Linear Algebra
Linear Operators and Matrices
$A (\Sigma a_i |v_i\rangle) = \Sigma a_iA|v_i\rangle$
Now, see a linear operator is just a matrix. Suppose $A: V \to W$﻿ and $|v_1\rangle, |v_2\rangle, ..., |v_m\rangle$﻿ are basis of $V$﻿ and $|w_1\rangle, |w_2\rangle, ..., |w_n\rangle$﻿ is a basis of $W$﻿ then,
$A|v_j\rangle = \Sigma_i A_{ij}|w_i\rangle$
Inner Products
Ok so imagine an operation $(\_ ,\_): V\times V \to \mathbb{C}$﻿ such that the following shit holds ok?
1.
$(|v\rangle, \Sigma_i \lambda_i|w_i\rangle) = \Sigma_i\lambda_i(|v\rangle, |w_i\rangle)$﻿
2.
$(|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*$﻿
3.
$(|v\rangle, |v\rangle) \geq 0$﻿ and $= 0$﻿ iff $|v\rangle$﻿
In finite dimensions, inner product space i.e., vector spaces equipped with inner prducts for all $|v\rangle \in$﻿ vector space $=$﻿ Hilbert Space
Consider $|i\rangle\ \&\ |j\rangle$﻿ to be orthonormal basis, we have —
$\langle v|w \rangle = (\Sigma_i v_i|i\rangle, \Sigma_jw_j|j\rangle) = \Sigma_i \Sigma_j v_i^*w_j\delta_{ij} = \Sigma_iv_i^*w_i = |v\rangle^\dagger |w\rangle$
Norm of a vector
$||v|| = \sqrt{\langle v|v \rangle}$
We can say that $|v\rangle$﻿ is normalized iff $||v|| = 1$﻿.
A set of $|a_i\rangle$﻿ vectors is orthonormal if $\langle a_i|a_j \rangle = \delta_{ij}$﻿ i.e., $\forall\ i \neq j\ \langle a_i|a_j \rangle = 0$﻿ and $\langle a_i|a_j \rangle = 1\ \forall\ i=j$﻿.
Gram Schmidt: for orthonormal basis
$|v_{k+1}\rangle = \frac{|w_{k+1}\rangle - \Sigma_{i=1}^k \langle v_i | w_{k+1}\rangle |v_i\rangle}{|||w_{k+1}\rangle - \Sigma_{i=1}^k \langle v_i | w_{k+1}\rangle |v_i\rangle||},\ |v_1\rangle = |w_1\rangle/|||w_1 \rangle||$
Outer Product
$|w\rangle \langle v|(|v'\rangle) = |w\rangle |v\rangle^\dagger |v'\rangle = |w\rangle \langle v|v' \rangle = \langle v|v' \rangle |w\rangle$
From this notion we obtain the completeness relation, $\Sigma_i |i\rangle \langle i| = I$﻿.
$A = I_wAI_v = \Sigma_{ij} |w_j\rangle\langle w_j|A|v_i\rangle\langle v_i| = \Sigma_{ij} \langle w_j|A|v_i\rangle|w_j\rangle\langle v_i|$﻿
Cauchy Schwarz: $\langle v|v \rangle \langle w|w \rangle \geq \langle v|w \rangle \langle w|v \rangle = |\langle v|w \rangle|^2$﻿
Hilbert Space
A Hilbert Space $\mathcal{H}$﻿ is complete which means that every Cauchy sequence of vectors admits in the space itself. Under this hypothesis there exist Hilbert bases also known as complete orthonormal systems of vectors in $\mathcal{H}$﻿.
For any orthonormal basis of $\mathcal{H}$﻿, we have the following.
$\text{Orthonormality} \equiv\langle \psi_i|\psi_j\rangle = \delta_{ij}\\ \text{Completeness} \equiv\sum_{i} |\psi_i\rangle\langle\psi_i| = I$
Eigenvectors and Eigenvalues
Under a given linear transformation $A$﻿, $A|v\rangle = \lambda|v\rangle$﻿ where $\exists\ |v\rangle$﻿ s.t. they do not get shifted off their span.
All such vectors are referred as eigenvectors and $(A - \lambda I)|v\rangle = 0 \implies det|A-\lambda I| = 0$﻿ gives all possible eigenvalue.
