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Chapter 1: Problem 3

Give an example where \(r(\mathbf{A B}) \neq r(\mathbf{B A})\). (Hint: Try some \(2 \times 2\) matrices.)

Short Answer

Expert verified
Example: \(\mathbf{A} = \begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix}\), \(\mathbf{B} = \begin{pmatrix} 1 & 0 \ 1 & 0 \end{pmatrix}\); \(\mathbf{r(A B)} = 1\) and \(\mathbf{r(B A)} = 2\).

Step by step solution

01

- Define Matrix A

Choose a simple 2x2 matrix for Matrix A. For example, let \(\mathbf{A} = \begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix}\).
02

- Define Matrix B

Choose another 2x2 matrix for Matrix B. For example, let \(\mathbf{B} = \begin{pmatrix} 1 & 0 \ 1 & 0 \end{pmatrix}\).
03

- Calculate AB

Multiply \(\mathbf{A}\) and \(\mathbf{B}\) to find \(\mathbf{A B}\). \(\mathbf{A} \mathbf{B} = \begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \ 0 & 0 \end{pmatrix}\). The rank of \(\mathbf{A B}\) is 1.
04

- Calculate BA

Next, multiply \(\mathbf{B}\) and \(\mathbf{A}\) to find \(\mathbf{B A}\). \(\mathbf{B} \mathbf{A} = \begin{pmatrix} 1 & 0 \ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix}\). The rank of \(\mathbf{B A}\) is 2.
05

- Compare Ranks

Observe the ranks calculated in Steps 3 and 4. \(\mathbf{r(A B)} = 1 \) and \(\mathbf{r(B A)} = 2 \). Therefore, \(\mathbf{r(A B)} eq \mathbf{r(B A)}\).

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Key Concepts

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

Matrix Rank
Matrix rank is a fundamental concept in linear algebra. The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. For example, the rank of a matrix can tell us about the solutions to a system of linear equations represented by that matrix. In our exercise, we found that \(\text{rank}(\textbf{A B}) = 1\) and \(\text{rank}(\textbf{B A}) = 2\). This difference in ranks highlights an important characteristics of matrix multiplication, which we will explore further later in this article.
Linear Algebra
Linear algebra is a branch of mathematics that studies vectors, vector spaces (also known as linear spaces), linear transformations and systems of linear equations. Matrix multiplication is a central operation in this field. When we talk about matrices, we're dealing with arrays of numbers that can represent linear equations, transformations, layers in neural networks, or countless other things.
In the given exercise, we are utilizing simple 2x2 matrices to demonstrate the concept of matrix rank in a straightforward manner. Understanding how these matrices interact through multiplication is crucial for deeper topics in linear algebra such as eigenvalues, eigenvectors, and determinants.
Matrix Operations
Matrix operations include addition, subtraction, and most importantly for our exercise, multiplication. Matrix multiplication is unique because it is not commutative, meaning that the order in which you multiply matrices matters.
To multiply two matrices, you take the dot product of rows and columns. For instance, if matrix \(\textbf{A}\) is multiplied by matrix \(\textbf{B}\), the element in the first row and the first column of the product matrix \(\textbf{A} \textbf{B}\) is the dot product of the first row of \(\textbf{A}\) and the first column of \(\textbf{B}\). We follow the steps of the exercise to find the product matrices \(\textbf{A B}\) and \(\textbf{B A}\) which resulted in different matrices, leading to different ranks. These results demonstrate that \(\textbf{r(A B)} eq \textbf{r(B A)}\), emphasizing that the placement and order of matrices in operations are critical.

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Most popular questions from this chapter

\(\mathbf{A}=\left(\begin{array}{lll}a & b & c \\ b & d & e \\ c & e & f\end{array}\right)\) has the eigenvectors \(\mathbf{v}_{1}=\left(\begin{array}{r}1 \\ 0 \\ -1\end{array}\right), \mathbf{v}_{2}=\left(\begin{array}{l}1 \\ 2 \\ 1\end{array}\right), \mathbf{v}_{3}=\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right)\) with associated eigenvalues \(\lambda_{1}=3, \lambda_{2}=1\), and \(\lambda_{3}=4 .\) Determine the matrix \(\mathbf{A}\).

