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A column matrix a has components \(a_{x}, a_{y}, a_{z}\) and \(\mathrm{A}\) is the matrix with elements \(A_{i j}=-\epsilon_{i j k} a_{k}\) (a) What is the relationship between column matrices \(\mathrm{b}\) and \(\mathrm{c}\) if \(\mathrm{Ab}=\mathrm{c}\) ? (b) Find the eigenvalues of \(A\) and show that a is one of its eigenvectors. Explain why this must be so.

Short Answer

Expert verified
The relationship is \( \mathbf{c} = \mathbf{a} \times \mathbf{b}. \) The eigenvalues include 0, and \( \mathbf{a} \) is an eigenvector.

Step by step solution

01

Understand the given matrices

Given is a column matrix \( \mathbf{a} = \begin{bmatrix} a_x \ a_y \ a_z \end{bmatrix} \) and a matrix \( \mathbf{A} \) with elements \( A_{ij} = -\epsilon_{ijk} a_k. \) Here \( \epsilon_{ijk} \) is the Levi-Civita symbol.
02

Matrix multiplication for part (a)

For part (a), \( \mathbf{A} \mathbf{b} = \mathbf{c} \) means \( c_i = A_{ij} b_j = -\epsilon_{ijk} a_k b_j. \) This represents a cross product in matrix form.
03

Relationship between b and c

Given the form \( c_i = -\epsilon_{ijk} a_k b_j, \) it implies \( \mathbf{c} = \mathbf{a} \times \mathbf{b}. \) Thus, \( \mathbf{c} \) is the cross product of \( \mathbf{a} \) and \( \mathbf{b}. \)
04

Eigenvalue equation for part (b)

The eigenvalue equation \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \) is used to find the eigenvalues. Since \( \mathbf{A} \mathbf{a} = \mathbf{0}, \) \( \mathbf{a} \) is an eigenvector associated with the eigenvalue \( \lambda = 0. \)
05

Show that \( \mathbf{a} \) is an eigenvector

Since \( A_{ij} a_j = -\epsilon_{ijk} a_k a_j = 0 \) (as the Levi-Civita symbol gives zero when two indices are the same), \( \mathbf{a} \) is indeed an eigenvector of \( \mathbf{A} \) corresponding to the eigenvalue 0.

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Key Concepts

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

Column Matrices
Column matrices are simply matrices with a single column. Each element in this column is called a component. For example, if we have a column matrix \( \mathbf{a} = \begin{bmatrix} a_x \ a_y \ a_z \end{bmatrix} \), \(a_x\), \(a_y\), and \(a_z\) are its components.
Understanding column matrices is essential because they represent vectors in higher dimensional spaces. In physical contexts, like in our problem, column matrices are typically used to represent quantities like position, velocity, or other vector quantities.
They are particularly useful in matrix multiplication, where they interact neatly with other matrices to form new vectors. In the matrix multiplication \( \mathbf{Ab} = \mathbf{c} \), \(\mathbf{A}\) represents a transformation that changes the vector \(\mathbf{b}\) into another vector \(\mathbf{c}\).
Levi-Civita Symbol
The Levi-Civita symbol, denoted as \(\epsilon_{ijk}\), is a mathematical symbol used primarily in tensor calculus, especially in three dimensions.
It is defined as:
  • \( \epsilon_{ijk} = +1 \) if \( (i,j,k) \) is an even permutation of (1,2,3)
  • \( \epsilon_{ijk} = -1 \) if \( (i,j,k) \) is an odd permutation of (1,2,3)
  • \( \epsilon_{ijk} = 0 \) if any two of \(i, j, k\) are the same.
The symbol is incredibly useful in defining the cross product operation in three dimensions through its relationship with the components of the cross product. In our exercise, the matrix \( \mathbf{A} \) has elements defined by \( A_{ij} = -\epsilon_{ijk} a_k \). This structure exploits the properties of the Levi-Civita symbol to embed the cross product operations within the matrix multiplication process.
Cross Product
The cross product is an operation on two three-dimensional vectors. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). The components of this resultant vector \( \mathbf{c} \) can be expressed using the Levi-Civita symbol as: \( c_i = -\epsilon_{ijk} a_k b_j \).
This means that matrix \( \mathbf{Ab} = \mathbf{c} \) translates to \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \), showing that \( \mathbf{A} \) effectively encodes the cross product operation. The cross product is significant because of its geometric and physical interpretations. For example, it's frequently used in physics to compute torque, angular momentum, and in expressions involving the magnetic force on a moving charge.
Eigenvalues
Eigenvalues are special numbers associated with a matrix that provide insight into its properties, especially in terms of its transformations. For a matrix \( \mathbf{A} \) and eigenvector \( \mathbf{v} \), the equation \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \) holds, where \(\lambda\) is the eigenvalue.
In our problem, since \( \mathbf{A} \mathbf{a} = \mathbf{0} \), we recognize \( \mathbf{a} \) as an eigenvector associated with the eigenvalue \( \lambda = 0 \).
Eigenvalues are crucial in many applications because they reveal important characteristics of the matrix. For instance, if the eigenvalue is zero, as in this case, it indicates that the matrix \( \mathbf{A} \) sends some vector to the zero vector, showcasing dependencies in the system. These concepts are widely applied in physics, engineering, and data science for simplifying complex problems and understanding dynamic systems.

