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Chapter 1: Problem 6

If \(A\) and \(B\) are \(n \times n\) matrices, is \(A^{2}-B^{2}\) necessarily equal to \((A-B)(A+B)\) ? When must this be true?

Short Answer

Expert verified
Answer: No, \(A^2 - B^2\) is not necessarily equal to \((A-B)(A+B)\) in general. The equality holds if and only if the matrices A and B commute, i.e., \(AB = BA\).

Step by step solution

01

Expand \((A-B)(A+B)\)

To check if \(A^2 - B^2\) is equal to \((A-B)(A+B)\), we need to expand the latter expression using matrix multiplication rules. So, let's multiply the matrices \((A-B)\) and \((A+B)\):
02

Multiply (A - B) by (A + B)

Using distributive law, we can expand this product and multiply the matrices: \((A-B)(A+B) = A(A+B) - B(A+B)\) Now, apply the distributive property again to get: \(A^2 + AB - BA - B^2\)
03

Compare the expanded form with \(A^2 - B^2\)

Now, let's compare the expanded form \(A^2 + AB - BA - B^2\) with the given \(A^2 - B^2\). We can see that they are not equal in general, since there are extra terms \(AB\) and \(-BA\). However, there is a specific case where this is true - when \(AB = BA\).
04

State the condition for the equality to hold

The equality \(A^2 - B^2 = (A-B)(A+B)\) will hold if and only if \(AB = BA\). In other words, the matrices A and B must commute. This is the condition that must be met for the given equality to be true.

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Key Concepts

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

Distributive Property
Understanding the distributive property in matrix multiplication is essential for recognizing how to simplify complex matrix expressions. In plain terms, the distributive property allows us to break down matrix multiplication involving sums or differences into more manageable pieces.

For instance, when we have the expression \( (A+B)C \), we can distribute the multiplication over the matrices A and B, effectively transforming the expression into \( AC + BC \). Similarly, this property is applied in our exercise when expanding \( (A-B)(A+B) \).

By distributing the matrices, we reach the expanded form \( A(A+B) - B(A+B) \), which can be further broken down into \( A^2 + AB - BA - B^2 \), showing that each matrix is multiplied individually. This step is crucial but it also leads us to a fundamental property of matrix operations: matrices do not generally commute 鈥� more on this in the next section.
Commutative Matrices
In arithmetic, we are used to the idea that the order in which we multiply numbers doesn't affect the result. This is known as the commutative property of multiplication. However, this property does not hold true when it comes to matrix multiplication. Generally, for two matrices A and B, it is not true that \( AB = BA \).

Matrices that do follow this property are known as commutative matrices. Our exercise illustrates a particular situation in which understand whether two matrices commute is important. The product \( AB \) differs from \( BA \) except in specific cases, and if we want the equation \( A^2 - B^2 = (A-B)(A+B) \) to hold, we must be dealing with commutative matrices.

It's crucial, especially in higher-level mathematics and physics, to be aware of the commuting nature of the matrices involved in your calculations, as this can greatly affect the outcome.
Matrix Equality
When we talk about matrix equality, we are saying that two matrices are equal if and only if their corresponding entries are equal. In other words, for two matrices A and B to be considered equal, every entry \( a_{ij} \) in matrix A must be equal to the corresponding entry \( b_{ij} \) in matrix B.

In the context of our problem, the equation \( A^2 - B^2 = (A-B)(A+B) \) suggests a form of matrix equality. However, by expanding and comparing the two forms, we can see that the matrices \( A^2 - B^2 \) and \( A^2 + AB - BA - B^2 \) are not equal in general because of the extra terms \( AB \) and \( -BA \).

The concept of matrix equality is straightforward as long as one remembers to compare every individual element, which can be especially important when verifying the outcomes of matrix operations like multiplication.

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Most popular questions from this chapter

Solve for \(X\) : a) \(C+X=D\) b) \(F-5 X=G\).

