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Linear Algebra
Linear Operators and Matrices
A(Σai∣vi⟩)=ΣaiA∣vi⟩A (\Sigma a_i |v_i\rangle) = \Sigma a_iA|v_i\rangle
Now, see a linear operator is just a matrix. Suppose A:V→WA: V \to W and ∣v1⟩,∣v2⟩,...,∣vm⟩|v_1\rangle, |v_2\rangle, ..., |v_m\rangle are basis of VV and ∣w1⟩,∣w2⟩,...,∣wn⟩|w_1\rangle, |w_2\rangle, ..., |w_n\rangle is a basis of WW then,
A∣vi⟩=ΣjAji∣wj⟩A|v_i\rangle = \Sigma_j A_{ji}|w_j\rangle
Inner Products
Ok so imagine an operation (_,_):V×V→C(\_ ,\_): V\times V \to \mathbb{C} such that the following shit holds ok?
1.
(∣v⟩,Σiλi∣wi⟩)=Σiλi(∣v⟩,∣wi⟩)(|v\rangle, \Sigma_i \lambda_i|w_i\rangle) = \Sigma_i\lambda_i(|v\rangle, |w_i\rangle)
2.
(∣v⟩,∣w⟩)=(∣w⟩,∣v⟩)∗(|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*
3.
(∣v⟩,∣v⟩)≥0(|v\rangle, |v\rangle) \geq 0 and =0= 0  iff ∣v⟩|v\rangle
In finite dimensions, inner product space i.e., vector spaces equipped with inner prducts for all ∣v⟩∈|v\rangle \in  vector space ==  Hilbert Space
Consider ∣i⟩ & ∣j⟩|i\rangle\ \&\ |j\rangle to be orthonormal basis, we have —
⟨v∣w⟩=(Σivi∣i⟩,Σjwj∣j⟩)=ΣiΣjvi∗wjδij=Σivi∗wi=∣v⟩†∣w⟩\langle v|w \rangle = (\Sigma_i v_i|i\rangle, \Sigma_jw_j|j\rangle) = \Sigma_i \Sigma_j v_i^*w_j\delta_{ij} = \Sigma_iv_i^*w_i = |v\rangle^\dagger |w\rangle
Norm of a vector
∣∣v∣∣=⟨v∣v⟩||v|| = \sqrt{\langle v|v \rangle}
We can say that ∣v⟩|v\rangle is normalized iff ∣∣v∣∣=1||v|| = 1.
A set of ∣ai⟩|a_i\rangle vectors is orthonormal if ⟨ai∣aj⟩=δij\langle a_i|a_j \rangle = \delta_{ij} i.e., ∀ i≠j ⟨ai∣aj⟩=0\forall\ i \neq j\ \langle a_i|a_j \rangle = 0 and ⟨ai∣aj⟩=1 ∀ i=j\langle a_i|a_j \rangle = 1\ \forall\ i=j.
Gram Schmidt: for orthonormal basis
∣vk+1⟩=∣wk+1⟩−Σi=1k⟨vi∣wk+1⟩∣vi⟩∣∣∣wk+1⟩−Σi=1k⟨vi∣wk+1⟩∣vi⟩∣∣, ∣v1⟩=∣w1⟩/∣∣∣w1⟩∣∣|v_{k+1}\rangle = \frac{|w_{k+1}\rangle - \Sigma_{i=1}^k \langle v_i | w_{k+1}\rangle |v_i\rangle}{|||w_{k+1}\rangle - \Sigma_{i=1}^k \langle v_i | w_{k+1}\rangle |v_i\rangle||},\ |v_1\rangle = |w_1\rangle/|||w_1 \rangle||
Outer Product
∣w⟩⟨v∣(∣v′⟩)=∣w⟩∣v⟩†∣v′⟩=∣w⟩⟨v∣v′⟩=⟨v∣v′⟩∣w⟩|w\rangle \langle v|(|v'\rangle) = |w\rangle |v\rangle^\dagger |v'\rangle = |w\rangle \langle v|v' \rangle = \langle v|v' \rangle |w\rangle
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From this notion we obtain the completeness relation, Σi∣i⟩⟨i∣=I\Sigma_i |i\rangle \langle i| = I.
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A=IwAIv=Σij∣wj⟩⟨wj∣A∣vi⟩⟨vi∣=Σij⟨wj∣A∣vi⟩∣wj⟩⟨vi∣A = I_wAI_v = \Sigma_{ij} |w_j\rangle\langle w_j|A|v_i\rangle\langle v_i| = \Sigma_{ij} \langle w_j|A|v_i\rangle|w_j\rangle\langle v_i|
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Cauchy Schwarz: ⟨v∣v⟩⟨w∣w⟩≥⟨v∣w⟩⟨w∣v⟩=∣⟨v∣w⟩∣2\langle v|v \rangle \langle w|w \rangle \geq \langle v|w \rangle \langle w|v \rangle = |\langle v|w \rangle|^2
Eigenvectors and Eigenvalues
Under a given linear transformation AA, A∣v⟩=λ∣v⟩A|v\rangle = \lambda|v\rangle  where ∃ ∣v⟩\exists\ |v\rangle  s.t. they do not get shifted off their span.
