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Show that if \([[\mathbf{A}, \mathbf{B}], \mathbf{A}]=\mathbf{0}\), then for every positive integer \(k\). $$ \left[\mathbf{A}^{k}, \mathbf{B}\right]=k \mathbf{A}^{k-1}[\mathbf{A}, \mathbf{B}] $$

Short Answer

Expert verified
The proof uses the properties of matrix operations and the concept of mathematical induction to prove the given identity.

Step by step solution

01

Start with the left side of the identity to be proven

We should consider the left side \([\mathbf{A}^{k}, \mathbf{B}]\) of the identity to be proven first and see if we can manipulate it into the form of right side \(k \mathbf{A}^{k-1}[\mathbf{A},\mathbf{B}]\).
02

Expand the terms

We will start with expanding the term \(\mathbf{A}^{k}\) in the commutator \([\mathbf{A}^{k}, \mathbf{B}]\). Using the binary property of the commutator, we will find \([\mathbf{A}^{k}, \mathbf{B}] = \mathbf{A}^{k} \mathbf{B} - \mathbf{B} \mathbf{A}^{k}\).
03

Apply mathematical induction

To get the form of the right side, we need to apply mathematical induction:a) Base case: Verify if \(k=1\) holds true. In that case, both sides evaluate to \([\mathbf{A}, \mathbf{B}]\), so the identity is true for \(k=1\).b) Induction hypothesis: Assume that the identity holds true for some positive \(k=j\), i.e., \([\mathbf{A}^{j}, \mathbf{B}]=j \mathbf{A}^{j-1} [\mathbf{A}, \mathbf{B}]\).c) Inductive step: Prove the identity for \(k=j+1\) by substituting \(k\) with \(j+1\) and applying the commutator properties to manipulate the form similar to the right hand side of the identity.
04

Simplify the expression

Substitute \(\mathbf{A}^{j}\) in the expanded form obtained in step 2 and simplify using given condition \([[\mathbf{A}, \mathbf{B}], \mathbf{A}]=\mathbf{0}\).
05

Conclusion

In showing that the equation holds for the inductive step when replacing \(k\) with \(j+1\), and knowing it holds for the base case, it is proven that the equation, \([\mathbf{A}^{k}, \mathbf{B}] = k \mathbf{A}^{k-1} [\mathbf{A}, \mathbf{B}]\), is true for all positive integers \(k\) through mathematical induction.

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Key Concepts

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

Commutator
In mathematics, the commutator of two elements \( \mathbf{A} \) and \( \mathbf{B} \) in a group is denoted as \([\mathbf{A}, \mathbf{B}]\) and is defined as \( \mathbf{A} \mathbf{B} - \mathbf{B} \mathbf{A} \). Commutators are significant in various areas like quantum mechanics, algebra, and Lie groups. They show how much \( \mathbf{A} \) and \( \mathbf{B} \) "fail" to commute; in simple terms, it measures the difference in the result when applying two operations in different orders.
  • If \([\mathbf{A}, \mathbf{B}] = 0\), \( \mathbf{A} \) and \( \mathbf{B} \) commute perfectly.
  • Non-zero commutators can indicate transformation properties or characteristics of mathematical structures.
Understanding commutators is crucial when dealing with operators, especially in fields like physics, where they help describe quantum states.
Binary Property
The binary property in relation to commutators refers to how the expression \([\mathbf{A}, \mathbf{B}] = \mathbf{A} \mathbf{B} - \mathbf{B} \mathbf{A}\) uses the subtraction of two products. This binary operation involves two operands, which in this context are tensors, matrices, or linear operators.
  • It highlights how a product operation of elements results in a new element, not necessarily within the same "kind".
  • In algebra, this is important for understanding how elements combine and interact with each other.
Through the binary property, one can analyze and simplify complex algebraic forms, especially when the problem involves multiple scales or layers of operations, like in long polynomials or matrix algebra.
Inductive Step
Mathematical induction is a powerful tool for proving the validity of statements indexed by positive integers. It generally involves two parts:
  • Base Case: Show that the statement is true for the initial value, often \( k=1 \).
  • Inductive Step: Assume the statement is true for some positive integer \( j \), and then show it holds for \( j+1 \).
In the exercise, the base case was verified. That proved the statement for \( k=1 \). The inductive step assumed the statement held for \( k=j \), and aimed to prove it for \( k=j+1 \). This step is crucial because if one assumes it is true for one case, its truth for the subsequent case often uses properties like the commutator laws or other algebraic expansions.
Positive Integers
Positive integers are the set of all whole numbers greater than zero. They are fundamental in mathematics and serve as the building blocks for discussing many concepts like number systems, counting, and sequences.
  • In mathematical induction, positive integers provide a framework for proving an infinite number of cases.
  • These integers help structure proofs where a statement is shown to be universally true, like in the exercise discussed.
When dealing with statements about infinite sequences or recursive formulas, positive integers guide the construction of proofs to ensure every case follows logically from the prior case, exhibiting a domino effect of correctness.

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Most popular questions from this chapter

For the vector $$ |a\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 0 \\ 1 \\ -1 \\ 0 \end{array}\right) $$ (a) find the associated projection matrix, \(\mathbf{P}_{a}\). (b) Verify that \(\mathbf{P}_{a}\) does project an arbitrary vector in \(\mathbb{C}^{4}\) along \(|a\rangle .\) (c) Verify directly that the matrix \(1-\mathbf{P}_{a}\) is also a projection operator.

Let the operator \(\mathbf{U}: \mathbb{C}^{2} \rightarrow \mathbb{C}^{2}\) be given by $$ \mathbf{U}\left(\begin{array}{l} \alpha_{1} \\ \alpha_{2} \end{array}\right)=\left(\begin{array}{l} i \frac{\alpha_{1}}{\sqrt{2}}-i \frac{\alpha_{2}}{\sqrt{2}} \\ \frac{\alpha_{1}}{\sqrt{2}}+\frac{\alpha_{2}}{\sqrt{2}} \end{array}\right) $$ Find \(\mathbf{U}^{\dagger}\) and test if \(\mathbf{U}\) is unitary.

Show that for any \(\alpha, \beta \in \mathbb{R}\) and any \(\mathbf{H} \in \operatorname{End}(\mathcal{V})\), we have $$ e^{\alpha \mathbf{H}} e^{\beta \mathbf{H}}=e^{(\alpha+\beta) \mathbf{H}} . $$

Consider a linear operator \(\mathbf{T}\) on a finite-dimensional vector space \(V\). (a) Show that there exists a polynomial \(p\) such that \(p(\mathbf{T})=\mathbf{0}\). Hint: Take a basis \(B=\left\\{\left|a_{i}\right\rangle\right\\}_{i=1}^{N}\) and consider the vectors \(\left\\{\mathbf{T}^{k}\left|a_{1}\right\rangle\right\\}_{k=0}^{M}\) for large enough \(M\) and conclude that there exists a polynomial \(p_{1}(\mathbf{T})\) such that \(p_{1}(\mathbf{T})\left|a_{1}\right\rangle=0\). Do the same for \(\left|a_{2}\right\rangle\), etc. Now take the product of all such polynomials. (b) From (a) conclude that for large enough \(n, \mathbf{T}^{n}\) can be written as a linear combination of smaller powers of \(\mathbf{T}\). (c) Now conclude that any infinite series in \(\mathbf{T}\) collapses to a polynomial in \(\mathbf{T}\).

Show that any two equivalent representations of any algebra have the same kernel.

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