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Find \(\mathbf{T}^{\dagger}\) for each of the following linear operators. (a) \(\mathbf{T}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) given by $$ \mathbf{T}\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} x+y \\ x-y \end{array}\right) $$ (b) \(\mathbf{T}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) given by $$ \mathbf{T}\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} x+2 y-z \\ 3 x-y+2 z \\ -x+2 y+3 z \end{array}\right) $$ (c) \(\mathbf{T}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) given by $$ \mathbf{T}\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} x \cos \theta-y \sin \theta \\ x \sin \theta+y \cos \theta \end{array}\right) $$ where \(\theta\) is a real number. What is \(\mathbf{T}^{\dagger} \mathbf{T}\) ? (d) \(\mathbf{T}: \mathbb{C}^{2} \rightarrow \mathbb{C}^{2}\) given by $$ \mathbf{T}\left(\begin{array}{l} \alpha_{1} \\ \alpha_{2} \end{array}\right)=\left(\begin{array}{l} \alpha_{1}-i \alpha_{2} \\ i \alpha_{1}+\alpha_{2} \end{array}\right) $$ (e) \(\mathbf{T}: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}\) given by $$ \mathbf{T}\left(\begin{array}{l} \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{array}\right)=\left(\begin{array}{c} \alpha_{1}+i \alpha_{2}-2 i \alpha_{3} \\ -2 i \alpha_{1}+\alpha_{2}+i \alpha_{3} \\ i \alpha_{1}-2 i \alpha_{2}+\alpha_{3} \end{array}\right) $$

Short Answer

Expert verified
Short answer would be the adjoints (computed from Step 2) and the product of the adjoint and the operator itself for part (c) computed in Step 3.

Step by step solution

01

Identify the Matrix Representation of Each Linear Operator

For each linear operator, identify the matrix representation. This matrix is found by evaluating the place of the operator on the standard basis for \( \mathbb{R}^{2} \), \( \mathbb{R}^{3} \), and \( \mathbb{C}^{2} \) respectively.
02

Take the Conjugate Transpose of Each Matrix

The adjoint of a linear operator is found by taking the conjugate transpose of the matrix representation. In the given operator cases, only part (d) and (e) have complex numbers. Thus, for (d) and (e), it will be necessary to take the conjugate of complex elements upon transposing. For the rest (a, b, c), it is a matter of simple matrix transposition.
03

Compute \( \mathbf{T}^{\dagger} \mathbf{T} \) for Part (c)

For part (c), we are also asked to compute \( \mathbf{T}^{\dagger} \mathbf{T} \). This is the product of the adjoint of the operator and the operator itself. We can find this by matrix multiplication of \( \mathbf{T}^{\dagger} \) and \( \mathbf{T} \).

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Key Concepts

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

Matrix Representation
When working with linear operators, one essential concept is their matrix representation. A linear operator in spaces like \( \mathbb{R}^{2} \) or \( \mathbb{C}^{3} \) can be represented as a matrix, which provides a concrete way to understand its transformations.
  • For each operator, the matrix is determined by how it transforms the standard basis vectors of the space. This involves calculating what happens to \( (1,0,0,\dots) \), \( (0,1,0,\dots) \), and so on, under the operator.
  • This process simplifies abstract linear maps into a form that’s easier to manipulate and visualize.
  • For example, if you have a transformation \( \mathbf{T} \), to find its representation, you determine how \( \mathbf{T} \) acts on these basis vectors, then express the results as columns of the matrix.
By converting linear operators into matrices, not only computational tasks like finding adjoints become straightforward, but it also allows for a systematic analysis of the transformations on vector spaces.
Conjugate Transpose
The concept of the conjugate transpose, also known as the adjoint, is crucial when dealing with matrices, especially in complex spaces like \( \mathbb{C}^{2} \) or \( \mathbb{C}^{3} \).
  • The conjugate transpose of a matrix \( A \), denoted as \( A^{\dagger} \), involves two main steps: taking the transpose of the matrix (flipping it over its diagonal) and then taking the complex conjugate of each entry.
  • For real matrices, this process is the same as taking a simple transpose since the complex conjugate of a real number is the number itself.
  • For complex matrices, you must consider the imaginary parts to ensure all entries are accurately transformed. For example, if \( a + bi \) is an entry, its conjugate would be \( a - bi \).
This operation is essential in many areas such as quantum mechanics and signal processing, where the structure of complex-valued linear transformations plays a significant role.
Linear Transformations
Linear transformations are functions between vector spaces that preserve the operations of vector addition and scalar multiplication.
  • They are the foundation of linear algebra and have applications across mathematics and engineering.
  • A transformation \( \mathbf{T} \) is defined as linear if for any vectors \( \mathbf{u} \) and \( \mathbf{v} \), and any scalar \( c \), the following hold: \( \mathbf{T}(\mathbf{u} + \mathbf{v}) = \mathbf{T}(\mathbf{u}) + \mathbf{T}(\mathbf{v}) \) and \( \mathbf{T}(c\mathbf{u}) = c\mathbf{T}(\mathbf{u}) \).
  • Finding the matrix representation of a linear operator helps us understand and apply these transformations effectively.
Understanding linear transformations is the key to solving systems of equations, performing geometric transformations, and more.
Vector Spaces
Vector spaces provide the framework within which linear transformations operate. They consist of a set of vectors, equipped with operations of vector addition and scalar multiplication.
  • Examples of vector spaces include \( \mathbb{R}^n \) (real coordinate spaces) and \( \mathbb{C}^n \) (complex coordinate spaces).
  • Vector spaces serve as a crucial setting for studying linear operators and transformations. They provide the 'playing field' where all linear algebra unfolds.
  • Understanding properties like subspaces, bases, dimension, and span is essential for deep diving into topics like matrix representation and transformations.
These properties of vector spaces ensure that operations on vectors, such as transformations, are well-defined and consistent across calculations and proofs.

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Most popular questions from this chapter

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

Let \(\mathbf{P}^{(m)}=\sum_{i=1}^{m}\left|e_{i}\right\rangle\left\langle e_{i}\right|\) be a projection operator constructed out of the first \(m\) orthonormal vectors of the basis \(B=\left\\{\left|e_{i}\right\rangle\right\\}_{i=1}^{N}\) of \(V .\) Show that \(\mathbf{P}^{(m)}\) projects into the subspace spanned by the first \(m\) vectors in \(B\).

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 \(\mathbf{a}_{1} \times \mathbf{a}_{2} ?\) What can you say about \(\mathbf{P}\) and your conclusion in (b)?

Show that the product of two unitary operators is always unitary, but the product of two hermitian operators is hermitian if and only if they commute.

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 \(\mathbf{T}\).

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