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Chapter 2: Problem 3

Prove Theorem 2.1 (i) and (v): (i) \((A+B)+C=A+(B+C),\) (v) \(k(A+B)=k A+k B\)

Short Answer

Expert verified
To prove the associative property (A+B)+C = A+(B+C) for matrices A, B, and C, we showed that the elements of the resulting matrices are equal to each other, i.e., \((a_{ij} + b_{ij}) + c_{ij} = a_{ij} + (b_{ij} + c_{ij})\) for all i and j. To prove the distributive property k(A+B) = kA + kB for a scalar k and matrices A and B, we showed that the elements of the resulting matrices are equal to each other, i.e., \(k (a_{ij} + b_{ij}) = k a_{ij} + k b_{ij}\) for all i and j.

Step by step solution

01

Proving Associative Property: (A+B)+C=A+(B+C)

We'll start by adding matrices A and B, and denote their sum as matrix D with elements \(d_{ij}\). By definition, the elements of D are: \[d_{ij} = a_{ij} + b_{ij}\] Now, matrix D is the same size as A, B, and C, so we can now add C to D element-wise. Let's denote the resulting matrix E with elements \(e_{ij}\). By definition, the elements of E are: \[e_{ij} = d_{ij} + c_{ij} = (a_{ij} + b_{ij}) + c_{ij}\] Now, we'll add B and C to obtain matrix F with elements \(f_{ij}\). By definition, the elements of F are: \[f_{ij} = b_{ij} + c_{ij}\] Now, add A to F element-wise to get matrix G with elements \(g_{ij}\). By definition, the elements of G are: \[g_{ij} = a_{ij} + f_{ij} = a_{ij} + (b_{ij} + c_{ij})\] To prove that (A+B)+C = A+(B+C), we need to show that the elements in E and G are equal, i.e., \(e_{ij} = g_{ij}\) for all i and j. By comparing the expressions for \(e_{ij}\) and \(g_{ij}\), we can see that they are equal: \[(a_{ij} + b_{ij}) + c_{ij} = a_{ij} + (b_{ij} + c_{ij})\] Thus, the associative property holds, and (A+B)+C = A+(B+C).
02

Proving Distributive Property: k(A+B)=kA+kB

Let's denote the sum of A and B as matrix H with elements \(h_{ij}\). By definition, the elements of H are: \[h_{ij} = a_{ij} + b_{ij}\] Now, multiply matrix H by scalar k to get matrix I with elements \(i_{ij}\). By definition, the elements of I are: \[i_{ij} = k h_{ij} = k (a_{ij} + b_{ij})\] Now, we'll multiply matrix A by scalar k to get matrix J with elements \(j_{ij}\). By definition, the elements of J are: \[j_{ij} = k a_{ij}\] Next, multiply matrix B by scalar k to get matrix K with elements \(k_{ij}\). By definition, the elements of K are: \[k_{ij} = k b_{ij}\] Now, add matrices J and K to get matrix L with elements \(l_{ij}\). By definition, the elements of L are: \[l_{ij} = j_{ij} + k_{ij} = k a_{ij} + k b_{ij}\] To prove that k(A+B) = kA + kB, we need to show that the elements in I and L are equal, i.e., \(i_{ij} = l_{ij}\) for all i and j. Thus, we have: \[k (a_{ij} + b_{ij}) = k a_{ij} + k b_{ij}\] So, the distributive property holds, and k(A+B) = kA + 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 concept in linear algebra. This process involves combining two matrices of the same dimensions by adding their corresponding elements.
For example, if you have two matrices, A and B, both sized 2x2, the sum is found by adding the elements in the same positions together.
  • If A matrix is:\[\begin{bmatrix}a_{11} & a_{12} \a_{21} & a_{22} \\end{bmatrix}\]
  • And B matrix is:\[\begin{bmatrix}b_{11} & b_{12} \b_{21} & b_{22} \\end{bmatrix}\]
The resulting matrix C, where C = A + B, will have elements:\[\begin{bmatrix}a_{11} + b_{11} & a_{12} + b_{12} \a_{21} + b_{21} & a_{22} + b_{22} \\end{bmatrix}\]
This method ensures that each element in the resulting matrix is the sum of the elements at the corresponding positions in matrices A and B. Matrix addition is vital for algorithms in multi-dimensional data analysis, computer graphics, and more.
Associative Property
The associative property, a key property in matrix operations, states that when adding matrices, the grouping of matrices does not affect the sum.
Essentially, if you have three matrices A, B, and C, the associative property guarantees that:
  • \[(A + B) + C = A + (B + C)\]
This property allows us to compute complex expressions efficiently since we can rearrange the matrices as needed.
To ensure this property holds, observe how any matrix elements are combined:
  • \((a_{ij} + b_{ij}) + c_{ij} = a_{ij} + (b_{ij} + c_{ij})\)
Each side will yield the same result as we are only performing addition, which is associative. This rule applies universally to any number of matrices, making mathematical computations simpler without affecting the final result.
Distributive Property
The distributive property in matrix operations bridges scalar multiplication and matrix addition. This principle asserts that a scalar value multiplied by the sum of two matrices is equal to the sum of the scalar multiplied by each individual matrix.
Symbolically, for any matrix A and B and scalar k:
  • \[k(A + B) = kA + kB\]
This property allows for versatile computations when the same scalar affects an entire sum of matrices.
For example, when this property is applied to matrix elements:
  • \[k(a_{ij} + b_{ij}) = k \cdot a_{ij} + k \cdot b_{ij}\]
Each term in the matrices is scaled before their summation, maintaining the integrity of the initial calculation. The distributive property is especially useful in computational algorithms and various matrix operations, ensuring efficiency and accuracy.

