Chapter 6

Spectral Theorem

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Problem 6.1.1

Consider the vectors in C2\mathbb {C}^{2} (sometimes called spinors)

v→1=(cos⁡(θ/2)e−iϕ/2sin⁡(θ/2)eiϕ/2),v→2=(sin⁡(θ/2)e−iϕ/2−cos⁡(θ/2)eiϕ/2). \overrightarrow {v}_{1} = \left(\begin{smallmatrix} \cos (\theta /2)e^{-i\phi /2} \\ \sin (\theta /2)e^{i\phi /2} \end{smallmatrix}\right), \qquad \overrightarrow {v}_{2} = \left(\begin{smallmatrix} \sin (\theta /2)e^{-i\phi /2} \\ -\cos (\theta /2)e^{i\phi /2} \end{smallmatrix}\right).
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(i)

First show that they are normalized and orthonormal.

(ii)

Assume that for the vector v→1\overrightarrow {v}_{1} is an eigenvector with the corresponding eigenvalue +1+1 and that the vector v→2\overrightarrow {v}_{2} is an eigenvector with the corresponding eigenvalue −1-1. Apply the spectral theorem to find the corresponding 2×22 \times 2 matrix.

(iii)

Since the vectors v→1\overrightarrow {v}_{1} and v→2\overrightarrow {v}_{2} form an orthonormal basis in C2\mathbb {C}^{2} we can form an orthonormal basis in C4\mathbb {C}^{4} via the Kronecker product

w→11=v→1⊗v→1,w→12=v→1⊗v→2,w→21=v→2⊗v→1,w→22=v→2⊗v→2. \overrightarrow {w}_{11} = \overrightarrow {v}_{1} \otimes \overrightarrow {v}_{1}, \quad \overrightarrow {w}_{12} = \overrightarrow {v}_{1} \otimes \overrightarrow {v}_{2}, \quad \overrightarrow {w}_{21} = \overrightarrow {v}_{2} \otimes \overrightarrow {v}_{1}, \quad \overrightarrow {w}_{22} = \overrightarrow {v}_{2} \otimes \overrightarrow {v}_{2}.

Assume that the eigenvalue for w→11\overrightarrow {w}_{11} is +1+1, for w→12\overrightarrow {w}_{12} −1-1, for w→21\overrightarrow {w}_{21} −1-1 and for w→22\overrightarrow {w}_{22} +1+1. Apply the spectral theorem to find the corresponding 4×44 \times 4 matrix.

Problem 6.1.2

Given a 4×44 \times 4 symmetric matrix AA over R\mathbb {R} with the eigenvalues λ1=0\lambda_{1} = 0, λ2=1\lambda_{2} = 1, λ3=2\lambda_{3} = 2, λ4=3\lambda_{4} = 3 and the corresponding normalized eigenvectors

v→1=12(1001),v→2=12(100−1),v→3=12(0110),v→4=12(01−10). \overrightarrow {v}_{1} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ 0 \\ 0 \\ 1 \end{smallmatrix}\right), \quad \overrightarrow {v}_{2} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ 0 \\ 0 \\ -1 \end{smallmatrix}\right), \quad \overrightarrow {v}_{3} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 0 \end{smallmatrix}\right), \quad \overrightarrow {v}_{4} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 0 \\ 1 \\ -1 \\ 0 \end{smallmatrix}\right).

Reconstruct the matrix AA using the spectral theorem. The eigenvectors given above are called Bell basis.

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Problem 6.1.3
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(i)

Consider the 2×22 \times 2 permutation matrix

σ1=(0110). \sigma _{1} = \left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right).

Find the spectral decomposition of σ1\sigma_{1}. Find a matrix KK with exp(K)=σ1\mathrm{exp}\left(K\right) = \sigma_{1}.

(ii)

Consider the permutation matrix

P=(001010100). P = \left(\begin{smallmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{smallmatrix}\right).

Find the spectral decomposition of PP. Find a matrix XX with exp(X)=P\mathrm{exp}\left(X\right) = P.