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Most popular questions from this chapter

The components of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) and a second-order tensor \(\mathbf{T}\) are given in one coordinate system by $$ \mathrm{A}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right), \quad \mathrm{B}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right), \quad \mathrm{T}=\left(\begin{array}{ccc} 2 & \sqrt{3} & 0 \\ \sqrt{3} & 4 & 0 \\ 0 & 0 & 2 \end{array}\right) $$ In a second coordinate system, obtained from the first by rotation, the components of \(\mathbf{A}\) and \(\mathbf{B}\) are $$ \mathrm{A}^{\prime}=\frac{1}{2}\left(\begin{array}{c} \sqrt{3} \\ 0 \\ 1 \end{array}\right), \quad \mathrm{B}^{\prime}=\frac{1}{2}\left(\begin{array}{c} -1 \\ 0 \\ \sqrt{3} \end{array}\right) $$ Find the components of \(\mathbf{T}\) in this new coordinate system and hence evaluate, with a minimum of calculation, $$ T_{i j} T_{j i}, \quad T_{k i} T_{j k} T_{i j}, \quad T_{i k} T_{m n} T_{n i} T_{k m} $$

Use tensor methods to establish the following vector identities: (a) \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}=(\mathbf{u} \cdot \mathbf{w}) \mathbf{v}-(\mathbf{v} \cdot \mathbf{w}) \mathbf{u}\) (b) \(\operatorname{curl}(\phi \mathbf{u})=\phi \operatorname{curl} \mathbf{u}+(\operatorname{grad} \phi) \times \mathbf{u}\) (c) \(\operatorname{div}(\mathbf{u} \times \mathbf{v})=\mathbf{v} \cdot \operatorname{curl} \mathbf{u}-\mathbf{u} \cdot \operatorname{curl} \mathbf{v}\) (d) \(\operatorname{curl}(\mathbf{u} \times \mathbf{v})=(\mathbf{v} \cdot \mathbf{g r a d}) \mathbf{u}-(\mathbf{u} \cdot \mathbf{g r a d}) \mathbf{v}+\mathbf{u} \operatorname{div} \mathbf{v}-\mathbf{v} \operatorname{div} \mathbf{u}\) (e) \(\operatorname{grad} \frac{1}{2}(\mathbf{u} \cdot \mathbf{u})=\mathbf{u} \times \operatorname{curl} \mathbf{u}+(\mathbf{u} \cdot \operatorname{grad}) \mathbf{u}\).

In a general coordinate system \(u^{i}, i=1,2,3\), in three-dimensional Euclidean space, a volume element is given by $$ d V=\left|\mathbf{e}_{1} d u^{1} \cdot\left(\mathbf{e}_{2} d u^{2} \times \mathbf{e}_{3} d u^{3}\right)\right| $$ Show that an alternative form for this expression, written in terms of the determinant \(g\) of the metric tensor, is given by $$ d V=\sqrt{g} d u^{1} d u^{2} d u^{3} $$ Show that under a general coordinate transformation to a new coordinate system \(u^{i}\) the volume element \(d V\) remains unchanged, i.e. show that it is a scalar quantity.

In a certain crystal the unit cell can be taken as six identical atoms lying at the corners of a regular octahedron. Convince yourself that these atoms can also be considered as lying at the centres of the faces of a cube and hence that the crystal has cubic symmetry. Use this result to prove that the conductivity tensor for the crystal, \(\sigma_{i j}\), must be isotropic.

The paramagnetic tensor \(\chi_{i j}\) of a body placed in a magnetic field, in which its energy density is \(-\frac{1}{2} \mu_{0} \mathbf{M} \cdot \mathbf{H}\) with \(M_{i}=\sum_{j} \chi_{i j} H_{j}\), is $$ \left(\begin{array}{ccc} 2 k & 0 & 0 \\ 0 & 3 k & k \\ 0 & k & 3 k \end{array}\right) $$ Assuming depolarizing effects are negligible, find how the body will orientate itself if the field is horizontal, in the following circumstances: (a) the body can rotate freely; (b) the body is suspended with the \((1,0,0)\) axis vertical; (c) the body is suspended with the \((0,1,0)\) axis vertical.

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