How could we define the factorial of a matrix?

Question

Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$.

How could we define the following operation? $$\mathsf{A}!$$

Maybe we could make some simple example, admitted it makes any sense, with

$$\mathsf{A} = \left(\begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix} \right) $$

Answer 1

For any holomorphic function $G$, we can define a corresponding matrix function $\tilde{G}$ via (a formal version of) the Cauchy Integral Formula: We set $$\tilde{G}(B) := \frac{1}{2 \pi i} \oint_C G(z) (z I - B)^{-1} \, dz ,$$ where $C$ is an (arbitrary) anticlockwise curve that encloses (once each) the eigenvalues of the (square) matrix $B$. Note that the condition on $C$ means that restrictions on the domain of $G$ determine restrictions on the domain of $\tilde{G}$.

So, we could make sense of the factorial of a matrix if we had a holomorphic function that restricted to the factorial function $n \mapsto n!$ on nonnegative integers. Fortunately, there is such a function: The function $$F: z \mapsto \Gamma(z + 1),$$ where $\Gamma$ denotes the Gamma function, satisfies $F(n) = n!$ for nonnegative integers $n$. (There is a sense in which $F$ is the best possible function extending the factorial function, but notice the target of that link really just discusses the real Gamma function, which our $\Gamma$ preferentially extends.) Thus, we may define factorial of a (square) matrix $B$ by substituting the second display equation above into the first: $$\color{#df0000}{\boxed{B! := \tilde{F}(B) = \frac{1}{2 \pi i} \oint_C \Gamma(z + 1) (z I - B)^{-1} \, dz}} .$$

The (scalar) Cauchy Integral Formula shows that this formulation has the obviously desirable property that for scalar matrices it recovers the usual factorial, or more precisely, that $\pmatrix{n}! = \pmatrix{n!}$ (for nonnegative integers $n$).

Alternatively, one could define a matrix function $\tilde G$ (and in particular define $B!$) by evaluating formally the power series $\sum_{i = 0}^{\infty} a_k (z - z_0)^k$ for $G$ about some point $z_0$, that is, declaring $\tilde G(B) := \sum_{i = 0}^{\infty} a_k (B - z_0 I)^k$, but in general this definition is more restrictive than the Cauchy Integral Formula definition, simply because the power series need not converge everywhere (where it does converge, it converges to the value given by the integral formula). Indeed, we cannot use a power series for $F$ to evaluate $A!$ directly for our particular $A$: The function $F$ has a pole on the line segment in $\Bbb C$ with endpoints the eigenvalues of $A$, so there is no open disk in the domain of $F$ containing all of the eigenvalues of $A$, and hence there is no basepoint $z_0$ for which the series for $\tilde F$ converges at $A$.

We can define $\tilde G$ in yet another way, which coincides appropriately with the above definitions but which is more amenable to explicit computation: If $B$ is diagonalizable, so that we can decompose $$B = P \pmatrix{\lambda_1 & & \\ & \ddots & \\ & & \lambda_n} P^{-1} ,$$ for eigenvalues $\lambda_a$ of $B$ and some matrix $P$, we define $$\tilde{G}(B) := P \pmatrix{G(\lambda_1) & & \\ & \ddots & \\ & & G(\lambda_n)} P^{-1} .$$ Indeed, by substituting and rearranging, we can see that this coincides, at least formally, with the power series characterization. There is a similar but more complicated formula for nondiagonalizable $B$ that I won't write out here but which is given in the Wikipedia article Matrix function.

Example The given matrix $A$ has distinct eigenvalues $\lambda_{\pm} = 1 \pm \sqrt{6}$, and so can be diagonalized as $$P \pmatrix{1 - \sqrt{6} & 0 \\ 0 & 1 + \sqrt{6}} P^{-1} ;$$ indeed, we can take $$P = \pmatrix{\tfrac{1}{2} & \tfrac{1}{2} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}}}.$$

