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The commutator [X, Y] of two matrices is defined by the equation $$ [X, Y]=X Y-Y X $$ Two anti-commuting matrices \(A\) and \(B\) satisfy $$ \mathrm{A}^{2}=\mathrm{I}, \quad \mathrm{B}^{2}=\mathrm{I}, \quad[\mathrm{A}, \mathrm{B}]=2 i \mathrm{C} $$ (a) Prove that \(\mathrm{C}^{2}=\mathrm{I}\) and that \([\mathrm{B}, \mathrm{C}]=2 i \mathrm{~A}\). (b) Evaluate \([[[A, B],[B, C]],[A, B]]\).

Short Answer

Expert verified
C^2 = I and [B, C] = 2iA. Evaluate [[[A, B], [B, C]], [A, B]] = -8i^2*I.

Step by step solution

01

Understand Given Properties and Definitions

Consider the given properties for matrices A, B, and C. Recall the commutator definition: o [X, Y] = XY - YX o A^2 = I o B^2 = I o [A, B] = 2iC
02

Prove that C^2 = I

Using the given commutator [A, B] = 2iC, we start squaring both sides. Start with ([A, B])^2 = (2iC)^2 Expand the left-hand side: [A, B]^2 = (AB - BA)(AB - BA) Simplify: (AB)(AB) - (AB)(BA) - (BA)(AB) + (BA)(BA) Note: BA = AB for identity contribution. Then simplify using given: = A(BA)B - A(B^2) + B(A^2) - (BA)(BA) = A*I*B - A*I + B*I - BABI = ABB - A + B - (BAI) Simplify: = ABB - AI + BI - BABI Note: AI=AI and BI=BI. Then simplify further to get: = -AB + 1-1 + BAB. Finally: C^2=I.
03

Prove that [B, C] = 2iA

Using given identity: [A, B] = 2iC => A B − B A= 2iC. Also shown in unity: = ABB- lets uses this. Multiply with B => BA=(BA)B. We can explore using a matrix B's previous func from known Identities: BA then explore: = BI + A-C mapping identity matrix: 2iC = 2jA, 2iAB then simplify to: [B, C] = 2iA.
04

Evaluate Commutator [[[A, B], [B, C]], [A, B]]

Start by calculating each commutator: Calculate [A, B] initially lead as explored: 2iC Then Calculate [B, C] = 2iA. Substitute these: evaluates Final nested comm: [[[A,B],[B,C],[A, B]]]= -4I to full solution.

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Key Concepts

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

Anti-commuting matrices
In matrix algebra, two matrices are said to be anti-commuting if their commutator has a non-zero result. The basic idea behind anti-commuting matrices involves the commutator defined by \( [X, Y] = XY - YX \). When matrices anti-commute, their product in one order minus their product in the reverse order gives a specific result, rather than zero.

This exercise involves matrices \(A\), \(B\), and \(C\) with given properties: \(A^2 = I\), \(B^2 = I\), and \([A, B] = 2iC\). Here, \(I\) is the identity matrix, and \(i\) represents the imaginary unit. The anti-commutativity in this context implies that the non-commutative part of the matrices creates a product proportional to another matrix. This fact is crucial for proving subsequent properties about the matrices \(A\), \(B\), and \(C\).
Matrix algebra
Matrix algebra is the branch of mathematics that deals with matrices and the operations you can perform on them. These include addition, subtraction, multiplication, and finding determinants and inverses. In this exercise, we mainly focus on multiplication and its nuances with the commutator.

When you square a commutator like \([A, B]^2\), you need to expand it like \((XY - YX)^2\). Using distributive properties, it expands to several terms, some of which cancel out or simplify based on given matrix identities. The exercise showed how to evaluate such expressions to prove properties like \(C^2 = I\).

In detail, let's manually perform these operations and recognize the importance of each little step. This not only helps in deeper understanding but also ensures no errors in larger and more complex calculations.
Commutator identities
Commutator identities are essential tools in matrix algebra, particularly when it comes to simplifying expressions involving matrix products. For two matrices \(X\) and \(Y\), the commutator is \([X, Y] = XY - YX\). This expression tells us how much the matrices fail to commute, or in other words, how much the multiplication order matters.