All such vectors are referred as eigenvectors and (A−λI)∣v⟩=0  ⟹  det∣A−λI∣=0(A - \lambda I)|v\rangle = 0 \implies det|A-\lambda I| = 0 gives all possible eigenvalue.
Eigenspace
It is the space of all vectors with a given eigenvalue λ\lambda. When an eigenspace is more than one dimensional, we call it degenerate.
Adjoints and Hermitian
Suppose A:V→VA: V \to V then ∃ A†:V→V\exists\ A^\dagger: V \to V such that ∀ ∣v⟩, ∣w⟩∈V\forall\ \vert v\rangle,\ \vert w\rangle \in V we have,
(∣v⟩,A∣w⟩)=(A†∣v⟩,∣w⟩)(\vert v\rangle, A\vert w\rangle) = (A^\dagger \vert v\rangle, \vert w\rangle)
This operator is called as the adjoint or Hermitian conjugate of the operator AA.
(∣v⟩,A∣w⟩)=⟨v∣A∣w⟩=v†Aw=(A†v)†w=(A†∣v⟩,∣w⟩)(\vert v\rangle, A\vert w\rangle) = \langle v\vert A\vert w\rangle = \bold{v}^\dagger A\bold{w} = (A^\dagger\bold{v})^\dagger\bold{w} = (A^\dagger\vert v\rangle, \vert w\rangle)
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We have, (AB)†=B†A†(AB)^\dagger = B^\dagger A^\dagger
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∣v⟩†=⟨v∣\vert v\rangle^\dagger = \langle v\vert
Some defintions
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Normal matrices: AA†=A†AAA^\dagger = A^\dagger A
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Hermitian matrices: A†=AA^\dagger = A
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Unitary matrices: AA†=IAA^\dagger = I
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A normal matrix is Hermitian if and only if it has real eigenvalues.
Spectral Decomposition
Definition: A linear operator is diagonalizable if and only if it is normal.
Some notes and derivation regarding the above:
Av⃗=λv⃗=Σiλijqi⃗A\vec{v} = \lambda\vec{v} = \Sigma_i \lambda_{ij}\vec{q_i}  where qiq_i's are linearly independent eigenvalues of AA.
AQ=QΛAQ = Q\Lambda where Q=[q1q2…qn]  ⟹  A=QΛQ−1Q = \begin{bmatrix} q_1&q_2 &\ldots&q_n \end{bmatrix} \implies A = Q{\Lambda}Q^{-1}
A=IAI=(P+Q)A(P+Q)=PAP+QAP+PAQ+QAQ  ⟹  A=λP2+0+0+QAQ  ⟹  A=λP2+QAQA = IAI = (P+Q)A(P+Q) = PAP + QAP + PAQ + QAQ\\ \implies A = \lambda{P^2} + 0 + 0 + QAQ\\ \implies A = \lambda{P^2} + QAQ
Matrices and Vectors
In the following statements we are dealing with {∣i⟩}\{\vert i\rangle\}  as a orthonormal basis set.
I=Σ∣i⟩⟨i∣∣ψ⟩=Σσi∣i⟩, where σi=⟨i∣ψ⟩I = \Sigma \vert i\rangle\langle i\vert\\ \vert \psi \rangle = \Sigma \sigma_i\vert i\rangle, \text{ where } \sigma_i = \langle i\vert \psi\rangle\\
Now, to represent a operator or linear transformation as matrix in orthonormal basis.
Aij=⟨i∣A∣j⟩A=Σi,j⟨i∣A∣j⟩∣i⟩⟨j∣tr(A)=Σi⟨i∣A∣i⟩A_{ij} = \langle i\vert A\vert j\rangle\\ A = \Sigma_{i, j} \langle i\vert A\vert j\rangle \vert i\rangle \langle j\vert\\ tr(A) = \Sigma_i \langle i\vert A\vert i\rangle
Now diagonalization for any normal matrix.
M=Σiλi∣i⟩⟨i∣=ΣiλiPi, where Pi†=Pif(M)=Σif(λi)∣i⟩⟨i∣M = \Sigma_i \lambda_i \vert i\rangle \langle i\vert = \Sigma_i \lambda_i P_i,\ \text{where}\ P_i^\dagger = P_i\\ f(M) = \Sigma_i f(\lambda_i) \vert i\rangle \langle i\vert
where λi\lambda_i are eigenvalues of MM under a given orthonormal basis set {∣i⟩}\{\vert i\rangle\} for vector space VV, each ∣i⟩\vert i \rangle is an eigenvector of MM with eigenvalue λi\lambda_i.