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Most popular questions from this chapter

Find \(x, y, z, t\) where \(3\left[\begin{array}{cc}x & y \\ z & t\end{array}\right]=\left[\begin{array}{rr}x & 6 \\ -1 & 2 t\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+t & 3\end{array}\right]\)

Compute \(A B\) using block multiplication, where $$ A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{llll} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$ Here \(A=\left[\begin{array}{cc}E & F \\ 0_{1 \times 2} & G\end{array}\right]\) and \(B=\left[\begin{array}{cc}R & S \\ 0_{1 \times 3} & T\end{array}\right]\), where \(E, F, G, R, S, T\) are the given blocks, and \(0_{1 \times 2}\) and \(0_{1 \times 3}\) are zero matrices of the indicated sites. Hence, $$ A B=\left[\begin{array}{cc} E R & E S+F T \\ 0_{1 \times 3} & G T \end{array}\right]=\left[\left[\begin{array}{rrr} 9 & 12 & 15 \\ 19 & 26 & 33 \end{array}\right]{\left[\begin{array}{l} 3 \\ 7 \end{array}\right]+\left[\begin{array}{l} 1 \\ 0 \end{array}\right]} \\ {\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right]} &{2} \end{array}\right]=\left[\begin{array}{rrrr} 9 & 12 & 15 & 4 \\ 19 & 26 & 33 & 7 \\ 0 & 0 & 0 & 2 \end{array}\right] $$

Let \(A\) and \(B\) be invertible matrices (with the same size). Show that \(A B\) is also invertible and \((A B)^{-1}=B^{-1} A^{-1} .\left[\text {Thus, by induction, }\left(A_{1} A_{2} \ldots A_{m}\right)^{-1}=A_{m}^{-1} \ldots A_{2}^{-1} A_{1}^{-1} .\right]\).

Calculate: (a) \([8,-4,5]\left[\begin{array}{r}3 \\ 2 \\\ -1\end{array}\right]\), (b) \([6,-1,7,5]\left[\begin{array}{r}4 \\ -9 \\ -3 \\ 2\end{array}\right]\) (c) \([3,8,-2,4]\left[\begin{array}{r}5 \\ -1 \\ 6\end{array}\right]\) (a) Multiply the corresponding entries and add: $$ [8,-4,5]\left[\begin{array}{r} 3 \\ 2 \\ -1 \end{array}\right]=8(3)+(-4)(2)+5(-1)=24-8-5=11 $$ (b) Multiply the corresponding entries and add: $$ [6,-1,7,5]\left[\begin{array}{r} 4 \\ -9 \\ -3 \\ 2 \end{array}\right]=24+9-21+10=22 $$ (c) The product is not defined when the row matrix and the column matrix have different numbers of elements.

Let \(A\) be a square matrix. Show that \((\mathrm{a}) \quad A+A^{H}\) is Hermitian, \((\mathrm{b}) \quad A-A^{H}\) is skew-Hermitian (c) \(A=B+C,\) where \(B\) is Hermitian and \(C\) is skew-Hermitian.
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