(iii)

Consider the permutation matrix

P=(0001001010000100). P = \left(\begin{smallmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{smallmatrix}\right).

Show that the eigenvalues form a group under multiplication. Find the spectral representation to find KK such that P=exp(K)P = \mathrm{exp}\left(K\right).

Problem 6.1.4

Let z∈Cz \in \mathbb {C} and AA be an n×nn \times n normal matrix with eigenvalues λ1,…,λn\lambda_{1}, \ldots , \lambda_{n} and pairwise orthonormal eigenvectors v→1,…,v→n\overrightarrow {v}_{1}, \ldots , \overrightarrow {v}_{n}. Use the spectral decomposition to calculate exp(zA)\mathrm{exp}\left(zA\right).

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Problem 6.1.5

Let AA be a hermitian n×nn \times n matrix. Assume that all the eigenvalues λ1,…,λn\lambda_{1}, \ldots , \lambda_{n} are pairwise different. Then the normalized eigenvectors u→j\overrightarrow {u}_{j} (j=1,…,nj = 1, \ldots , n) satisfy u→j∗u→k=0\overrightarrow {u}_{j}^{*}\overrightarrow {u}_{k} = 0 for j≠kj \neq k and u→j∗u→j=1\overrightarrow {u}_{j}^{*}\overrightarrow {u}_{j} = 1. We have

A=∑j=1nλju→ju→j∗. A = \sum _{j=1}^{n} \lambda _{j}\overrightarrow {u}_{j}\overrightarrow {u}_{j}^{*}.

Let e→k\overrightarrow {e}_{k} (k=1,…,nk = 1, \ldots , n) be the standard basis in Cn\mathbb {C}^{n}. Calculate U∗AUU^{*}AU, where U=∑k=1nu→ke→k∗U = \sum_{k=1}^{n} \overrightarrow {u}_{k}\overrightarrow {e}_{k}^{*}.

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Problem 6.1.6

Let AA be a positive definite n×nn \times n matrix. Thus all the eigenvalues are real and positive. Assume that all the eigenvalues λ1,…,λn\lambda_{1}, \ldots , \lambda_{n} are pairwise different. Then the normalized eigenvectors u→j\overrightarrow {u}_{j} (j=1,…,nj = 1, \ldots , n) satisfy u→j∗u→k=0\overrightarrow {u}_{j}^{*}\overrightarrow {u}_{k} = 0 for j≠kj \neq k and u→j∗u→j=1\overrightarrow {u}_{j}^{*}\overrightarrow {u}_{j} = 1. We have (spectral theorem)

A=∑j=1nλju→ju→j∗. A = \sum _{j=1}^{n} \lambda _{j}\overrightarrow {u}_{j}\overrightarrow {u}_{j}^{*}.

Let e→k\overrightarrow {e}_{k} (k=1,…,nk = 1, \ldots , n) be the standard basis in Cn\mathbb {C}^{n}. Calculate ln⁡(A)\ln (A). Note that the unitary matrix

U=∑k=1nu→ke→k∗ U = \sum _{k=1}^{n} \overrightarrow {u}_{k}\overrightarrow {e}_{k}^{*}

transforms AA into a diagonal matrix, i.e. A~=U∗AU\widetilde{A} = U^{*}AU is a diagonal matrix.

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Problem 6.1.7

Consider the normal noninvertible 3×33 \times 3 matrix

A=(010101010). A = \left(\begin{smallmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix}\right).
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(i)

Find the spectral decomposition.

(ii)

Use this result to calculate exp(zA)\mathrm{exp}\left(zA\right), where z∈Cz \in \mathbb {C}.

Problem 6.1.8

Consider the two 3×33 \times 3 permutation matrices with P1=P2TP_{1} = P_{2}^{T}

P1=(010001100),P2=(001100010). P_{1} = \left(\begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{smallmatrix}\right), \qquad P_{2} = \left(\begin{smallmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{smallmatrix}\right).
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(i)

Find the spectral decomposition of the permutation matrices.