Determine the ranks of the following matrices for all values of the parameters: (a) \(\left(\begin{array}{ccc}x & 0 & x^{2}-2 \\ 0 & 1 & 1 \\ -1 & x & x-1\end{array}\right)\) (b) \(\left(\begin{array}{ccc}t+3 & 5 & 6 \\ -1 & t-3 & -6 \\ 1 & 1 & t+4\end{array}\right)\) (c) \(\left(\begin{array}{cccc}1 & x & y & 0 \\ 0 & z & w & 1 \\ 1 & x & y & 0 \\ 0 & z & w & 1\end{array}\right)\)

Let \(\mathbf{A x}=\mathbf{b}\) be a linear system of equations in matrix form. Prove that if \(\mathbf{x}_{1}\) and \(\mathrm{x}_{2}\) are both solutions of the system, then so is \(\lambda \mathbf{x}_{1}+(1-\lambda) \mathbf{x}_{2}\) for every real number \(\lambda\). Use this fact to prove that a linear system of equations that is consistent has either one solution or infinitely many solutions. (For instance, it cannot have exactly 3 solutions.)

Use partitioning to compute the inverses of the following matrices: (a) \(\left(\begin{array}{lllll}1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1\end{array}\right)\) (b) \(\left(\begin{array}{llll}2 & 3 & 0 & 0 \\ 5 & 2 & 0 & 0 \\ 0 & 0 & 4 & 3 \\ 0 & 0 & 3 & 2\end{array}\right)\) (c) \(\left(\begin{array}{llll}2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0\end{array}\right)\)

(a) Suppose \(\mathbf{A}\) is \(n \times m\) and \(\mathbf{B}\) is \(m \times n .\) Prove that \(\left|\mathbf{I}_{n}+\mathbf{A B}\right|=\left|\mathbf{I}_{m}+\mathbf{B} \mathbf{A}\right| .\) (Hint: Define \(\mathbf{D}=\left(\begin{array}{cc}\mathbf{I}_{n} & \mathbf{A} \\ \mathbf{0} & \mathbf{I}_{m}\end{array}\right), \mathbf{E}=\left(\begin{array}{cc}\mathbf{I}_{n} & -\mathbf{A} \\ \mathbf{B} & \mathbf{I}_{m}\end{array}\right) .\) Then \(\left|\mathbf{I}_{n}+\mathbf{A} \mathbf{B}\right|=|\mathbf{D} \mathbf{E}|=|\mathbf{D}||\mathbf{E}|=\) \(\left.|\mathbf{E}||\mathbf{D}|=|\mathbf{E D}|=\left|\mathbf{I}_{m}+\mathbf{B} \mathbf{A}\right| .\right)\) (b) Use the result in (a) to prove that if \(a_{1}, \ldots, a_{n}\) are all different from 1 , then $$ \left|\begin{array}{ccccc} a_{1} & 1 & 1 & \ldots & 1 \\ 1 & a_{2} & 1 & \ldots & 1 \\ 1 & 1 & a_{3} & \ldots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \ldots & a_{n} \end{array}\right|=\left(a_{1}-1\right)\left(a_{2}-1\right) \cdots\left(a_{n}-1\right)\left[1+\sum_{i=1}^{n} \frac{1}{a_{i}-1}\right] $$ (Hint: Let \(\mathbf{F}=\mathbf{I}_{n}+\mathbf{A}_{n \times 1} \mathbf{B}_{1 \times n}\) where \(\mathbf{A}_{n \times 1}=\left(\frac{1}{a_{1}-1}, \frac{1}{a_{2}-1}, \cdots, \frac{1}{a_{n}-1}\right)^{\prime}\) and \(\left.\mathbf{B}_{1 \times n}=(1,1, \ldots, 1) .\right)\)
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