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Chapter 6: Problem 5

Which of the following pairs of matrices are unitarily equivalent? (a) \(\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) and \(\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) (b) \(\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) and \(\left(\begin{array}{ll}0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{array}\right)\) (c) \(\left(\begin{array}{rrr}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right) \quad\) and \(\quad\left(\begin{array}{rrr}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0\end{array}\right)\) (d) \(\left(\begin{array}{rrr}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right) \quad\) and \(\quad\left(\begin{array}{rrr}1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i\end{array}\right)\) (e) \(\left(\begin{array}{lll}1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{array}\right)\) and \(\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\\ 0 & 0 & 3\end{array}\right)\)

Short Answer

Expert verified
The only pair of matrices unitarily equivalent is pair (d).

Step by step solution

01

Pair (a)

We will check if these matrices are unitarily equivalent: A = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) and B = \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) Step 1: Both matrices A and B have dimensions 2x2. Step 2: Check if they are normal matrices: - A*A.conjugate() = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) * \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) - A.conjugate()*A = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) * \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) - B*B.conjugate() = \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) * \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) - B.conjugate()*B = \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) * \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\) Both matrices A and B are normal matrices. Step 3: Check if they have the same eigenvalues by comparing their characteristic polynomials: - The eigenvalues for A are {1,1}. - The eigenvalues for B are {1,-1}. Since their eigenvalues are different, the matrices in pair (a) are not unitarily equivalent.
02

Pair (b)

Similarly, we will check if these matrices are unitarily equivalent: A = \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\) and B = \(\begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \\ \end{pmatrix}\) Step 1: Both matrices A and B have dimensions 2x2. Step 2: Both matrices A and B are normal matrices (similarly to the pair (a)). Step 3: Check if they have the same eigenvalues by comparing their characteristic polynomials: - The eigenvalues for A are {1,-1}. - The eigenvalues for B are {0.5,-0.5}. Since their eigenvalues are different, the matrices in pair (b) are not unitarily equivalent. #continue_checking_pairs#Follow the same procedure for the remaining pairs (c), (d), and (e). After analysis, we can conclude that: - Pair (c) is not unitarily equivalent since they have different eigenvalues. - Pair (d) is actually unitarily equivalent since they have the same dimensions, are both normal matrices, and have the same eigenvalues. - Pair (e) is not unitarily equivalent since they have different eigenvalues. Thus, the only pair of matrices that are unitarily equivalent is pair (d).

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental to understanding matrix behaviors. They are values that give us special insight into the matrix's properties and how it transforms a vector.
  • The eigenvalue of a matrix is associated with a vector such that when the matrix multiplies this vector, the result is a scalar multiple of the vector.
  • This concept is captured in the equation: \(Av = \lambda v\), where \(A\) is the matrix, \(v\) the eigenvector, and \(\lambda\) the eigenvalue.
  • In the context of unitary equivalence, two matrices must have the same eigenvalues to be considered equivalent, meaning their eigenvalues must match exactly, including their multiplicities.
Finding eigenvalues involves solving the characteristic polynomial equation, a topic detailed in its own section. Understanding eigenvalues helps simplify complex problems like determining if matrices are unitarily equivalent. Even though two matrices might have the same size or similar shapes, differences in eigenvalues directly prevent them from being equivalent unitary matrices.
Normal Matrices
Normal matrices are a class of matrices that hold specific properties making them easier to work with, especially when considering unitary equivalence.
  • A matrix \(A\) is considered normal if it commutes with its conjugate transpose, i.e., \(A^*A = AA^*\).
  • This condition ensures that the eigenvectors of the matrix form an orthogonal basis, simplifying many computations.
  • Normal matrices encompass several familiar types of matrices, such as Hermitian, unitary, and orthogonal matrices.
In terms of unitary equivalence, both matrices in a pair must be normal matrices to potentially be unitarily equivalent.
Since normal matrices have such symmetry in their eigenvectors and eigenvalues, they ensure that matrix operations behave predictably. This makes them an essential part of linear algebra and its applications, particularly for matrices looking to be unitarily equivalent as seen in the exercise.
Characteristic Polynomial
The characteristic polynomial of a matrix is a tool that helps find the eigenvalues by transforming the matrix into a polynomial expression.
  • It is derived from the determinant of the matrix \(A\) minus an unknown scalar multiplied by the identity matrix, expressed as \(det(A - \lambda I)\).
  • The roots of this polynomial give the eigenvalues of the matrix, revealing critical insights into the matrix's behavior.
  • Characteristic polynomials are pivotal in problems involving unitary equivalence, ensuring the matrices have the same set of eigenvalues.
For matrices to be unitarily equivalent, it's not just their eigenvalues that must match but their characteristic polynomials must reflect this equivalence too.
Given the nature of characteristic polynomials, working out these values and their multiplicities confirms potential equivalence in properties between matrices, thereby determining unitary equivalence. This essential relationship between polynomials and matrices underpins much of linear algebra.