Now, $F(\lambda_{\pm}) = \Gamma(\lambda_{\pm} + 1) = \Gamma (2 {\pm} \sqrt{6}),$ and putting this all together gives that \begin{align*}\pmatrix{1 & 3 \\ 2 & 1} ! = \bar{F}(A) &= P \pmatrix{F(\lambda_-) & 0 \\ 0 & F(\lambda_+)} P^{-1} \\ &= \pmatrix{\tfrac{1}{2} & \tfrac{1}{2} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}}} \pmatrix{\Gamma (2 - \sqrt{6}) & 0 \\ 0 & \Gamma (2 + \sqrt{6})} \pmatrix{1 & -\frac{\sqrt{3}}{\sqrt{2}} \\ 1 & \frac{\sqrt{3}}{\sqrt{2}}} .\end{align*} Multiplying this out gives $$\color{#df0000}{\boxed{\pmatrix{1 & 3 \\ 2 & 1} ! = \pmatrix{\frac{1}{2} \alpha_+ & \frac{\sqrt{3}}{2 \sqrt{2}} \alpha_- \\ \frac{1}{\sqrt{6}} \alpha_- & \frac{1}{2} \alpha_+}}} ,$$ where $$\color{#df0000}{\alpha_{\pm} = \Gamma(2 + \sqrt{6}) \pm \Gamma(2 - \sqrt{6})}. $$

It's perhaps not very illuminating, but $A!$ has numerical value $$ \pmatrix{1 & 3 \\ 2 & 1}! \approx \pmatrix{3.62744 & 8.84231 \\ 5.89488 & 3.62744} . $$

To carry out these computations, one can use Maple's built-in MatrixFunction routine (it requires the LinearAlgebra package) to write a function that computes the factorial of any matrix:

MatrixFactorial := X -> LinearAlgebra:-MatrixFunction(X, GAMMA(z + 1), z);

To evaluate, for example, $A!$, we then need only run the following:

A := Matrix([[1, 3], [2, 1]]);
MatrixFactorial(A);

(NB executing this code returns an expression for $A!$ different from the one above: Their values can be seen to coincide using the the reflection formula $-z \Gamma(z) \Gamma(-z) = \frac{\pi}{\sin \pi z} .$ We can further simplify the expression using the identity $\Gamma(z + 1) = z \Gamma(z)$ extending the factorial identity $(n + 1)! = (n + 1) \cdot n!$ to write $\Gamma(2 \pm \sqrt{6}) = (6 \pm \sqrt{6}) \Gamma(\pm \sqrt{6})$ and so write the entries as expressions algebraic in $\pi$, $\sin(\pi \sqrt{6})$, and $\Gamma(\sqrt{6})$ alone. One can compel Maple to carry out these substitutions by executing simplify(map(expand, %)); immediately after executing the previous code.) To compute the numerical value, we need only execute evalf(%); immediately after the previous code.

By the way, we need not have that $\det B \neq 0$ in order to define $B!$. In fact, proceeding as above we find that the factorial of the (already diagonal) zero matrix is the identity matrix: $$0! = \pmatrix{\Gamma(1) \\ & \ddots \\ & & \Gamma(1)} = I .$$ Likewise using the formula for nondiagonalizable matrices referenced above together with a special identity gives that the factorial of the $2 \times 2$ Jordan block of eigenvalue $0$ is, somewhat amusingly, $$\pmatrix{0 & 1\\0 & 0} ! = \pmatrix{1 & -\gamma \\ 0 & 1} ,$$ where $\gamma$ is the Euler-Mascheroni constant.

Answer 2

The gamma function is analytic. Use the power series of it.

EDIT: already done: Some properties of Gamma and Beta matrix functions (maybe paywalled).

Answer 3

I would start from the logical definition of the matrix factorial, without assuming that we want to cover all properties that we know from factorial in set of reals.

We define standard factorial as $1 \cdot (1+1) \cdot (1+1+1) \cdot ... \cdot (1+1+...+1+1)$

So first let us define $[n]!$ using the same logic replacing 1 with identity matrix. The obvious way to define it is

$$[n]!=\prod\limits_{k=1}^{n}\begin{bmatrix} k & 0\\ 0 & k \end{bmatrix}=\begin{bmatrix} n! & 0\\ 0 & n! \end{bmatrix}$$

All properties of the standard factorial are there. Now, we were defining Gamma function by simple extension $\Gamma (x+1)=x\Gamma (x)$ where $n!=\Gamma (n+1)$. That is all that is required. So we want to find matrix Gamma $\Gamma ([x]+I)=[x]\Gamma ([x])$