Eigenspace
It is the space of all vectors with a given eigenvalue $\lambda$﻿. When an eigenspace is more than one dimensional, we call it degenerate.
Suppose $A: V \to V$﻿ then $\exists\ A^\dagger: V \to V$﻿ such that $\forall\ \vert v\rangle,\ \vert w\rangle \in V$﻿ we have,
$(\vert v\rangle, A\vert w\rangle) = (A^\dagger \vert v\rangle, \vert w\rangle)$
This operator is called as the adjoint or Hermitian conjugate of the operator $A$﻿.
$(\vert v\rangle, A\vert w\rangle) = \langle v\vert A\vert w\rangle = \bold{v}^\dagger A\bold{w} = (A^\dagger\bold{v})^\dagger\bold{w} = (A^\dagger\vert v\rangle, \vert w\rangle)$
We have, $(AB)^\dagger = B^\dagger A^\dagger$﻿
$\vert v\rangle^\dagger = \langle v\vert$﻿
Some defintions
Normal matrices: $AA^\dagger = A^\dagger A$﻿
Hermitian matrices: $A^\dagger = A$﻿
Unitary matrices: $AA^\dagger = I$﻿
A normal matrix is Hermitian if and only if it has real eigenvalues.
If $\langle x| A|x\rangle \geq 0, \forall\ |x\rangle$﻿ then $A$﻿ is positive semi-definite and has positive eigenvalues.
Some properties
If a Hermitian matrix has positive eigenvalues then it is positive semi-definite.
If $M = AA^\dagger$﻿ then it is both Hermitian and positive semi-definite.
All positive semi-definite operators are Hermitian, by definition.
Spectral Decomposition
Definition: A linear operator is diagonalizable if and only if it is normal.
Some notes and derivation regarding the above:
$A\vec{v} = \lambda\vec{v} = \Sigma_i \lambda_{ij}\vec{q_i}$﻿ where $q_i$﻿'s are linearly independent eigenvalues of $A$﻿.
$AQ = Q\Lambda$﻿ where $Q = \begin{bmatrix} q_1&q_2 &\ldots&q_n \end{bmatrix} \implies A = Q{\Lambda}Q^{-1}$﻿
$A = IAI = (P+Q)A(P+Q) = PAP + QAP + PAQ + QAQ\\ \implies A = \lambda{P^2} + 0 + 0 + QAQ\\ \implies A = \lambda{P^2} + QAQ$﻿
Matrices and Vectors
In the following statements we are dealing with $\{\vert i\rangle\}$﻿ as a orthonormal basis set.
$I = \Sigma \vert i\rangle\langle i\vert\\ \vert \psi \rangle = \Sigma \sigma_i\vert i\rangle, \text{ where } \sigma_i = \langle i\vert \psi\rangle\\$
Now, to represent a operator or linear transformation as matrix in orthonormal basis.
$A_{ij} = \langle i\vert A\vert j\rangle\\ A = \Sigma_{i, j} \langle i\vert A\vert j\rangle \vert i\rangle \langle j\vert\\ \text{tr}(A) = \Sigma_i \langle i\vert A\vert i\rangle$
Now diagonalization for any normal matrix.
$M = \Sigma_i \lambda_i \vert i\rangle \langle i\vert = \Sigma_i \lambda_i P_i,\ \text{where}\ P_i^\dagger = P_i\\ f(M) = \Sigma_i f(\lambda_i) \vert i\rangle \langle i\vert$
where $\lambda_i$﻿ are eigenvalues of $M$﻿ under a given orthonormal basis set $\{\vert i\rangle\}$﻿ for vector space $V$﻿, each $\vert i \rangle$﻿ is an eigenvector of $M$﻿ with eigenvalue $\lambda_i$﻿.
If $M$﻿ is Hermitian, all eigenvalues $(\lambda_i\text{ s})$﻿ are non-negative.
