Positive Operators

Andrew Thangaraj

Aug-Nov 2020

Recap

  • Vector space \(V\) over a scalar field \(F= \mathbb{R}\) or \(\mathbb{C}\)
  • \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
    • dim null \(T+\) dim range \(T=\) dim \(V\)
    • Solution to \(Ax=b\) (if it exists): \(u+\) null\((A)\)
  • Four fundamental subspaces of a matrix
    • Column space, row space, null space, left null space
  • Eigenvalue \(\lambda\) and Eigenvector \(v\): \(Tv=\lambda v\)
    • There is a basis w.r.t. which a linear map is upper-triangular
    • If there is a basis of eigenvectors, linear map is diagonal w.r.t. it
  • Inner products, norms, orthogonality and orthonormal basis
    • There is an orthonormal basis w.r.t. which a linear map is upper-triangular
    • Orthogonal projection: distance from a subspace
  • Adjoint of a linear map: \(\langle Tv,w\rangle=\langle v,T^*w\rangle\)
    • null \(T=\) \((\)range \(T^*)^{\perp}\)
  • Self-adjoint: \(T=T^*\), Normal: \(TT^*=T^*T\)
    • Eigenvectors corresponding to different eigenvalues are orthogonal
  • Complex spectral theorem: \(T\) is normal \(\leftrightarrow\) orthonormal basis of eigenvectors
  • Real spectral theorem: \(T\) is self-adjoint \(\leftrightarrow\) orthonormal basis of eigenvectors

Definition of positive operators

\(V\): inner product space over \(F= \mathbb{R}\) or \(\mathbb{C}\)

An operators \(T:V\to V\) is said to be positive if \(T\) is self-adjoint and \[\langle Tv,v\rangle\ge 0,\quad v\in V\]

Necessary condition for being positive

\(\langle Tv,v\rangle\in\mathbb{R},\quad v\in V\)

\(F=\mathbb{C}\): above implies \(T\) is self-adjoint

\(F=\mathbb{R}\): \(T\) self-adjoint is not implied

In terms of matrices

\(A\): \(n\times n\) matrix with, possibly, complex entries

\(A\) is positive if \(A=A^H\) and \(x^HAx\ge0\), \(x\in F^n\)

Notation: \(x^H\) denotes conjugate-transpose

Written as \(A\succeq0\)

  1. \(A=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\): \(A\succeq0\) if \(\lambda_1,\lambda_2\ge0\)

    \(x^HAx=\begin{bmatrix}\overline{x_1}&\overline{x_2}\end{bmatrix}\begin{bmatrix}\lambda_1x_1\\\lambda_2x_2\end{bmatrix}=\lambda_1\lvert x_1\rvert^2+\lambda_2\lvert x_2\rvert^2\)

  2. \(A=\begin{bmatrix}2&i\\-i&2\end{bmatrix}\): \(A\succeq0\) (exercise)

    \(x^HAx=2\lvert x_1\rvert^2+i\overline{x_1}x_2-ix_1\overline{x_2}+2\lvert x_2\rvert^2\)

  3. \(A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\): not positive

Is self-adjoint necessary in real spaces?

In real spaces, there are some interesting examples.

\(A=\begin{bmatrix}1&1\\-1&1\end{bmatrix}\)

Not symmetric

\(x^TAx=x_1^2+x_2^2\ge0\)

\(x^TAx=x^T\dfrac{A+A^T}{2}x\)

So, w.r.t. positivity, it is sufficient to consider symmetric matrices.

Orthogonal Projections and Positivity

\(U\subseteq V\) subspace, \(P_U\): orthogonal projection onto \(U\)

\(P_U\) is positive.

Proof

\(P_Uv=u\) implies \(v=u+w\) with \(u\in U\) and \(w\in U^{\perp}\)

\(\langle v,u\rangle=\langle u,u\rangle+\langle w,u\rangle=\langle u,u\rangle\ge0\)

So, \(\langle v,P_Uv\rangle=\langle P_Uv,P_Uv\rangle\)

Is \(P_U\) self-adjoint?

\(0=\langle P_Ux,y-P_Uy\rangle=\langle P_Ux,y\rangle - \langle P_Ux,P_Uy\rangle\)

\(0=\langle x-P_Ux,P_Uy\rangle=\langle x,P_Uy\rangle - \langle P_Ux,P_Uy\rangle\)

So, \(\langle P_Ux,y\rangle=\langle x,P_Uy\rangle\)

Square root of an operator

\(R\) is said to be a square root of \(T\) is \(T=R^2\).

Square root is said to be positive if \(R\succeq0\).

Examples

  1. \(A=\begin{bmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{bmatrix}\), \(B=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}\), \(B^2=A\)

  2. \(P=\begin{bmatrix}0&0\\ \alpha&1\end{bmatrix}\), \(P^2=P\) (is \(P\) an orthogonal projection?)

  3. \(A=\begin{bmatrix}4&0\\ 0&9\end{bmatrix}\), \(B=\begin{bmatrix}2&0\\ 0&3\end{bmatrix}\), \(B\): positive square root of \(A\)

Operator-adjoint product and positivity

\(T:V\to W\), \(T^*:W\to V\)

\(TT^*\) and \(T^*T\) are positive operators.

Proof

\(T^*T\) and \(TT^*\) are clearly self-adjoint

\(\langle T^*Tv,v\rangle=\langle Tv,Tv\rangle=\lVert Tv\rVert^2\ge0\)

\(\langle TT^*v,v\rangle=\langle T^*v,T^*v\rangle=\lVert T^*v\rVert^2\ge0\)

Characterising positive operators

The following are equivalent:

  1. \(T\) is positive
  2. \(T\) is self-adjoint with non-negative eigenvalues
  3. \(T\) has a positive square root
  4. \(T\) has a self-adjoint square root
  5. There is an operator \(R\) such that \(T=RR^*\)

Proof

(1) implies (2)

\(\lambda\): eigenvalue of \(T\) with eigenvector \(v\)

\(0\le\langle Tv,v\rangle=\langle\lambda v,v\rangle=\lambda\langle v,v\rangle\)

Proof (continued)

(2) implies (3), (4), (5)

\(\{e_1,\ldots,e_n\}\): orthonormal eigenvector basis of \(T\) (coordinates in standard basis)

\(T\leftrightarrow\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_ne_n\overline{e^T_n}\), \(\lambda_i\ge0\)

Let \(R=\sqrt{\lambda_1}e_1\overline{e^T_1}+\cdots+\sqrt{\lambda_n}e_n\overline{e^T_n}\)

Verify \(T=R^2\), \(R=R^*\)

(5) implies (1)

\(RR^*\): operator-adjoint product

Partial ordering of operators

Condition Notation Terminology
\(x^HAx>0\) \(A\succ0\) positive definite (pd)
\(x^HAx\ge0\) \(A\succeq0\) positive semidefinite (psd)
\(x^HAx<0\) \(A\prec0\) negative definite (nd)
\(x^HAx\le0\) \(A\preceq0\) negative semidefinite (nsd)

\(A\succ B\) if \(A-B\succ0\)
(and similarly for other relations)

Why partial ordering?: There are matrices that are neither positive nor negative.

Example: \(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\)

So, \(A\) and \(B\) may not be comparable using \(\succ\) or \(\prec\)

Positive: short for positive semidefinite