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Chapter 4: Problem 40

The absolute value of a matrix \(A=\left[a_{i j}\right]\) is defined to be the matrix \(|A|=\left[\left|a_{i j}\right|\right]\). Let \(A\) and \(B\) be \(n \times n\) matrices, \(x\) a vector in \(\mathbb{R}^{n},\) and a scalar. Prove the following matrix inequalities: (a) \(|c A|=|c||A|\) (b) \(|A+B| \leq|A|+|B|\) (c) \(|A \mathbf{x}| \leq|A||\mathbf{x}|\) (d) \(|A B| \leq|A||B|\)

Short Answer

Expert verified
The inequalities hold for absolute values of matrices due to properties of absolute values and sums.

Step by step solution

01

Prove |cA| = |c||A|

Start by considering the definition of the absolute value of a matrix. For \[ |cA| = [|c a_{ij}|] \]This is equivalent to applying absolute value to the constant c and to each individual element of the matrix A separately, \[ = [|c||a_{ij}|] \]Notice that this is equal to\[ |c||A| \]Hence, we have proven that \[ |cA| = |c||A| \]
02

Prove |A+B| 鈮� |A| + |B|

Consider the matrices A and B with their entries \[ A = [a_{ij}] \]and \[ B = [b_{ij}] \]Then, the absolute value of their sum is \[ |A+B| = [|a_{ij} + b_{ij}|] \]By properties of absolute values, we have:\[ |a_{ij} + b_{ij}| \leq |a_{ij}| + |b_{ij}| \]So,\[ [|a_{ij} + b_{ij}|] \leq [|a_{ij}| + |b_{ij}|] = |A| + |B| \]Thus, \[ |A+B| \leq |A| + |B| \]
03

Prove |Ax| 鈮� |A||x|

Consider a matrix A and vector x. The product Ax gives another vector whose ith component is:\[ (Ax)_i = \sum_{j=1}^{n} a_{ij} x_j \]Taking absolute values, we get:\[ |(Ax)_i| = \left| \sum_{j=1}^{n} a_{ij} x_j \right| \leq \sum_{j=1}^{n} |a_{ij}| |x_j| \]Therefore,\[ |Ax| \leq |A||x| \]
04

Prove |AB| 鈮� |A||B|

Consider A and B where each entry of the product AB is given by:\[ (AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj} \]Taking absolute value gives:\[ |(AB)_{ij}| = \left| \sum_{k=1}^{n} a_{ik} b_{kj} \right| \leq \sum_{k=1}^{n} |a_{ik}| |b_{kj}| \]Thus, this shows:\[ |AB| \leq [ \sum_{k=1}^{n} |a_{ik}| |b_{kj}| ] = |A||B| \]