(Complex case) For each of the following choices of matrix \(A\), find all cigenvalues and associated eigenvectors, and find a diagonal matrix \(B\) such that \(B=C^{-1} A C\) : a) \(\left[\begin{array}{rr}1 & -1 \\ 4 & 1\end{array}\right]\). b) \(\left[\begin{array}{rr}3 & -2 \\ 1 & 5\end{array}\right]\). c) \(\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 1 & -1\end{array}\right]\).

Let the linear mapping \(T\) have the equation \(\mathbf{y}=C \mathbf{x}\). For general \(\mathbf{x}\), find the angle between \(\mathbf{x}\) and \(T(\mathbf{x})=\mathbf{y}\), as vectors in \(V^{2}\), and also compare \(|\mathbf{x}|\) and \(|T(\mathbf{x})|\). From these results, interpret \(T\) geometrically. [Hint: Consider \(\mathbf{x}\) as \(\overrightarrow{O P}\) and \(\mathbf{y}\) as \(\overrightarrow{O Q}\), where \(O\) is the origin of \(E^{2}\) and \(P\) and \(Q\) are points of \(E^{2}\).]

In these problems the following matrices are given: $$ \begin{aligned} &A=\left[\begin{array}{l} 1 \\ 3 \end{array}\right], \quad B=\left[\begin{array}{l} 2 \\ 0 \end{array}\right], \quad C=\left[\begin{array}{ll} 2 & 3 \\ 4 & 1 \end{array}\right] . \quad D=\left[\begin{array}{rr} 1 & -1 \\ 2 & 0 \end{array}\right] . \quad E=\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right], \\ &F=\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 0 & 7 \end{array}\right], \quad G=\left[\begin{array}{rrr} 3 & 1 & 4 \\ -1 & 0 & -1 \end{array}\right], \quad H=(1,0,1), \quad J=(3,5,2), \quad K=(3,5), \\ &L=\left[\begin{array}{lll} 3 & 1 & 0 \\ 2 & 5 & 6 \\ 1 & 4 & 3 \end{array}\right], \quad M=\left[\begin{array}{rrr} 2 & -1 & 0 \\ 1 & 2 & 1 \\ 3 & 2 & -1 \end{array}\right], \quad N=\left[\begin{array}{ll} 1 & 4 \\ 0 & 3 \\ 7 & 1 \end{array}\right], \quad P=\left[\begin{array}{rr} 2 & 2 \\ -1 & -1 \\ 3 & 3 \end{array}\right] . \end{aligned} $$ a) Give the number of rows and columns for each of the matrices \(A, F, H, L\), and \(P\). b) Writing \(A=\left(a_{i j}\right), B=\left(b_{i j}\right)\), and so on, give the values of the following entries: \(a_{11}, a_{21}, c_{21}, c_{22}, d_{12}, e_{21}, f_{11}, g_{23}, g_{21}, h_{12}, m_{23}\). c) Give the row vectors of \(C, G, L\), and \(P\). d) Give the column vectors of \(D, F, L\), and \(N\).

Let \(A\) be a real square matrix, let \(\lambda\) be real, and let \(A \mathbf{v}=\lambda \mathbf{v}\) for a nonzero complex column vector \(\mathbf{v}\). Show that \(A \mathbf{u}=\lambda \mathbf{u}\) for a real nonzero vector \(\mathbf{u}\), so that \(\lambda\) is an eigenvalue of \(A\) considered as a real matrix. [Hint: Let \(\mathbf{v}=\mathbf{p}+i \mathbf{q}\). where \(\mathbf{p}\) and \(\mathbf{q}\) are real and not both zero. Show, with the aid of the results of Problem 10, that \(A \mathbf{p}=\lambda \mathbf{p}\) and \(A \mathbf{q}=\lambda \mathbf{q}\) and hence that \(\mathbf{u}\) can be chosen as one of \(\mathbf{p}, \mathbf{q}\).]
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