(ii)

Use the spectral decomposition to find the matrices K1K_{1} and A2A_{2} such that P1=exp(K1)P_{1} = \mathrm{exp}\left(K_{1}\right), P2=exp(A2)P_{2} = \mathrm{exp}\left(A_{2}\right).

(iii)

We want to find efficiently K1K_{1} and K2K_{2} such that P1=eK1P_{1} = e^{K_{1}} and P2=eK2P_{2} = e^{K_{2}}. We would apply the spectral decomposition theorem to find K1K_{1}, i.e.

K1=∑j=13ln⁡(λj)v→jv→j∗ K_{1} = \sum _{j=1}^{3} \ln (\lambda _{j})\overrightarrow {v}_{j}\overrightarrow {v}_{j}^{*}

where λj\lambda_{j} are the eigenvalues of U1U_{1} and v→j\overrightarrow {v}_{j} are the corresponding normalized eigenvectors. But then to find K2K_{2} we would apply the property that U12=U2U_{1}^{2} = U_{2}. Or could we actually apply that U2=U1TU_{2} = U_{1}^{T}? Note that U1U_{1}, U2U_{2}, I3I_{3} form a commutative subgroup of the group of 3×33 \times 3 permutation under matrix multiplication.

Problem 6.1.9

Consider the unitary matrix

U=(0001001001001000)=(0110)⊗(0110). U = \left(\begin{smallmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{smallmatrix}\right) = \left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right) \otimes \left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right).

Find the skew-hermitian matrix KK such that U=exp(K)U = \mathrm{exp}\left(K\right). Utilize the spectral representation.

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Problem 6.1.10

Consider the Hadamard matrix

U=12(111−1)≡12(σ3+σ1) U = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 & 1 \\ 1 & -1 \end{smallmatrix}\right) \equiv \frac{1}{\sqrt{2}}(\sigma _{3} + \sigma _{1})

with eigenvalues +1+1 and −1-1. Find the square roots of this matrix applying the spectral theorem. Since 1=±1\sqrt{1} = \pm 1 and −1=±i\sqrt{-1} = \pm i four cases have to be considered

(1,i),(1,−i),(−1,i),(−1,−i). (1, i), \quad (1, -i), \quad (-1, i), \quad (-1, -i).
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Problem 6.1.11

Let AA be an n×nn \times n matrix over C\mathbb {C}. An n×nn \times n matrix BB is called a square root of AA if B2=AB^{2} = A. Find the square roots of the 2×22 \times 2 identity matrix applying the spectral theorem. The eigenvalues of I2I_{2} are λ1=1\lambda_{1} = 1 and λ2=1\lambda_{2} = 1. As normalized eigenvectors choose

(eiϕ1cos⁡(θ)eiϕ2sin⁡(θ)),(eiϕ1sin⁡(θ)−eiϕ2cos⁡(θ)) \left(\begin{smallmatrix} e^{i\phi _{1}}\cos (\theta ) \\ e^{i\phi _{2}}\sin (\theta ) \end{smallmatrix}\right), \qquad \left(\begin{smallmatrix} e^{i\phi _{1}}\sin (\theta ) \\ -e^{i\phi _{2}}\cos (\theta ) \end{smallmatrix}\right)

which form an orthonormal basis in C2\mathbb {C}^{2}. Four cases

(λ1,λ2)=(1,1),(λ1,λ2)=(1,−1), (\sqrt{\lambda _{1}}, \sqrt{\lambda _{2}}) = (1, 1), \quad (\sqrt{\lambda _{1}}, \sqrt{\lambda _{2}}) = (1, -1), (λ1,λ2)=(−1,1),(λ1,λ2)=(−1,−1) (\sqrt{\lambda _{1}}, \sqrt{\lambda _{2}}) = (-1, 1), \quad (\sqrt{\lambda _{1}}, \sqrt{\lambda _{2}}) = (-1, -1)

have to be studied. The first and last cases are trivial. So study the second case (λ1,λ2)=(1,−1)(\sqrt{\lambda_{1}}, \sqrt{\lambda_{2}}) = (1, -1). The second case and the third case are ``equivalent''.