If we define

$$\Gamma (\begin{bmatrix} x & 0\\ 0 & x \end{bmatrix})=\begin{bmatrix} \Gamma (x) & 0\\ 0 & \Gamma (x) \end{bmatrix}$$

we are totally fine because

$$\begin{bmatrix} x & 0\\ 0 & x \end{bmatrix}\begin{bmatrix} \Gamma (x) & 0\\ 0 & \Gamma (x) \end{bmatrix}=\begin{bmatrix} x\Gamma (x) & 0\\ 0 & x\Gamma (x) \end{bmatrix}=\begin{bmatrix} \Gamma (x+1) & 0\\ 0 & \Gamma (x+1) \end{bmatrix}$$

There is nothing amiss if we start from $\begin{bmatrix} x & 0\\ 0 & y \end{bmatrix}$ because

$$\begin{bmatrix} x & 0\\ 0 & y \end{bmatrix}\begin{bmatrix} \Gamma (x) & 0\\ 0 & \Gamma (y) \end{bmatrix}=\begin{bmatrix} x\Gamma (x) & 0\\ 0 & y\Gamma (y) \end{bmatrix}=\begin{bmatrix} \Gamma (x+1) & 0\\ 0 & \Gamma (y+1) \end{bmatrix}$$

The remaining part is the other diagonal. What to do with $A=\begin{bmatrix} x_{0} & x_{1}\\ x_{2} & x_{3} \end{bmatrix}$?

So we start from what we would like to have $\Gamma([A]+I)=[A]\Gamma([A])$.

If we are able to diagonalize $A=P^{-1}\overline{A}P$ and to express in the same manner $\Gamma([A]) = P^{-1}\Gamma(\overline{A})P$ then we have

$$\Gamma([A]+I) = P^{-1} \overline{A} P P^{-1} \Gamma(\overline{A}) P = P^{-1} \overline{A} \Gamma(\overline{A}) P = P^{-1} \Gamma(\overline{A+I}) P=\Gamma(A+I)$$

so all should be fine.

Since $\overline{A}$ is diagonal with eigenvalues on the main diagonal $\lambda_{1},\lambda_{2}$ and we know how to deal with that type of matrix, we have the full definition of $\Gamma(A)$ even for matrices.

$$\Gamma(A)=P^{-1}\begin{bmatrix} \Gamma (\lambda_{1}) & 0\\ 0 & \Gamma (\lambda_{2}) \end{bmatrix}P$$

and now $A!=\Gamma(A+I)$ making it all

$$A!=P^{-1}\begin{bmatrix} \Gamma (\lambda_{1}+1) & 0\\ 0 & \Gamma (\lambda_{2}+1) \end{bmatrix}P$$

Instead of giving the solution just to the example I will give a general form for 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Take discriminant $D=\sqrt{(a-d)^2+4bc} \neq 0, c \neq 0$. Then

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} ! = \begin{bmatrix} \frac{a-d-D}{2c} & \frac{a-d+D}{2c} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \Gamma (\frac{a+d-D}{2}+1 ) & 0 \\ 0 & \Gamma ( \frac{a+d+D}{2} +1)\end{bmatrix} \begin{bmatrix} -\frac{c}{D} & \frac{a-d+D}{2D} \\ \frac{c}{D} & -\frac{a-d-D}{2D} \end{bmatrix}$$

From here you can nicely conclude that the factorial matrix can be expressed using classical integer factorial if $a+d \pm D$ are even positive integers (including $0$).

For other values we use the extension of $\Gamma(x)$ itself.

Answer 4

I use the well known (and simple) definitions

$$n!=\Gamma (n+1)$$

and

$$\Gamma (A+1)=\int_0^{\infty } \exp (-t) \exp (A \log (t)) \, dt$$

Now, if A is a (square) matrix, all we need is to define the exponential function for a matrix.

This can always be done (in principle) via the power series which only requires to calculate powers of the matrix and adding matrices.

