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Chapter 2: Problem 6

Prove or give a counterexample. Assume all the matrices are \(n \times n\). a. If \(A B=C B\) and \(B \neq \mathrm{O}\), then \(A=C\). b. If \(A^{2}=A\), then \(A=\mathrm{O}\) or \(A=I\). c. \((A+B)(A-B)=A^{2}-B^{2}\). d. If \(A B=C B\) and \(B\) is nonsingular, then \(A=C\). e. If \(A B=B C\) and \(B\) is nonsingular, then \(A=C\).

Short Answer

Expert verified
a. Counterexample: \(AB = CB\), but \(A \neq C\). b. Counterexample: \(A^2=A\), but \(A \neq \mathrm{O}\) and \(A \neq I\). c. Counterexample: \((A+B)(A-B) \neq A^2-B^2\). d. True, proof: Since \(B\) is nonsingular, \(BB^{-1}=I\), thus \(AI=CI\), which implies \(A=C\). e. True, proof: Since \(B\) is nonsingular, \(BB^{-1}=I\), thus \(AI=CI\), which implies \(A=C\).

Step by step solution

01

a. If \(AB=CB\) and \(B \neq \mathrm{O}\), then \(A=C\).\

We cannot directly prove this statement for all matrices, so we need to provide a counterexample. Let's consider the following matrices: \(A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\). Clearly, \(B \neq \mathrm{O}\). Now let's check if \(AB = CB\): \(A B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \(C B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\). Hence, we have found a counterexample: \(AB = CB\), but \(A \neq C\).
02

b. If \(A^2=A\), then \(A=\mathrm{O}\) or \(A=I\).\

Let's provide a counterexample for this as well: Consider the following matrix \(A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\). Now, let's compute \(A^2\): \(A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = A\). Thus, we have a counterexample: \(A^2=A\), but \(A \neq \mathrm{O}\) and \(A \neq I\).
03

c. \((A+B)(A-B)=A^2-B^2\).\

Let's demonstrate that this statement is false with a counterexample. Consider two arbitrary matrices \(A\) and \(B\): \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\). Now, compute \((A+B)(A-B)\): \((A+B)(A-B)= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix}\). And compute \(A^2-B^2\): \(A^2 - B^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\). As we can see, \((A+B)(A-B) \neq A^2-B^2\).
04

d. If \(AB=CB\) and \(B\) is nonsingular, then \(A=C\).\

This statement is true. Let's prove it: Since \(B\) is nonsingular, it means that \(B^{-1}\) exists. Now, multiply both sides of the equation \(AB=CB\) by \(B^{-1}\) from the right: $$A(BB^{-1})=C(BB^{-1}).$$ According to the definition of the inverse matrix, \(BB^{-1}=I\). Consequently, $$AI=CI,$$ which implies \(A=C\).
05

e. If \(AB=BC\) and \(B\) is nonsingular, then \(A=C\).\

This statement is also true. Let's prove it: Since \(B\) is nonsingular, it means that there exists an inverse matrix \(B^{-1}\). Now, multiply both sides of the equation \(AB=BC\) by the inverse matrix \(B^{-1}\) from the right: $$A(BB^{-1})=C(BB^{-1}).$$ As we know, the inverse matrix of \(B\) ensures that \(BB^{-1}=I\). Consequently, $$AI=CI,$$ which implies \(A=C\).

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Key Concepts

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

Counterexample in Linear Algebra
In linear algebra, a counterexample is used to disprove a statement by providing an instance where the statement does not hold true. This method is especially powerful because it only requires one valid case to show that a statement is not universally true, irrespective of how many instances it works for. For example, consider the assertion that if two matrices multiplied by a third matrix result in the same product, then the two matrices must be equal. To disprove this, we can construct specific matrices where the products are equal, but the matrices themselves are not. The exercise provides such an example with matrices A, B, and C, where AB = CB yet A ≠ C. This clearly illustrates that the original claim does not hold for all cases in linear algebra.