If MM is Hermitian, all eigenvalues (λi s)(\lambda_i\text{ s}) are non-negative.
Tensor Products
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z∣vw⟩=(z∣v⟩)⊗(∣w⟩)=(∣v⟩)⊗(z∣w⟩)z\vert{vw}\rangle = (z\vert{v}\rangle) \otimes (\vert{w}\rangle) =(\vert{v}\rangle) \otimes (z\vert{w}\rangle) 
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(∣v1⟩+∣v2⟩)⊗∣w⟩=∣v1w⟩+∣v2w⟩(\vert v_1 \rangle + \vert v_2 \rangle) \otimes \vert w \rangle = \vert{v_1w}\rangle + \vert{v_2w}\rangle 
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∣v⟩⊗(∣w1⟩+∣w2⟩)=∣vw1⟩+∣vw2⟩\vert v \rangle \otimes (\vert w_1 \rangle + \vert w_2 \rangle) = \vert{vw_1}\rangle + \vert{vw_2}\rangle 
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∣ψ⟩⊗k=∣ψ⟩⊗…⊗∣ψ⟩ k times\vert \psi \rangle^{\otimes^k} = \vert \psi \rangle \otimes \ldots \otimes \vert \psi \rangle \text{ k times}
Linear Product
A⊗BA\otimes{B} forms the linear operator that acts on V⊗WV\otimes W vector space givern that AA acts on VV and BB acts on WW.
(A⊗B)(Σiai∣vi⟩⊗∣wi⟩)=ΣiaiA∣vi⟩⊗B∣wi⟩(A\otimes B)(\Sigma_{i} a_i\vert v_i\rangle \otimes \vert w_i \rangle) = \Sigma_{i} a_iA\vert v_i\rangle \otimes B\vert w_i \rangle
Inner Product
(Σiai∣vi⟩⊗∣wi⟩,Σiaj∣vj′⟩⊗∣wj′⟩)=Σijai∗bj⟨vi∣vj′⟩⟨wi∣wj′⟩(\Sigma_{i} a_i\vert v_i\rangle \otimes \vert w_i \rangle, \Sigma_{i} a_j\vert v'_j\rangle \otimes \vert w'_j \rangle) = \Sigma_{ij} a_i^*b_j \langle v_i\vert v'_j\rangle\langle w_i\vert w'_j\rangle
Trace
Properties of trace are given below as follows.
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tr(A)=ΣiAiitr(A) = \Sigma_i A_{ii}
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tr(A)=Σi⟨i∣A∣i⟩tr(A) = \Sigma_i \langle i\vert A\vert i\rangle for orthonormal basis
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tr(AB)=tr(BA)tr(AB) = tr(BA)
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tr(zA+B)=z⋅tr(A)+tr(B)tr(zA+B) = z\cdot tr(A) + tr(B)
The above properties yield certain implications as follows.
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tr(UAU†)=tr(A)tr(UAU^\dagger) = tr(A)
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tr(A∣ψ⟩⟨ψ∣)=Σi⟨i∣A∣ψ⟩⟨ψ∣i⟩tr(A\vert \psi\rangle\langle\psi\vert) = \Sigma_i \langle i\vert A\vert \psi\rangle\langle\psi\vert i \rangle
Hilbert-Schimdt Inner Product
LvL_v forms the vector space of operators over the Hilbert space VV. Then, we can show that LvL_v  is also a Hilbert space with tr(A†B)tr(A^\dagger B) as the inner product operator on Lv×LvL_v \times L_v.
Also, we have div(Lv)=dim(V)2div(L_v) = dim(V)^2.
Commutator and Anti-commutator
[A,B]=AB−BA{A,B}=AB+BA[A, B] = AB - BA\\ \{A, B\} = AB + BA
Theorem of Simultaneous Diagonalization
Suppose AA and BB are both Hermitian matrices, then [A,B]=0[A, B] = 0 iff ∃\exists orthonormal basis such that both AA and BB are diagonal with respect to that basis.
Polar Value Decomposition
If AA is any linear operator and UU is a unitary then J,KJ, K are positive operators, such that
A=UJ=KU, where J=A†A and K=AA†A = UJ = KU, \text{ where } J = \sqrt{A^\dagger A} \text{ and } K = \sqrt{AA^\dagger}
Moreover, if A−1A^{-1} exists, then UU is unique.
Singular Value Decomposition
If AA is a square matrix and ∃ U,V\exists\ U, V unitaries then DD is a diagonal matrix, such that
A=UDVA = UDV
where DD has non-negative values.