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Show that (a) if \(\lambda\) is an eigenvalue of an antisymmetric operator, then so is \(-\lambda\), and (b) antisymmetric operators (matrices) of odd dimension cannot be invertible.

Short Answer

Expert verified
For antisymmetric operators, if \(\lambda\) is an eigenvalue then \(-\lambda\) is also an eigenvalue. Antisymmetric operators of odd dimension cannot be invertible because they must have a zero eigenvalue.

Step by step solution

01

Define Antisymmetric Matrices

An antisymmetric (or skew-symmetric) matrix is a square matrix whose transpose equals its negative. That is, for a matrix A, \(A^{T} = -A\) holds true.
02

Understand Eigenvalues of an Operator

An eigenvalue of an operator (or matrix) is a scalar, \(\lambda\), such that when the operator acts on a non-zero vector (eigenvector), the vector is only scaled and not rotated. That is, \(A\mathbf{v} = \lambda\mathbf{v}\) holds true.
03

Eigenvalue is Negative

Suppose \(\lambda\) is an eigenvalue associated with an eigenvector \(\mathbf{v}\), of an antisymmetric operator A. Then, \(\mathbf{v}^T A \mathbf{v} = \lambda \mathbf{v}^T \mathbf{v}\). Taking the transpose of both sides leads to: \(\mathbf{v}^T A^T \mathbf{v} = \lambda \mathbf{v}^T \mathbf{v}\). Since A is antisymmetric, this implies: \(-\mathbf{v}^T A \mathbf{v} = \lambda \mathbf{v}^T \mathbf{v}\). Therefore, \(-\lambda\) is also an eigenvalue of A.
04

Antisymmetric Operator of Odd Dimension

Let's consider an odd-dimension antisymmetric matrix A of size \(n \times n\) where \(n\) is odd. The determinant of A is equal to the product of its eigenvalues. Due to the conclusion in part (a), each positive eigenvalue \(\lambda\) has a corresponding negative eigenvalue \(-\lambda\), and these yield a product of \(\lambda^2\), which is positive. However, because \(n\) is odd, there must be an extra eigenvalue which is zero, this is because eigenvalues come in negative and positive pairs. Therefore, det(A) = 0, which means A cannot be invertible.
05

Conclusion

This shows that firstly, if \(\lambda\) is an eigenvalue of an antisymmetric operator, \(-\lambda\) is also an eigenvalue. Secondly, antisymmetric operators (matrices) of odd dimension cannot be invertible due to having a zero eigenvalue.

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Key Concepts

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

Eigenvalues of Operators
In the realm of mathematics, particularly in linear algebra, the concept of eigenvalues is crucial when dealing with operators, which are often represented as matrices. But what exactly are eigenvalues? Imagine you have a transformation that squishes, stretches, flips, or rotates shapes in some fashion. Eigenvalues are special numbers that tell you how much the transformation stretches or squishes along particular directions, which are defined by vectors called eigenvectors.

For an operator or matrix A, if there is a vector v that is not the zero vector, and a scalar λ such that Av = λv, then λ is an eigenvalue of A, and v is the corresponding eigenvector. This relationship becomes fascinating for antisymmetric operators where each eigenvalue λ brings along a twin, , essentially reflecting the inherent symmetry of these operators. Understanding eigenvalues is fundamental as they play a key role in determining properties like stability, resonance, and the invertibility of operators.
Invertibility of Matrices
One might wonder, when faced with a square matrix, can it be inverted? The invertibility of a matrix is akin to asking whether a particular puzzle can be reversed to its starting position. For a matrix to be invertible, it must have an inverse matrix such that when the two are multiplied together, the result is the identity matrix.

The ability to find this inverse is heavily tied to the concept of eigenvalues. If a matrix has any eigenvalue that is zero, it's like having a puzzle piece that cannot return to its original position, rendering the entire matrix non-invertible. In the specific case of antisymmetric matrices of odd dimensions, the presence of a zero eigenvalue is inevitable, as demonstrated in our exercise solution; hence, such matrices cannot find their way back to the identity matrix, and thus, are not invertible.
Determinants of Matrices
The determinant of a matrix is a value that can be computed from its elements and has a wide array of applications. It serves as a mathematical indicator for various properties of the matrix, like volume distortion during transformation, and is crucial in solving systems of linear equations, finding inverses, and more.

Quite interestingly, the determinant can inform us whether a matrix is invertible or not. If the determinant is zero, the matrix cannot be inverted, in a similar vein to how a locked door with a missing key cannot be opened. The determinant of an antisymmetric matrix is particularly intriguing because if the matrix has an odd dimension, the determinant will always be zero. This zero determinant is a clear signal; no inverse will be found for this matrix, as illustrated by the exercise example where the odd-dimension antisymmetric matrix cannot escape its zero determinant fate.

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Most popular questions from this chapter

Let \(\pi\) be a permutation of the integers \(\\{1,2, \ldots, n\\} .\) Find the spectrum of \(\mathbf{A}_{\pi}\), if for \(|x\rangle=\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\right) \in \mathbb{C}^{n}\), we define $$\mathbf{A}_{\pi}|x\rangle=\left(\alpha_{\pi(1)}, \ldots, \alpha_{\pi(n)}\right) .$$

Show that \(\|\mathbf{A} x\|=\left\|\mathbf{A}^{\dagger} x\right\|\) if and only if \(\mathbf{A}\) is normal.

Three equal point masses are located at \((a, a, 0),(a, 0, a)\), and \((0, a, a) .\) Find the moment of inertia matrix as well as its eigenvalues and the corresponding eigenvectors.

Let \(\left|a_{1}\right\rangle \equiv \mathbf{a}_{1}=(1,1,-1)\) and \(\left|a_{2}\right\rangle \equiv \mathbf{a}_{2}=(-2,1,-1)\). (a) Construct (in the form of a matrix) the projection operators \(\mathbf{P}_{1}\) and \(\mathbf{P}_{2}\) that project onto the directions of \(\left|a_{1}\right\rangle\) and \(\left|a_{2}\right\rangle\), respectively. Verify that they are indeed projection operators. (b) Construct (in the form of a matrix) the operator \(\mathbf{P}=\mathbf{P}_{1}+\mathbf{P}_{2}\) and verify directly that it is a projection operator. (c) Let \(\mathbf{P}\) act on an arbitrary vector \((x, y, z)\). What is the dot product of the resulting vector with the vector \(\mathbf{a}_{1} \times \mathbf{a}_{2}\) ? Is that what you expect?

Show that a \(2 \times 2\) rotation matrix does not have a real eigenvalue (and, therefore, eigenvector) when the rotation angle is not an integer multiple of \(\pi\). What is the physical interpretation of this?

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