We skip here a possible diagonalization procedure (which was shown by others before) and use the function MatrixExp[] of Mathematica. For the Matrix given in the OP

$$A=\left( \begin{array}{cc} 1 & 3 \\ 2 & 1 \\ \end{array} \right);$$

we have

Ax = MatrixExp[A Log[t]]

which gives

$$\left( \begin{array}{cc} \frac{t^{1-\sqrt{6}}}{2}+\frac{t^{1+\sqrt{6}}}{2} & \frac{1}{2} \sqrt{\frac{3}{2}} t^{1+\sqrt{6}}-\frac{1}{2} \sqrt{\frac{3}{2}} t^{1-\sqrt{6}} \\ \frac{t^{1+\sqrt{6}}}{\sqrt{6}}-\frac{t^{1-\sqrt{6}}}{\sqrt{6}} & \frac{t^{1-\sqrt{6}}}{2}+\frac{t^{1+\sqrt{6}}}{2} \\ \end{array} \right)$$

We observe that in some cases the exponent of t is less than -1 ($1-\sqrt{6}=-1.44949$). This leads to a divergent integral which will then be understood as the analytic continuation.

This is accomplished simply by replacing each $t^q$ by $\Gamma (q+1)$.

fA = Ax /. t^q_ -> Gamma[q + 1]

giving

$$\left( \begin{array}{cc} \frac{\Gamma \left(2-\sqrt{6}\right)}{2}+\frac{\Gamma \left(2+\sqrt{6}\right)}{2} & \frac{1}{2} \sqrt{\frac{3}{2}} \Gamma \left(2+\sqrt{6}\right)-\frac{1}{2} \sqrt{\frac{3}{2}} \Gamma \left(2-\sqrt{6}\right) \\ \frac{\Gamma \left(2+\sqrt{6}\right)}{\sqrt{6}}-\frac{\Gamma \left(2-\sqrt{6}\right)}{\sqrt{6}} & \frac{\Gamma \left(2-\sqrt{6}\right)}{2}+\frac{\Gamma \left(2+\sqrt{6}\right)}{2} \\ \end{array} \right)$$

Numerically this is

$$\left( \begin{array}{cc} 3.62744 & 8.84231 \\ 5.89488 & 3.62744 \\ \end{array} \right)$$

This is in agreement with the result of hans and travis.

Discussion

(1) Let me point out that the definitions presented here do not need any specific property of the matrix. For example, is does not matter whether the eigenvalues and eigenvectors are singular or not, the matrix might well be deficient.

(2) I have used Mathematica here just to facilitate things. In the end we all use some tools at a certain stage to come to terms. The main ideas are idependent.

(3) The procedure desribed here obviously generalizes to other more or less complicated analytic functions.

As a more exotic example let us take the harmonic number H(A) of a matrix A. This function can be defined using the integral representation (see e.g. Relation between binomial coefficients and harmonic numbers)

$$H(\text{A})\text{=}\int_0^1 \frac{1-(1-x)^A}{x} \, dx$$

This definition also needs only the exponential function of the matrix.

The result for our matrix A is (after some analytic continuation)

$$\left( \begin{array}{cc} \frac{1}{2} \left(H_{-\sqrt{6}}+H_{\sqrt{6}}+\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) & \frac{1}{2} \sqrt{\frac{3}{2}} \left(-H_{-\sqrt{6}}+H_{\sqrt{6}}-\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) \\ \frac{-H_{-\sqrt{6}}+H_{\sqrt{6}}-\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}}{\sqrt{6}} & \frac{1}{2} \left(H_{-\sqrt{6}}+H_{\sqrt{6}}+\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) \\ \end{array} \right)$$

Numerically,

$$\left( \begin{array}{cc} 1.51079 & 0.542134 \\ 0.361423 & 1.51079 \\ \end{array} \right)$$

Answer 5

Travis' answer is very nice.

It would be good to mention that (almost) any matrix function can be made into a power-series expansion, which eventually involves the values of the function on the eigenvalues of the matrix multiplied by the eigenvectors.

In other words the matrix function is completely characterised by the values it takes on the eigenvalues of the matrix (even if a power-series expansion may be needed).

The above hold for matrices which are diagonalisable (i.e. the number of linearly independent eigenvectors is equal to the matrix dimension). There are ways to expand an arbitrary matrix into what is referred to as generalised eigenvectors, but this will not be pursued further here.