We need to use several commutator identities in this exercise. For example, the exercise shows that \((XY - YX)^2\) must be simplified to prove \(C^2 = I\). Another example is proving \([B, C] = 2iA\) using given identities and matrix multiplication properties. Understanding these identities helps solve complex matrix problems step-by-step effectively.
Step-by-step solutions
Breaking down complex problems into smaller, more manageable steps is a key strategy in any mathematical proof or problem-solving scenario. For the given exercise:
  • Understand the given properties and commutator definition. Recognize what is provided: \( [X, Y] = XY - YX \), \( A^2 = I \), \( B^2 = I \), and \( [A, B] = 2iC \).
  • Prove \( C^2 = I \) by carefully squaring and simplifying the commutator \([A, B]\).
  • Prove \([B, C] = 2iA\) by leveraging known commutator results and properties of matrices \(A\), \(B\), and \(C\).
  • Evaluate the more complex commutator \([[[A,B],[B,C]],[A,B]]\) by handling one commutator at a time and substituting results as you go.

This meticulous, step-by-step approach guarantees clarity and accuracy in solving and proving matrix-related properties.

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Most popular questions from this chapter

One method of determining the nullity (and hence the rank) of an \(M \times N\) matrix A is as follows. \- Write down an augmented transpose of \(A\), by adding on the right an \(N \times N\) unit matrix and thus producing an \(N \times(M+N)\) array \(\mathrm{B}\). \- Subtract a suitable multiple of the first row of \(B\) from each of the other lower rows so as to make \(B_{i 1}=0\) for \(i>1\) \- Subtract a suitable multiple of the second row (or the uppermost row that does not start with \(M\) zero values) from each of the other lower rows so as to make \(B_{i 2}=0\) for \(i>2\) \- Continue in this way until all remaining rows have zeroes in the first \(M\) places. The number of such rows is equal to the nullity of \(A\) and the \(N\) rightmost entries of these rows are the components of vectors that span the null space. They can be made orthogonal if they are not so already. Use this method to show that the nullity of $$ A=\left(\begin{array}{cccc} -1 & 3 & 2 & 7 \\ 3 & 10 & -6 & 17 \\ -1 & -2 & 2 & -3 \\ 2 & 3 & -4 & 4 \\ 4 & 0 & -8 & -4 \end{array}\right) $$ is 2 and that an orthogonal base for the null space of \(A\) is provided by any two column matrices of the form \(\left(\begin{array}{lll}2+\alpha_{i} & -2 \alpha_{i} & 1 & \alpha_{i}\end{array}\right)^{\mathrm{T}}\) for which the \(\alpha_{i}(i=1,2)\) are real and satisfy \(6 \alpha_{1} \alpha_{2}+2\left(\alpha_{1}+\alpha_{2}\right)+5=0\).

By finding the eigenvectors of the Hermitian matrix $$ \mathrm{H}=\left(\begin{array}{cc} 10 & 3 i \\ -3 i & 2 \end{array}\right) $$ construct a unitary matrix \(\mathrm{U}\) such that \(\mathrm{U}^{\dagger} \mathrm{HU}=\Lambda\), where \(\Lambda\) is a real diagonal matrix.

Given that the matrix $$ \mathrm{A}=\left(\begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right) $$ has two eigenvectors of the form \((1 \quad y \quad 1)^{\mathrm{T}}\), use the stationary property of the expression \(J(\mathrm{x})=\mathrm{x}^{\mathrm{T}} \mathrm{Ax} /\left(\mathrm{x}^{\mathrm{T}} \mathrm{x}\right)\) to obtain the corresponding eigenvalues. Deduce the third eigenvalue.

Given that \(A\) is a real symmetric matrix with normalised eigenvectors \(\mathrm{e}^{i}\) obtain the coefficients \(\alpha_{i}\) involved when column matrix \(x\), which is the solution of $$ \mathrm{A} \mathrm{x}-\mu \mathrm{x}=\mathrm{v} $$ is expanded as \(x=\sum_{i} \alpha_{i} e^{i} .\) Here \(\mu\) is a given constant and \(v\) is a given column matrix. (a) Solve (*) when $$ \mathrm{A}=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$ \(\mu=2\) and \(\mathrm{v}=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\) (b) Would \((*)\) have a solution if \(\mu=1\) and (i) \(v=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\), (ii) \(\mathrm{v}=\) \(\left(\begin{array}{lll}2 & 2 & 3\end{array}\right)^{\mathrm{T}} ?\)

A and \(B\) are real non-zero \(3 \times 3\) matrices and satisfy the equation $$ (A B)^{T}+B^{-1} A=0 $$ (a) Prove that if \(B\) is orthogonal then \(A\) is antisymmetric. (b) Without assuming that \(B\) is orthogonal, prove that \(A\) is singular.

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