Matrix function

If the real function f has the Taylor expansion

${\displaystyle f(x)=f(0)+f'(0)\cdot x+f''(0)\cdot {\frac {x^{2}}{2!}}+\cdots }$

then a matrix function can be defined by substituting x by a matrix: the powers become matrix powers, the additions become matrix sums and the multiplications become scaling operations. If the real series converges for ${\displaystyle |x|<r}$, then the corresponding matrix series will converge for matrix argument A if ${\displaystyle \|A\|<r}$ for some matrix norm ${\displaystyle \|\cdot \|}$ which satisfies ${\displaystyle \|AB\|\leq \|A\|\cdot \|B\|}$.

If the matrix A is diagonalizable, the problem may be reduced to an array of the function on each eigenvalue. This is to say we can find a matrix P and a diagonal matrix D such that ${\displaystyle A=P~D~P^{-1}}$. Applying the power series definition to this decomposition, we find that f(A) is defined by

${\displaystyle f(A)=P{\begin{bmatrix}f(d_{1})&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &f(d_{n})\end{bmatrix}}P^{-1}~,}$

where ${\displaystyle d_{1},\dots ,d_{n}}$ denote the diagonal entries of D.

For example, suppose one is seeking ${\displaystyle A!=\Gamma (A+1)}$ for

${\displaystyle A={\begin{bmatrix}1&3\\2&1\end{bmatrix}}~.}$

One has

${\displaystyle A=P{\begin{bmatrix}1-{\sqrt {6}}&0\\0&1+{\sqrt {6}}\end{bmatrix}}P^{-1}~,}$

for

${\displaystyle P={\begin{bmatrix}1/2&1/2\\-{\frac {1}{\sqrt {6}}}&{\frac {1}{\sqrt {6}}}\end{bmatrix}}~.}$

Application of the formula then simply yields

${\displaystyle A!={\begin{bmatrix}1/2&1/2\\-{\frac {1}{\sqrt {6}}}&{\frac {1}{\sqrt {6}}}\end{bmatrix}}\cdot {\begin{bmatrix}\Gamma (2-{\sqrt {6}})&0\\0&\Gamma (2+{\sqrt {6}})\end{bmatrix}}\cdot {\begin{bmatrix}1&-{\sqrt {6}}/2\\1&{\sqrt {6}}/2\end{bmatrix}}\approx {\begin{bmatrix}3.6274&8.8423\\5.8949&3.6274\end{bmatrix}}~.}$

Likewise,

${\displaystyle A^{4}={\begin{bmatrix}1/2&1/2\\-{\frac {1}{\sqrt {6}}}&{\frac {1}{\sqrt {6}}}\end{bmatrix}}\cdot {\begin{bmatrix}(1-{\sqrt {6}})^{4}&0\\0&(1+{\sqrt {6}})^{4}\end{bmatrix}}\cdot {\begin{bmatrix}1&-{\sqrt {6}}/2\\1&{\sqrt {6}}/2\end{bmatrix}}={\begin{bmatrix}73&84\\56&73\end{bmatrix}}~.}$

All complex matrices, whether they are diagonalizable or not, have a Jordan normal form ${\displaystyle A=P\,J\,P^{-1}}$, where the matrix J consists of Jordan blocks. Consider these blocks separately and apply the power series to a Jordan block:

${\displaystyle f\left({\begin{bmatrix}\lambda &1&0&\ldots &0\\0&\lambda &1&\vdots &\vdots \\0&0&\ddots &\ddots &\vdots \\\vdots &\ldots &\ddots &\lambda &1\\0&\ldots &\ldots &0&\lambda \end{bmatrix}}\right)={\begin{bmatrix}{\frac {f(\lambda )}{0!}}&{\frac {f'(\lambda )}{1!}}&{\frac {f''(\lambda )}{2!}}&\ldots &{\frac {f^{(n)}(\lambda )}{n!}}\\0&{\frac {f(\lambda )}{0!}}&{\frac {f'(\lambda )}{1!}}&\vdots &{\frac {f^{(n-1)}(\lambda )}{(n-1)!}}\\0&0&\ddots &\ddots &\vdots \\\vdots &\ldots &\ddots &{\frac {f(\lambda )}{0!}}&{\frac {f'(\lambda )}{1!}}\\0&\ldots &\ldots &0&{\frac {f(\lambda )}{0!}}\end{bmatrix}}.}$

This definition can be used to extend the domain of the matrix function beyond the set of matrices with spectral radius smaller than the radius of convergence of the power series. Note that there is also a connection to divided differences.

A related notion is the Jordan–Chevalley decomposition which expresses a matrix as a sum of a diagonalizable and a nilpotent part.

A Hermitian matrix has all real eigenvalues and can always be diagonalized by a unitary matrix P, according to the spectral theorem. In this case, the Jordan definition is natural. Moreover, this definition allows one to extend standard inequalities for real functions:

If ${\displaystyle f(a)\leq g(a)}$ for all eigenvalues of ${\displaystyle A}$, then ${\displaystyle f(A)\preceq g(A)}$. (As a convention, ${\displaystyle X\preceq Y\Leftrightarrow Y-X}$ is a positive-semidefinite matrix.) The proof follows directly from the definition.