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Most popular questions from this chapter

(a) Let \(A\) and \(B\) be square matrices that are unitarily equivalent. Prove that \(\|A\|_{E}=\|B\|_{E^{*}}\) (b) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional inner product space V. Define $$ \|\mathrm{T}\|_{E}=\max _{x \neq 0} \frac{\|\mathrm{T}(x)\|}{\|x\|} . $$ Prove that \(\|\mathrm{T}\|_{E}=\left\|[\mathrm{T}]_{\beta}\right\|_{E}\), where \(\beta\) is any orthonormal basis for \(\mathrm{V}\). (c) Let \(V\) be an infinite-dimensional inner product space with an orthonormal basis \(\left\\{v_{1}, v_{2}, \ldots\right\\}\). Let \(\mathrm{T}\) be the linear operator on \(\mathrm{V}\) such that \(\mathrm{T}\left(v_{k}\right)=k v_{k}\). Prove that \(\|\mathrm{T}\|_{E}\) (defined in (b)) does not exist. Visit goo.gl/B8Uw33 for a solution. The next exercise assumes the definitions of singular value and pseudoinverse and the results of Section \(6.7\).

Let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be a linear transformation, where \(\mathrm{V}\) and \(\mathrm{W}\) are finitedimensional inner product spaces with inner products \(\langle\cdot, \cdot\rangle_{1}\) and \(\langle\cdot, \cdot\rangle_{2}\), respectively. Prove the following results. (a) There is a unique adjoint \(\mathrm{T}^{*}\) of \(\mathrm{T}\), and \(\mathrm{T}^{*}\) is linear. (b) If \(\beta\) and \(\gamma\) are orthonormal bases for \(\mathbf{V}\) and \(\mathbf{W}\), respectively, then \(\left[\mathrm{T}^{*}\right]_{\gamma}^{\beta}=\left([\mathrm{T}]_{\beta}^{\gamma}\right)^{*} .\) (c) \(\operatorname{rank}\left(T^{*}\right)=\operatorname{rank}(T)\). (d) \(\left\langle\mathrm{T}^{*}(x), y\right\rangle_{1}=\langle x, \mathrm{~T}(y)\rangle_{2}\) for all \(x \in \mathrm{W}\) and \(y \in \mathrm{V}\). (e) For all \(x \in \mathrm{V}, \mathrm{T}^{*} \mathrm{~T}(x)=0\) if and only if \(\mathrm{T}(x)=0\).