Furthermore, since any square, finite-dimensional, matrix satisfies its characteristic polynomial equation (if seen as a matrix function), aka Cayley-Hamilton theorem, the powers of $A^k$ for $k \ge n$ ($n$ is the dimension) can be expressed as a function of the powers of $A$ up to $n$. So eventually the matrix function power-series expansion collapses to polynomial expansion (for square matrices). Finally, this polynomial expansion, for a given function, can be found more easily by methods such as variation of parameters or polynomial modeling.

Answer 6

A factorial of an element of a group $n$ can be interpreted in an abstract algebraic sense, which can then be used with matrices. I will try to motivate the intuition here rather than ONLY giving a definition. Consider the expression $n!$ this is $ 1 \times 2 \times 3 \times ... n $

We now consider an arbitrary group $G$ with (not necessarily commutative operation +) then we can consider a particular element of $G$ (with generators $g_1, g_2 ... g_r$) which we express minimally as the word

$$W = g_{\mu_1} + g_{\mu_2}+.... g_{\mu_s} $$

To be concrete if we let $G$ be the integers then there are two generators $1, -1$ and a positive integer number $N$ can be expressed in the way above as

$$N = N(1) = \underbrace{1+1+1...}_{N \ \text{times}} $$

Here $g_{\mu_k}$ is just positive $1$ the whole time.

If there was a a multiplication operation $\times$ defined now on this group as well (We haven't done it yet!) then we could define $W!$ as

$$ g_{\mu_1} \times \left( g_{\mu_1} + g_{\mu_2} \right) \times \ (g_{\mu_1} + g_{\mu_2} + g_{\mu_3} ) \times \ ...\times \ (g_{\mu_1} + g_{\mu_2} + ... g_{\mu_s}) $$

You can verify that if the group $G$ is the integers this results in

$$ 1 \times 2 \times 3 ... n $$

Which is what we would expect.

So the problem of defining a factorial on a group reduces very naturally to: "How to add a multiplication to a group"?

If $G$ is abelian then you can make $G$ into a ring (with possibly non commutative multiplication). If $G$ is not abelian then either a left or right near ring is the way to go.

Once you add your multiplication then you get a natural factorial.

Answer 7

For a matrix $A$ the value $f(A)$ depends only on the restriction of $f$ on the spectrum of $A$ ( with multiplicities).

If the spectrum is simple, consider $P$ the Lagrange interpolation polynomial of the function $f$ on the spectrum of $A$. Then $f(A)= P(A)$. ( if the spectrum of $A$ is not simple, consider the Hermite interpolation polynomial instead).

Examples:

1. Assume that the $2\times 2$ matrix $A$ has a simple spectrum $\{\alpha, \beta\}$. The Lagrange interpolation polynomial for the function $f$ at the nodes $\alpha$, $\beta$ equals

$$P(x) = \frac{f(\alpha)- f(\beta)}{\alpha - \beta}\cdot x + \frac{\alpha f(\beta) - \beta f(\alpha)}{\alpha- \beta}$$ and

$$f(A) = P(A) = \frac{f(\alpha)- f(\beta)}{\alpha - \beta}\cdot A + \frac{\alpha f(\beta) - \beta f(\alpha)}{\alpha- \beta}\cdot I_2$$

2. Let t $q(x)$ be a polynomial such that $q(A) = 0$ (for instance $q(x)$ is the characteristic polynomial, or the minimal polynomial). Then $f(A) = P(A)$, where $P$ is the Lagrange (Hermite) interpolation polynomial of $f$ at the roots of $q$.

For instance,

Assume $A$ is a projector ($A^2 = A$) then

$$f(A) = (f(1)-f(0))\cdot A + f(0) \cdot I_2$$

Assume $A$ is nilponent, $A^m =0$. Then

$$f(A) = \sum_{k=0}^{m-1} \frac{f^{k}(0)}{k!} A^k$$

3. $f(A)= P(A)$ for any polynomial that agrees with $f$ at the roots of some $q(x)$ annihilating $A$. It follows that

$$(f\cdot g) (A) = f(A) \cdot g(A)$$

$$f(g(A)) = (f\circ g)(A)$$