Positive Operators

Andrew Thangaraj

Aug-Nov 2020


  • 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.


Not symmetric



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.


\(P_Uv=u\) 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 \(R\succeq0\).


  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.


\(T^*T\) 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 \(T=RR^*\)


(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)

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

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