Using counterexamples is a fundamental aspect of mathematical proof and is crucial for developing a deep understanding of algebraic concepts, as it encourages critical thinking and a thorough examination of the conditions under which a mathematical statement is valid.
Nonsingular Matrix
A nonsingular matrix鈥攁lso known as an invertible or nondegenerate matrix鈥攊s central to many operations in linear algebra because it means that the matrix has an inverse. In practical terms, a nonsingular matrix B does not compress different inputs into the same outputs when it transforms vectors. The implications of a matrix being nonsingular are significant: if we multiply a nonsingular matrix by another matrix and get the same result as when we multiply it by a third matrix, we can infer that the latter two matrices must be equal. The existence of an inverse allows us to 'cancel' the nonsingular matrix out of an equation, simplifying the expression and often leading to a proof of matrix equality, as seen in the exercises. In step four, the statement 'if AB = CB and B is nonsingular, then A = C' is shown to be true through such a process. The inverse property simplifies algebraic manipulation and is a powerful tool in solving linear equations and understanding system behavior.
Matrix Equality
The concept of matrix equality is a fundamental one: two matrices are considered equal if and only if every corresponding entry in one matrix is equal to the corresponding entry in the other. In the context of equation solving, this may seem straightforward, but it becomes more complex in proofs or when dealing with matrix product relationships. In the exercise regarding the equality A = C, when pre-multiplied or post-multiplied by a nonsingular matrix B, we see that this equality holds under specific conditions. When matrices are not equal, even if their products are the same (as in the counterexample), the situation underscores the importance of considering matrix properties like singularity or nonsingularity. In matrix theory, equality is robust; it does not hold if even one element differs between the matrices. This precision makes working with matrices particularly exacting, but also powerful as a language and tool within linear algebra.

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Most popular questions from this chapter

If \(P\) is a permutation matrix (see Exercise \(2.1 .12\) for the definition), show that \(P\) is invertible and find \(P^{-1}\).

We say an \(n \times n\) matrix \(A\) is orthogonal if \(A^{\top} A=I_{n}\). a. Prove that the column vectors \(\mathbf{a}_{1}, \ldots, \mathbf{a}_{n}\) of an orthogonal matrix \(A\) are unit vectors that are orthogonal to one another, i.e., \(\mathbf{a}_{i} \cdot \mathbf{a}_{j}= \begin{cases}1, & i=j \\ 0, & i \neq j\end{cases}\) b. Fill in the missing columns in the following matrices to make them orthogonal: $$ \left[\begin{array}{cc} \frac{\sqrt{3}}{2} & ? \\ -\frac{1}{2} & ? \end{array}\right],\left[\begin{array}{rrr} 1 & 0 & ? \\ 0 & -1 & ? \\ 0 & 0 & ? \end{array}\right], \quad\left[\begin{array}{rrr} \frac{1}{3} & ? & \frac{2}{3} \\ \frac{2}{3} & ? & -\frac{2}{3} \\ \frac{2}{3} & ? & \frac{1}{3} \end{array}\right] $$ c. Show that any \(2 \times 2\) orthogonal matrix \(A\) must be of the form $$ \left[\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \text { or }\left[\begin{array}{rr} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right] $$ for some real number \(\theta\). (Hint: Use part \(a\), rather than the original definition.) *d. Show that if \(A\) is an orthogonal \(2 \times 2\) matrix, then \(\mu_{A}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is either a rotation or the composition of a rotation and a reflection. e. Prove that the row vectors \(\mathbf{A}_{1}, \ldots, \mathbf{A}_{n}\) of an orthogonal matrix \(A\) are unit vectors that are orthogonal to one another. (Hint: Corollary 3.3.)

Suppose \(A\) is an invertible matrix and \(A^{-1}\) is known. a. Suppose \(B\) is obtained from \(A\) by switching two columns. How can we find \(B^{-1}\) from \(A^{-1}\) ? (Hint: Since \(A^{-1} A=I\), we know the dot products of the rows of \(A^{-1}\) with the columns of \(A\). So rearranging the columns of \(A\) to make \(B\), we should be able to suitably rearrange the rows of \(A^{-1}\) to make \(B^{-1}\).) b. Suppose \(B\) is obtained from \(A\) by multiplying the \(j^{\text {th }}\) column by a nonzero scalar. How can we find \(B^{-1}\) from \(A^{-1}\) ? c. Suppose \(B\) is obtained from \(A\) by adding a scalar multiple of one column to another. How can we find \(B^{-1}\) from \(A^{-1}\) ? d. Suppose \(B\) is obtained from \(A\) by replacing the \(j^{\text {th }}\) column by a different vector. Assuming \(B\) is still invertible, how can we find \(B^{-1}\) from \(A^{-1}\) ?

Prove or give a counterexample: Every invertible matrix can be written as a product of elementary matrices.

Show that the inverse of every elementary matrix is again an elementary matrix. Indeed, give a simple prescription for determining the inverse of each type of elementary matrix. (See the proof of Theorem \(4.1\) of Chapter 1.)
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