## Linear Operators and Matrices

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,

## Inner Products

Ok so imagine an operation \((\_ ,\_): V\times V \to \mathbb{C}\) such that the following shit holds ok?

- \((|v\rangle, \Sigma_i \lambda_i|w_i\rangle) = \Sigma_i\lambda_i(|v\rangle, |w_i\rangle)\)
- \((|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*\)
- \((|v\rangle, |v\rangle) \geq 0\) 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 —

## Norm of a vector

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

## Outer Product

- 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.

## 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.

If all \(λ_i ≥ 0\), it is positive semidefinite and if they are \(> 0\), it is positive definite.

## 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.

## Adjoints and Hermitian

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,

This operator is called as the adjoint or Hermitian conjugate of the operator \(A\).

- 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 \(Q = \begin{bmatrix} q_1&q_2 &\ldots&q_n \end{bmatrix} \implies A = Q{\Lambda}Q^{-1}\)

## Matrices and Vectors

In the following statements we are dealing with \(\{\vert i\rangle\}\) as a ** orthonormal** basis set.

Now, to represent a operator or linear transformation as matrix in orthonormal basis.

Now diagonalization for any ** normal** matrix.

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 real.

## 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 \(W\).

### Inner Product

## 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.

## 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

## 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

Moreover, if \(A^{-1}\) exists, then \(U\) is unique.

## Singular Value Decomposition

SVD in general is given as follows.

It generalizes the eigen decomposition of a square normal matrix with an orthonormal eigen basis to any \(m\times n\) matrix.

\({\Sigma}\) is an \({m\times n}\) rectangular diagonal matrix with non-negative real numbers on the diagonal (called singular values).

Corollary: If \(A\) is a square matrix and \(\exists\ U, V\) unitaries then \(D\) is a diagonal matrix, such that

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.

- If \(A\) is square both SVD and EVD exist but might not be same.
- If \(A\) is a square symmetric matrix both SVD and EVD exist and are equivalent.
- If \(A\) is non-square only SVD is possible.

## Rank of a matrix

Rank \(=\) number of dimensions in column space.

- The row rank is the largest number of rows of \(A\) that constitute a linearly independent set.
- The column rank is the largest number of columns of \(A\) that constitute a linearly independent set. Moreover, column-rank \(=\) row-rank for \(A \in \mathbb{R}^{m \times n}\).

Matrix is called full rank if equality holds.

- \(rank(A^T) = rank(A)\)
- \(rank(AB) ≤ min(rank(A), rank(B))\)
- \(rank(A + B) ≤ rank(A) + rank(B)\)

## Projection and Spaces

- \(\mathcal{N} (A) = \{x ∈ \mathbb R^n : Ax = 0\}\) denotes all vectors in \(\mathbb R^n\) that land at the origin after transformation. It is also called kernel.
- \(\mathcal R(A) = \{v ∈ \mathbb R^m : v = Ax,~x ∈ \mathbb R^n \}\) denotes the space spanned by the transformed basis vectors in \(\mathbb R^n\).
- \(\mathcal R(A^T)\) and \(\mathcal N (A)\) are orthogonal spaces which together span \(\mathbb R^n\).

Determinant \(\neq 0\) implies that the matrix has an inverse.

## Quadratic Forms

Reminder: we are in real \(\mathbb R\) space.

## Moore-Penrose Pseudoinverse

This is a pseudo inverse formalism with left and right inverses.

## Row Echelon Forms

A matrix is in row echelon form if:

- All rows consisting of only zeroes are at the bottom.
- The leading coefficient (also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it.

A matrix is in reduced row echelon form if:

- It is in row echelon form.
- The leading entry in each nonzero row is a 1 (called a leading 1).
- Each column containing a leading 1 has zeros in all its other entries.

## Spectral Decomposition

Any normal operator \(M\) on a vector space \(V\) is diagonal with respect to some orthonormal basis for \(V\).

Conversely, any diagonalizable operator is normal.

### Proof

Now we have

Thus, if \(M\) is normal then \(M = PMP + QMQ\) where \(PMP = \lambda P^2 = \lambda P\) and thus is diagonalizable wrt orthonormal basis for \(P\). Similarly, \(QMQ\) is also diagonalizable wrt some orthonormal basis for \(Q\).

Thus, \(M\) is diagonalizable for orthonormal basis of the entire vector space.

## Polar Value Decomposition

Let \(A\) be a matrix on vector space \(V\). Then there exists unitary \(U\) and positive operators \(J\) and \(K\) such that, \(A= UJ=KU\) where the unique positive operators shall satisfy the equations \(J \equiv \sqrt{A^\dagger A}\) and \(K \equiv \sqrt{AA^\dagger}\).

Moreover, if \(A\) is invertible then \(U\) is unique.

## Singular Value Decomposition

Let \(A\) be a square matrix. Then there exist unitary matrices \(U\) and \(V\), and a diagonal matrix \(D\) with non-negative entries such that \(A = UDV\).

The diagonal elements of \(D\) are called the singular values of \(A\).

### Proof

From polar value decomposition we have, \(A = SJ = STDT^\dagger = (ST)D(T^\dagger) = UDV\).