Cauchy's integral formula from complex analysis can also be used to generalize scalar functions to matrix functions. Cauchy's integral formula states that for any analytic function f defined on a set D ⊂ ℂ, one has

${\displaystyle f(x)={\frac {1}{2\pi i}}\oint _{C}\!{\frac {f(z)}{z-x}}\,\mathrm {d} z~,}$

where C is a closed simple curve inside the domain D enclosing x.

Now, replace x by a matrix A and consider a path C inside D that encloses all eigenvalues of A. One possibility to achieve this is to let C be a circle around the origin with radius larger than ‖A‖ for an arbitrary matrix norm ‖•‖. Then, f(A) is definable by

${\displaystyle f(A)={\frac {1}{2\pi i}}\oint _{\!\!\!\!\!C}\!{f(z)(zI-A)^{-1}}\,\mathrm {d} z~.}$

This integral can readily be evaluated numerically using the trapezium rule, which converges exponentially in this case. That means that the precision of the result doubles when the number of nodes is doubled. In routine cases, this is bypassed by Sylvester's formula.

This idea applied to bounded linear operators on a Banach space, which can be seen as infinite matrices, leads to the holomorphic functional calculus.

The above Taylor power series allows the scalar ${\displaystyle x}$ to be replaced by the matrix. This is not true in general when expanding in terms of ${\displaystyle A(\eta )=A+\eta B}$ about ${\displaystyle \eta =0}$ unless ${\displaystyle [A,B]=0}$. A counterexample is ${\displaystyle f(x)=x^{3}}$, which has a finite length Taylor series. We compute this in two ways,

• Distributive law:

${\displaystyle f(A+\eta B)=(A+\eta B)^{3}=A^{3}+\eta (A^{2}B+ABA+BA^{2})+\eta ^{2}(AB^{2}+BAB+B^{2}A)+\eta ^{3}B^{3}}$

• Using scalar Taylor expansion for ${\displaystyle f(a+\eta b)}$ and replacing scalars with matrices at the end :

${\displaystyle {\begin{array}{rcl}f(a+\eta b)&=&f(a)+f'(a){\frac {\eta b}{1!}}+f''(a){\frac {(\eta b)^{2}}{2!}}+f'''(a){\frac {(\eta b)^{3}}{3!}}\\[.5em]&=&a^{3}+3a^{2}(\eta b)+3a(\eta b)^{2}+(\eta b)^{3}\\[.5em]&\to &A^{3}+3A^{2}(\eta B)+3A(\eta B)^{2}+(\eta B)^{3}\end{array}}}$

The scalar expression assumes commutativity while the matrix expression does not, and thus they cannot be equated directly unless ${\displaystyle [A,B]=0}$. For some f(x) this can be dealt with using the same method as scalar Taylor series. For example, ${\displaystyle f(x)={\frac {1}{x}}}$. If ${\displaystyle A^{-1}}$ exists then ${\displaystyle f(A+\eta B)=f(\mathbb {I} +\eta A^{-1}B)f(A)}$. The expansion of the first term then follows the power series given above,

${\displaystyle f(\mathbb {I} +\eta A^{-1}B)=\mathbb {I} -\eta A^{-1}B+(-\eta A^{-1}B)^{2}+\ldots =\sum _{n=0}^{\infty }(-\eta A^{-1}B)^{n}}$

The convergence criteria of the power series then apply, requiring ${\displaystyle \Vert \eta A^{-1}B\Vert }$ to be sufficiently small under the appropriate matrix norm. For more general problems, which cannot be rewritten in such a way that the two matrices commute, the ordering of matrix products produced by repeated application of the Leibniz rule must be tracked.

An arbitrary function f(A) of a 2×2 matrix A has its Sylvester's formula simplify to

${\displaystyle f(A)={\frac {f(\lambda _{+})+f(\lambda _{-})}{2}}I+{\frac {A-\left({\frac {tr(A)}{2}}\right)I}{\sqrt {\left({\frac {tr(A)}{2}}\right)^{2}-|A|}}}{\frac {f(\lambda _{+})-f(\lambda _{-})}{2}}~,}$

where ${\displaystyle \lambda _{\pm }}$ are the eigenvalues of its characteristic equation, |A-λI|=0, and are given by

${\displaystyle \lambda _{\pm }={\frac {tr(A)}{2}}\pm {\sqrt {\left({\frac {tr(A)}{2}}\right)^{2}-|A|}}.}$

• Matrix polynomial
• Matrix root
• Matrix logarithm
• Matrix exponential
• Matrix sign function[1]

A function ${\displaystyle f}$ is called operator monotone if and only if ${\displaystyle 0\prec A\preceq H\Rightarrow f(A)\preceq f(H)}$ for all self-adjoint matrices ${\displaystyle A,H}$ with spectra in the domain of f. This is analogous to monotone function in the scalar case.

A function ${\displaystyle f}$ is called operator concave if and only if

${\displaystyle \tau f(A)+(1-\tau )f(H)\preceq f\left(\tau A+(1-\tau )H\right)}$

for all self-adjoint matrices ${\displaystyle A,H}$ with spectra in the domain of f and ${\displaystyle \tau \in [0,1]}$. This definition is analogous to a concave scalar function. An operator convex function can be defined be switching ${\displaystyle \preceq }$ to ${\displaystyle \succeq }$ in the definition above.

The matrix log is both operator monotone and operator concave. The matrix square is operator convex. The matrix exponential is none of these. Loewner's theorem states that a function on an open interval is operator monotone if and only if it has an analytic extension to the upper and lower complex half planes so that the upper half plane is mapped to itself.[2]