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Problem 6.1.12

Let λ1,λ2,μ1,μ2∈C\lambda_{1}, \lambda_{2}, \mu_{1}, \mu_{2} \in \mathbb {C}. Let v→1\overrightarrow {v}_{1}, v→2\overrightarrow {v}_{2} (column vectors) be an orthonormal basis in C2\mathbb {C}^{2}. We define the 2×22 \times 2 matrices (spectral theorem)

A=λ1v→1v→1∗+λ2v→2v→2∗,B=μ1v→1v→1∗+μ2v→2v→2∗. A = \lambda _{1}\overrightarrow {v}_{1}\overrightarrow {v}_{1}^{*} + \lambda _{2}\overrightarrow {v}_{2}\overrightarrow {v}_{2}^{*}, \qquad B = \mu _{1}\overrightarrow {v}_{1}\overrightarrow {v}_{1}^{*} + \mu _{2}\overrightarrow {v}_{2}\overrightarrow {v}_{2}^{*}.

Find the commutator [A,B][A, B]. Find the conditions on λ1\lambda_{1}, λ2\lambda_{2}, μ1\mu_{1}, μ2\mu_{2} such that [A,B]=02[A, B] = 0_{2}.

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Problem 6.1.13

Let v→0,v→1,…,v→2n−1\overrightarrow {v}_{0}, \overrightarrow {v}_{1}, \ldots , \overrightarrow {v}_{2^{n}-1} be an orthonormal basis in C2n\mathbb {C}^{2^{n}}. We define

U:=12n∑j=02n−1∑k=02n−1e−i2πkj/2nv→kv→j∗. U := \frac{1}{\sqrt{2^{n}}} \sum _{j=0}^{2^{n}-1} \sum _{k=0}^{2^{n}-1} e^{-i2\pi kj/2^{n}} \overrightarrow {v}_{k}\overrightarrow {v}_{j}^{*}.

Show that UU is unitary. In other words show that UU∗=I2nUU^{*} = I_{2^{n}}, using the completeness relation

I2n=∑j=02n−1v→jv→j∗. I_{2^{n}} = \sum _{j=0}^{2^{n}-1} \overrightarrow {v}_{j}\overrightarrow {v}_{j}^{*}.
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Problem 6.2.1

The star product of the Hadamard matrix UHU_{H} with itself provides the Bell matrix

B=UH⋆UH=12(1001011001−10100−1) B = U_{H} \star U_{H} = \frac{1}{\sqrt{2}} \left(\begin{smallmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{smallmatrix}\right)

which is a unitary matrix. Note that the eigenvalues of the Bell matrix are +1+1 (twice) and −1-1 (twice).

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(i)

Find the square roots of the Bell matrix applying the spectral theorem.

(ii)

Apply the spectral theorem to find the skew-hermitian matrix KK such that B=eKB = e^{K}.

Problem 6.2.2

Let AA be a normal matrix with eigenvalues λ1,…,λn\lambda_{1}, \ldots , \lambda_{n} and pairwise orthonormal eigenvectors a→j\overrightarrow {a}_{j} (column vectors). Let BB be a normal matrix with eigenvalues μ1,…,μn\mu_{1}, \ldots , \mu_{n} and pairwise orthonormal eigenvectors b→j\overrightarrow {b}_{j} (column vectors). Then AA and BB can be written as

A=∑j=1nλja→ja→j∗,B=∑k=1nμkb→kb→k∗. A = \sum _{j=1}^{n} \lambda _{j}\overrightarrow {a}_{j}\overrightarrow {a}_{j}^{*}, \qquad B = \sum _{k=1}^{n} \mu _{k}\overrightarrow {b}_{k}\overrightarrow {b}_{k}^{*}.

Let z∈Cz \in \mathbb {C}. Use the spectral decomposition to calculate ezABe−zAe^{zA}Be^{-zA}.

