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Prove that for any two matrices \(A\) and \(B\), where \(A\) has as many columns as \(B\) has rows, the transpose of \(A B\) satisfies \((A B)=\tilde{B} \tilde{A}\)

Short Answer

Expert verified
The transpose of a matrix product \((AB)^T\) equals \(B^T A^T\).

Step by step solution

01

Understand the Problem

We need to show that the transpose of the product of two matrices, \(A\) and \(B\), is equal to the product of their transposes in reverse order, i.e., \((AB)^T = B^T A^T\).
02

Definition of Matrix Transpose

Recall that the transpose of a matrix \(C\), denoted \(C^T\), is formed by swapping its rows with columns. Therefore, if \(C = AB\), then \(C^T\) will have its \((i,j)\)-th element equal to the \((j,i)\)-th element of \(C\).
03

Expression for Matrix Product

The element \((i, j)\) of the matrix product \(AB\) is given by \((AB)_{ij} = \sum_{k} a_{ik} b_{kj}\).
04

Transpose of Product \((AB)^T\)

The element \((j, i)\) of \((AB)^T\) is \((AB)_{ij}\)'s transpose, which yields \((AB)^T_{ji} = (AB)_{ij} = \sum_{k} a_{ik} b_{kj}\).
05

Expression for Transposed Product \(B^T A^T\)

The element \((j, i)\) of \(B^T A^T\) is \(\sum_{k} (B^T)_{jk} (A^T)_{ki} = \sum_{k} b_{kj} a_{ik}\).
06

Comparing Transpose Expressions

Compare \( (AB)^T_{ji} = \sum_{k} a_{ik} b_{kj} \) and \( (B^T A^T)_{ji} = \sum_{k} b_{kj} a_{ik} \). They are identical, showing that \( (AB)^T = B^T A^T \).
07

Conclusion

Since both expressions for \((AB)^T\) and \(B^T A^T\) are equal for any matrix element \((j, i)\), we conclude that \((AB)^T = B^T A^T\), proving the statement.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra, allowing us to combine matrices in meaningful ways. It involves combining rows of the first matrix with columns of the second matrix. To multiply two matrices, you need to satisfy the condition where the number of columns in the first matrix equals the number of rows in the second matrix.

When multiplying matrices, every element of the resulting matrix is computed through a dot product. This means, for each element in the resulting matrix, you take the corresponding row from the first matrix and the corresponding column from the second matrix, multiply them element-wise, and sum up the results.
  • For example, if you are multiplying a 2x3 matrix A by a 3x2 matrix B, the resulting matrix will be a 2x2 matrix.
  • The element in the first row and first column of the resulting matrix is determined by multiplying the first row of A by the first column of B.
Understanding matrix multiplication is crucial because it is the groundwork for many other operations in linear algebra, such as finding the determinant, inverse, and solving systems of equations.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations in vector spaces and matrices. It provides an essential framework for solving various mathematical problems and is heavily used in fields like computer science, physics, and engineering.

Several key concepts are part of linear algebra:
  • Vectors and Vector Spaces: A vector is a quantity defined by both magnitude and direction, and a vector space is a collection of vectors.
  • Matrices and Matrix Operations: Matrices represent linear transformations and can be added, subtracted, and multiplied.
  • Determinants and Inverses: The determinant provides information about the matrix's properties, and the inverse helps in solving matrix equations.
Linear algebra's power lies in its ability to abstract and simplify complex problems by allowing the use of matrix operations and transformations.
Mathematical Proof
Mathematical proofs are logical arguments that verify the truth of mathematical statements. Proofs are the bedrock of mathematics, ensuring that the results we use and trust are correct. In the context of matrix theory, you often need to prove properties and theorems that apply to matrices and operations like transposition and multiplication.

Consider the proof task: showing that the transpose of a matrix product equals the product of transposes in reverse order, i.e., \((AB)^T = B^T A^T\). This involves several steps:
  • Understanding the definitions and properties of matrix transpose and multiplication.
  • Deriving general expressions for both \((AB)^T\) and \(B^T A^T\).
  • Logical comparison of expressions to show equivalence.
Good proofs check every logical step, ensuring that the derived expressions truly match, validating the theorem's claims beyond simple intuition.

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Most popular questions from this chapter

A particle of mass \(m_{a}\) decays at rest into two identical particles each of mass \(m_{b} .\) Use conservation of momentum and energy to find the speed of the outgoing particles.

Two particles \(a\) and \(b\) with masses \(m_{a}=0\) and \(m_{b}>0\) approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.]

A rocket traveling at speed \(\frac{1}{2} c\) relative to frame \(\mathcal{S}\) shoots forward bullets traveling at speed \(\frac{3}{4} c\) relative to the rocket. What is the speed of the bullets relative to \(\mathcal{S} ?\)

Let \(\Lambda_{\mathrm{B}}(\theta)\) denote the 4 \(\times 4\) matrix that gives a pure boost in the direction that makes an angle \(\theta\) with the \(x_{1}\) axis in the \(x_{1} x_{2}\) plane. Explain why this can be found as \(\Lambda_{\mathrm{B}}(\theta)=\Lambda_{\mathrm{R}}(-\theta) \Lambda_{\mathrm{B}}(0) \Lambda_{\mathrm{R}}(\theta)\) where \(\Lambda_{\mathrm{R}}(\theta)\) denotes the matrix that rotates the \(x_{1} x_{2}\) plane through angle \(\theta\) and \(\Lambda_{\mathrm{B}}(0)\) is the standard boost along the \(x_{1}\) axis. Use this result to find \(\Lambda_{\mathrm{B}}(\theta)\) and check your result by finding the motion of the spatial origin of the frame \(\mathcal{S}\) as observed in \(\mathcal{S}^{\prime}\).

As measured in the inertial frame \(\mathcal{S},\) a proton has four-momentum \(p\). Also as measured in \(\mathcal{S}\), an observer at rest in a frame \(\mathcal{S}^{\prime}\) has four-velocity \(u\). Show that the proton's energy, as measured by this observer, is - \(u \cdot p\).

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