Recall the moving space vehicle considered in the study of time contraction. Suppose that the vehicle is moving toward a fixed star located on the \(x\)-axis of \(S\) at a distance \(b\) units from the origin of \(S\). If the space vehicle moves toward the star at velocity \(v\), Earthlings (who remain "almost" stationary relative to \(S\) ) compute the time it takes for the vehicle to reach the star as \(t=b / v\). Due to the phenomenon of time contraction, the astronaut perceives a time span of \(t^{\prime}=t \sqrt{1-v^{2}}=(b / v) \sqrt{1-v^{2}}\). A paradox appears in that the astronaut perceives a time span inconsistent with a distance of \(b\) and a velocity of \(v\). The paradox is resolved by observing that the distance from the solar system to the star as measured by the astronaut is less than \(b\). (a) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S\) and \(C\) are $$ \left(\begin{array}{l} b \\ 0 \\ t \end{array}\right) \text {. } $$ (b) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S^{\prime}\) and \(C^{\prime}\) are $$ \left(\begin{array}{c} \frac{b-v t}{\sqrt{1-v^{2}}} \\ 0 \\ \frac{t-b v}{\sqrt{1-v^{2}}} \end{array}\right) $$ (c) For $$ x^{\prime}=\frac{b-t v}{\sqrt{1-v^{2}}} \quad \text { and } \quad t^{\prime}=\frac{t-b v}{\sqrt{1-v^{2}}}, $$ we have \(x^{\prime}=b \sqrt{1-v^{2}}-t^{\prime} v\). This result may be interpreted to mean that at time \(t^{\prime}\) as measured by the astronaut, the distance from the astronaut to the star, as measured by the astronaut, (see Figure 6.9) is $$ b \sqrt{1-v^{2}}-t^{\prime} v . $$ Assuming that the coordinate systems \(S\) and \(S^{\prime}\) and clocks \(C\) and \(C^{\prime}\) are as in the discussion of time contraction, prove the following results. (d) Conclude from the preceding equation that (1) the speed of the space vehicle relative to the star, as measured by the astronaut, is \(v\); (2) the distance from Earth to the star, as measured by the astronaut, is \(b \sqrt{1-v^{2}}\). Thus distances along the line of motion of the space vehicle appear to be contracted by a factor of \(\sqrt{1-v^{2}}\).

An \(n \times n\) real matrix \(A\) is said to be a Gramian matrix if there exists a real (square) matrix \(B\) such that \(A=B^{t} B\). Prove that \(A\) is a Gramian matrix if and only if \(A\) is symmetric and all of its eigenvalues are nonnegative. Hint: Apply Theorem \(6.17\) to \(\mathrm{T}=\mathrm{L}_{A}\) to obtain an orthonormal basis \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) of eigenvectors with the associated eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} .\) Define the linear operator \(\mathrm{U}\) by \(\mathrm{U}\left(v_{i}\right)=\sqrt{\lambda_{i}} v_{i}\).

Consider three coordinate systems \(S, S^{\prime}\), and \(S^{\prime \prime}\) with the corresponding axes \(\left(x, x^{\prime}, x^{\prime \prime} ; y, y^{\prime}, y^{\prime \prime} ;\right.\) and \(\left.z, z^{\prime}, z^{\prime \prime}\right)\) parallel and such that the \(x-, x^{\prime}-\), and \(x^{\prime \prime}\)-axes coincide. Suppose that \(S^{\prime}\) is moving past \(S\) at a velocity \(v_{1}>0\) (as measured on \(S\) ), \(S^{\prime \prime}\) is moving past \(S^{\prime}\) at a velocity \(v_{2}>0\) (as measured on \(S^{\prime}\) ), and \(S^{\prime \prime}\) is moving past \(S\) at a velocity \(v_{3}>0\) (as measured on \(S\) ), and that there are three clocks \(C, C^{\prime}\), and \(C^{\prime \prime}\) such that \(C\) is stationary relative to \(S, C^{\prime}\) is stationary relative to \(S^{\prime}\), and \(C^{\prime \prime}\) is stationary relative to \(S^{\prime \prime}\). Suppose that when measured on any of the three clocks, all the origins of \(S, S^{\prime}\), and \(S^{\prime \prime}\) coincide at time 0 . Assuming that \(\mathrm{T}_{v_{3}}=\mathrm{T}_{v_{2}} \mathrm{~T}_{v_{1}}\) (i.e., \(B_{v_{3}}=B_{v_{2}} B_{v_{1}}\) ), prove that $$ v_{3}=\frac{v_{1}+v_{2}}{1+v_{1} v_{2}} . $$ Note that substituting \(v_{2}=1\) in this equation yields \(v_{3}=1\). This tells us that the speed of light as measured in \(S\) or \(S^{\prime}\) is the same. Why would we be surprised if this were not the case?
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