Tensor Products $z\vert{vw}\rangle = (z\vert{v}\rangle) \otimes (\vert{w}\rangle) =(\vert{v}\rangle) \otimes (z\vert{w}\rangle)$﻿
$(\vert v_1 \rangle + \vert v_2 \rangle) \otimes \vert w \rangle = \vert{v_1w}\rangle + \vert{v_2w}\rangle$﻿
$\vert v \rangle \otimes (\vert w_1 \rangle + \vert w_2 \rangle) = \vert{vw_1}\rangle + \vert{vw_2}\rangle$﻿
$\vert \psi \rangle^{\otimes^k} = \vert \psi \rangle \otimes \ldots \otimes \vert \psi \rangle \text{ k times}$﻿
$(A \otimes B)^\dagger = A^\dagger \otimes B^\dagger$﻿
Linear Product
$A\otimes{B}$﻿ forms the linear operator that acts on $V\otimes W$﻿ vector space givern that $A$﻿ acts on $V$﻿ and $B$﻿ acts on $W$﻿.
$(A\otimes B)(\Sigma_{i} a_i\vert v_i\rangle \otimes \vert w_i \rangle) = \Sigma_{i} a_iA\vert v_i\rangle \otimes B\vert w_i \rangle$
Inner Product
$(\Sigma_{i} a_i\vert v_i\rangle \otimes \vert w_i \rangle, \Sigma_{i} a_j\vert v'_j\rangle \otimes \vert w'_j \rangle) = \Sigma_{ij} a_i^*b_j \langle v_i\vert v'_j\rangle\langle w_i\vert w'_j\rangle$
Trace
Properties of trace are given below as follows.
$\text{tr}(A) = \Sigma_i A_{ii}$﻿
$\text{tr}(A) = \Sigma_i \langle i\vert A\vert i\rangle$﻿ for orthonormal basis
$\text{tr}(AB) = \text{tr}(BA)$﻿
$\text{tr}(zA+B) = z\cdot \text{tr}(A) + \text{tr}(B)$﻿
The above properties yield certain implications as follows.
$\text{tr}(UAU^\dagger) = tr(A)$﻿
$\text{tr}(A\vert \psi\rangle\langle\psi\vert) = \Sigma_i \langle i\vert A\vert \psi\rangle\langle\psi\vert i \rangle$﻿
$\text{tr}(A) = \sum_i \lambda_i$﻿, $\text{det}(A) = \prod_i \lambda_i$﻿ with algebraic multiplicities
Partial Trace
Entanglement excludes the possibility of associating state vectors with individual subsystems. Therefore, we introduce density matrices and the corresponding idea of reduction preformed with partial trace.
$\text{tr}(A \otimes B) = \text{ tr}(A) \cdot \text{tr}(B)$
$\rho_{AB} : \mathcal{H_A}\otimes\mathcal{H_B} \xrightarrow{\text{ tr}_B} \rho_A : \mathcal{H_A}$
$\text{ tr}_B(AB) = A \text{ tr}(B)$
Hilbert-Schimdt Inner Product
$L_v$﻿ forms the vector space of operators over the Hilbert space $V$﻿. Then, we can show that $L_v$﻿ is also a Hilbert space with $\text{tr}(A^\dagger B)$﻿ as the inner product operator on $L_v \times L_v$﻿.
Also, we have $div(L_v) = dim(V)^2$﻿.
Commutator and Anti-commutator
$[A, B] = AB - BA\\ \{A, B\} = AB + BA$
Theorem of Simultaneous Diagonalization
Suppose $A$﻿ and $B$﻿ are both Hermitian matrices, then $[A, B] = 0$﻿ iff $\exists$﻿ orthonormal basis such that both $A$﻿ and $B$﻿ are diagonal with respect to that basis.
Polar Value Decomposition
If $A$﻿ is any linear operator and $U$﻿ is a unitary then $J, K$﻿ are positive operators, such that
$A = UJ = KU, \text{ where } J = \sqrt{A^\dagger A} \text{ and } K = \sqrt{AA^\dagger}$
Moreover, if $A^{-1}$﻿ exists, then $U$﻿ is unique.
Singular Value Decomposition
If $A$﻿ is a square matrix and $\exists\ U, V$﻿ unitaries then $D$﻿ is a diagonal matrix, such that
$A = UDV$
where $D$﻿ has non-negative values.
Corollary: If $A$﻿ has non-negative eigenvalues then, $A = U^\dagger DU$﻿ is possible where $D$﻿ has non-negative values.