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Problem 6.2.3

Consider the normalized vectors

v→1=12(101),v→2=(010),v→3=12(10−1) \overrightarrow {v}_{1} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right), \quad \overrightarrow {v}_{2} = \left(\begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right), \quad \overrightarrow {v}_{3} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ 0 \\ -1 \end{smallmatrix}\right)

in R3\mathbb {R}^{3}. Find the eigenvalues and eigenvectors of the 3×33 \times 3 matrix M=1⋅v→1v→1∗+0⋅v→2v→2∗−1⋅v→3v→3∗M = 1 \cdot \overrightarrow {v}_{1}\overrightarrow {v}_{1}^{*} + 0 \cdot \overrightarrow {v}_{2}\overrightarrow {v}_{2}^{*} - 1 \cdot \overrightarrow {v}_{3}\overrightarrow {v}_{3}^{*} without calculating MM.

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Problem 6.2.4

Let AA be an n×nn \times n normal matrix with eigenvalues λ1,…,λn\lambda_{1}, \ldots , \lambda_{n} and pairwise normalized orthogonal eigenvectors v→j\overrightarrow {v}_{j} (j=1,…,nj = 1, \ldots , n). Then

A=∑j=1nλjv→jv→j∗. A = \sum _{j=1}^{n} \lambda _{j}\overrightarrow {v}_{j}\overrightarrow {v}_{j}^{*}.

Show that cos⁡(A)=∑j=1ncos⁡(λj)v→jv→j∗\cos (A) = \sum_{j=1}^{n} \cos (\lambda_{j})\overrightarrow {v}_{j}\overrightarrow {v}_{j}^{*}, sin⁡(A)=∑j=1nsin⁡(λj)v→jv→j∗\sin (A) = \sum_{j=1}^{n} \sin (\lambda_{j})\overrightarrow {v}_{j}\overrightarrow {v}_{j}^{*}.

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Problem 6.2.5

The spectral theorem for n×nn \times n normal matrices over C\mathbb {C} is as follows: A matrix AA is normal if and only if there exists an n×nn \times n unitary matrix UU and a diagonal matrix DD such that D=U∗AUD = U^{*}AU. Use this theorem to prove that the matrix

A=(011001000) A = \left(\begin{smallmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{smallmatrix}\right)

is nonnormal.

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Problem 6.2.6

Find all 2×22 \times 2 matrices over C\mathbb {C} which admit the normalized eigenvectors

v→1=12(11),v→2=12(1−1) \overrightarrow {v}_{1} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right), \qquad \overrightarrow {v}_{2} = \frac{1}{\sqrt{2}}\left(\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right)

with the corresponding eigenvalues λ1\lambda_{1} and λ2\lambda_{2}.

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Problem 6.2.7

Consider the nonnormal 2×22 \times 2 matrix

A=(110−1) A = \left(\begin{smallmatrix} 1 & 1 \\ 0 & -1 \end{smallmatrix}\right)

with eigenvalues λ1=+1\lambda_{1} = +1 and λ2=−1\lambda_{2} = -1 and normalized eigenvectors

v→1=(10),v→2=25(1/2−1). \overrightarrow {v}_{1} = \left(\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right), \qquad \overrightarrow {v}_{2} = \frac{2}{\sqrt{5}}\left(\begin{smallmatrix} 1/2 \\ -1 \end{smallmatrix}\right).

The eigenvectors are linearly independent, but not orthonormal. Calculate v→1∗v→2\overrightarrow {v}_{1}^{*}\overrightarrow {v}_{2} and

B=λ1v→1v→1∗+λ2v→2v→2∗. B = \lambda _{1}\overrightarrow {v}_{1}\overrightarrow {v}_{1}^{*} + \lambda _{2}\overrightarrow {v}_{2}\overrightarrow {v}_{2}^{*}.

Find ∥A−B∥\left\| A - B\right\|. Discuss. Is the matrix BB normal?

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