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Key Concepts

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

Matrix Absolute Value
The concept of a matrix's absolute value is similar to that of single numbers. Given a matrix \(A = [a_{ij}]\), the absolute value of this matrix, denoted as \(|A|\), is the matrix formed by taking the absolute value of each element within \(A\). So, if you have a matrix:
  • For an element \(a_{ij}\), the corresponding element in \(|A|\) is \(|a_{ij}|\).
This operation turns every element into its absolute value equivalent, creating a non-negative matrix. This is crucial when dealing with inequalities involving matrices because it standardizes the comparison of values across the matrix.
Matrix Addition
Matrix addition is an operation where two matrices of the same dimensions are added element-wise. For example, consider matrices \(A = [a_{ij}]\) and \(B = [b_{ij}]\), both of dimension \(n \times n\). The matrix sum \(A + B\) is calculated as:
  • \( (A + B)_{ij} = a_{ij} + b_{ij} \)
Each element of the resulting matrix \(A + B\) is simply the sum of the corresponding elements in matrices \(A\) and \(B\). When considering the inequality \(|A+B| \leq |A| + |B|\), it shows that the absolute value of the sum is not greater than the sum of individual absolute values. This is analogous to the triangle inequality in traditional arithmetic.
Matrix Scalar Multiplication
Scalar multiplication involves multiplying every element of a matrix by a scalar (a constant). If \(c\) is a scalar and \(A = [a_{ij}]\) is a matrix, then the scalar multiplication \(cA\) results in:
  • \((cA)_{ij} = c \cdot a_{ij}\)
Every element of the original matrix \(A\) is multiplied by \(c\). When considering the inequality \(|cA| = |c||A|\), it indicates that taking the absolute value of a scalar multiplied matrix equals multiplying the absolute value of the scalar by the absolute value of the matrix. This expression signifies that scalar multiplication distributes across the absolute value operation and is fundamental in understanding how scalars interact with matrices.
Matrix Vector Multiplication
Matrix-vector multiplication is a process where a matrix \(A\) multiplies a vector \(x\), producing another vector. If \(A = [a_{ij}]\) is an \(n \times n\) matrix and \(x\) is a vector in \(\mathbb{R}^n\), then the product \(Ax\) is a vector whose components come from the formula:
  • \((Ax)_i = \sum_{j=1}^{n} a_{ij} x_j\)
This operation involves taking dot products of the rows of the matrix with the vector \(x\). The inequality \(|Ax| \leq |A||x|\) shows how the result in terms of magnitude (absolute value) is controlled by the separate magnitudes of the matrix \(A\) and vector \(x\). This idea helps in establishing limit boundaries in practical applications, safeguarding the potential size of the multiplication result.

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Most popular questions from this chapter

It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>\) O.Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form $$\left[\begin{array}{ll} B & C \\ O & D\end{array}\right]$$. $$A=\left[\begin{array}{llll} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$

Let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\) be a complete set of eigenvalues (repetitions included) of the \(n \times n\) matrix \(A\). Prove that $$\begin{array}{l} \operatorname{det}(A)=\lambda_{1} \lambda_{2} \cdots \lambda_{n} \quad \text { and } \\ \operatorname{tr}(A)=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n} \end{array}$$ [Hint: The characteristic polynomial of \(A\) factors as \\[ \operatorname{det}(A-\lambda I)=(-1)^{n}\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right) \cdots\left(\lambda-\lambda_{n}\right) \\] Find the constant term and the coefficient of \(\lambda^{n-1}\) on the left and right sides of this equation.

Let \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\). (a) Prove that \(A\) is diagonalizable if \((a-d)^{2}+\) \(4 b c>0\) and is not diagonalizable if \((a-d)^{2}+\) \(4 b c<0\) (b) Find two examples to demonstrate that if \((a-d)^{2}+4 b c=0,\) then \(A\) may or may not be diagonalizable.

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of t.) $$\begin{array}{l} x^{\prime}=\quad y-z, \quad x(0)=1 \\ y^{\prime}=x+\quad z, \quad y(0)=0 \\ z^{\prime}=x+y, \quad z(0)=-1 \end{array}$$

Verify the Cayley-Hamilton Theorem for $$A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]$$ The Cayley-Hamilton Theorem can be used to calculate powers and inverses of matrices. For example, if \(A\) is a \(2 \times 2\) matrix with characteristic polynomial \(c_{A}(\lambda)=\lambda^{2}+a \lambda+b\) \(\operatorname{then} A^{2}+a A+b I=0,\) so $$A^{2}=-a A-b I$$ and $$\begin{aligned} A^{3} &=A A^{2}=A(-a A-b I) \\ &=-a A^{2}-b A \\ &=-a(-a A-b I)-b A \\ &=\left(a^{2}-b\right) A+a b I \end{aligned}$$ It is easy to see that by continuing in this fashion we can express any positive power of \(A\) as a linear combination of I and A. From \(A^{2}+a A+b I=O,\) we also obtain \(A(A+a I)=-b I, s o\) $$A^{-1}=-\frac{1}{b} A-\frac{a}{b} I$$ provided \(b \neq 0\).
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