, matrix multiplication involves the rows of the first matrix and the columns of the second matrix.
  • For two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
  • The resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.
Given matrices \( A \) and \( B \), the element \( (AB)_{ij} \) of the resulting matrix is the summation of the products of elements from row \( i \) of matrix \( A \) and column \( j \) of matrix \( B \):\[(AB)_{ij} = \sum_{k=1}^n a_{ik}b_{kj}\]This requires multiplying elements at respective positions and summing them up across each row and column. This can be visualized as the dot product of the row vector from the first matrix and column vector from the second matrix.Matrix multiplication is a key operation in linear transformations and provides the foundation for many advanced topics in matrix algebra.
Distributive Property
The distributive property is a significant feature of matrix algebra that connects different operations. This property states that matrix multiplication is distributive over addition.For any three matrices, if dimensions match appropriately, the equation holds:\[(B+C)A = BA + CA\]Here's how it works:- First, add matrices \( B \) and \( C \) to form a new matrix.- Multiply matrix \( A \) by this new resulting matrix.- Simultaneously, you can separately multiply \( A \) by \( B \) and \( A \) by \( C \), then add the results.The distributive property confirms that both methods yield the same final matrix. This allows flexibility in calculations and problem-solving, especially when simplifying complex matrix expressions. Constant practice with manipulation of expressions will strengthen understanding and speed in applying the distributive property.
Scalar Multiplication
In matrix algebra, scalar multiplication involves multiplying every element of a matrix by a scalar value. This operation changes the magnitude of the matrix without altering its direction in a geometric sense.To perform scalar multiplication, follow these easy steps:- Choose a scalar value, \( k \).- Multiply each element of the matrix by \( k \).For example, if matrix \( A \) is:\[A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}\]and \( k \) is a scalar, then the scalar multiplication \( kA \) is:\[kA = \begin{bmatrix} k \times a_{11} & k \times a_{12} \ k \times a_{21} & k \times a_{22} \end{bmatrix}\]Scalar multiplication is often used in conjunction with matrix addition or matrix multiplication to scale matrices and is linear.The associative property ensures that for matrices \( A \), \( B \), and scalar \( k \), \[k(AB) = (kA)B = A(kB)\] This property allows interchangeable computation, providing versatility in solving equations and simplifying matrix expressions. Understanding scalar multiplication paves the way for exploring more intricate matrix transformations.

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Most popular questions from this chapter

For each of the following parts, determine whether the statement is true or false. Justify your claim with either a proof or a counterexample, whichever is appropriate. (a) Any finite-dimensional subspace of \(C^{\infty}\) is the solution space of a homogeneous linear differential equation with constant coefficients. (b) There exists a homogeneous linear differential equation with constant coefficients whose solution space has the basis \(\left\\{t, t^{2}\right\\}\). (c) For any homogeneous linear differential equation with constant coefficients, if \(x\) is a solution to the equation, so is its derivative \(x^{\prime}\). Given two polynomials \(p(t)\) and \(q(t)\) in \(\mathrm{P}(C)\), if $x \in \mathrm{N}(p(\mathrm{D}))\( and \)y \in\( \)\mathrm{N}(q(\mathrm{D}))$, then (d) \(x+y \in \mathrm{N}(p(\mathrm{D}) q(\mathrm{D}))\). (e) \(x y \in \mathrm{N}(p(\mathrm{D}) q(\mathrm{D}))\).

Find \(x, y, z\) such that \(A\) is symmetric, where (a) \(A=\left[\begin{array}{lll}2 & x & 3 \\ 4 & 5 & y \\ z & 1 & 7\end{array}\right]\) (b) \(A=\left[\begin{array}{rrr}7 & -6 & 2 x \\ y & z & -2 \\ x & -2 & 5\end{array}\right]\)

Compute \(A B\) using block multiplication, where $$A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cccc} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right].$$

Label the following statements as true or false. In each part, \(V\) and \(W\) are finite-dimensional vector spaces (over \(F\) ), and \(\mathrm{T}\) is a function from \(\mathrm{V}\) to \(\mathrm{W}\). (a) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) preserves sums and scalar products. (b) If \(\mathrm{T}(x+y)=\mathrm{T}(x)+\mathrm{T}(y)\), then \(\mathrm{T}\) is linear. (c) \(\mathrm{T}\) is one-to-one if and only if the only vector \(x\) such that \(\mathrm{T}(x)=0\) is \(x=0\). (d) If \(T\) is linear, then \(T\left(0_{v}\right)=0_{w}\). (e) If \(T\) is linear, then nullity \((T)+\operatorname{rank}(T)=\operatorname{dim}(W)\). (f) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) carries linearly independent subsets of \(\mathrm{V}\) onto linearly independent subsets of W. (g) If \(\mathrm{T}, \mathrm{U}: \mathrm{V} \rightarrow \mathrm{W}\) are both linear and agree on a basis for \(\mathrm{V}\), then \(T=U\). (h) Given \(x_{1}, x_{2} \in \mathrm{V}\) and \(y_{1}, y_{2} \in \mathrm{W}\), there exists a linear transformation $\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\( such that \)\mathrm{T}\left(x_{1}\right)=y_{1}$ and \(\mathrm{T}\left(x_{2}\right)=y_{2}\). For Exercises 2 through 6, prove that \(\mathrm{T}\) is a linear transformation, and find bases for both \(N(T)\) and \(R(T)\). Then compute the nullity and rank of \(T\), and verify the dimension theorem. Finally, use the appropriate theorems in this section to determine whether \(\mathrm{T}\) is one-to-one or onto.

Let \(A\) be invertible. Prove that \(A^{t}\) is invertible and \(\left(A^{t}\right)^{-1}=\left(A^{-1}\right)^{t}\). Visit goo.gl/suFm6V for a solution.
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