Matrix Factorial

This page explains the factorial of a matrix in some detail.

Here's what I did.

It is known that $n! = \Gamma(n+1)$ , therefore, $A! = \Gamma(A + I)$ .

Now $A + I = \begin{bmatrix} 2 && 2 \\ 2 && 5 \end{bmatrix}$ can be diagonalized as $A + I = P D P^{-1}$ where $D = \begin{bmatrix} 1 && 0 \\ 0 && 6 \end{bmatrix}$

Hence, it follows that $\Gamma(A + I) = P \Gamma(D) P^{-1}$ , and we have

$\Gamma(D) = \begin{bmatrix} \Gamma(1) && 0 \\ 0 && \Gamma(6) \end{bmatrix} = \begin{bmatrix} 0! && 0 \\ 0 && 5! \end{bmatrix} = \begin{bmatrix} 1 && 0 \\ 0 && 120 \end{bmatrix}$

And since similar matrices have the same trace, it follows that $\text{Trace}({ A! }) = \text{Trace}({ \Gamma(A + I) }) = \text{Trace}({\Gamma(D)}) = 1 + 120 = \boxed{121}$

I'll provide a Laplace solution:

It's a well-known fact that, for a real number $r$ :

$\large \displaystyle e^{rt} = \mathcal{L}^{-1} \{ (s-r)^{-1} \}$

Where $\mathcal{L}^{-1}$ is the Inverse Laplace Transform . This reasoning extends to the matrix case as:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \{ (sI-A)^{-1} \}$

So for a matix $A$ such as:

$\large \displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

We have:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \begin{bmatrix} s-a & -b \\ -c & s-d \end{bmatrix} ^{-1} \right \}$

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \frac{1}{p(s)} \begin{bmatrix} s-d & b \\ c & s-a \end{bmatrix} \right \}$

Where $p(s) = s^2 - (a+d)s + (ad-bc)$ is the characteristic polynomial of matrix $A$ . Let us sat that the roots of $p(s)$ are $k$ and $l$ . So, expanding in partial fractions:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \frac{1}{k-l} \begin{bmatrix} \frac{k-d}{s-k} + \frac{d-l}{s-l} & \frac{b}{s-k} - \frac{b}{s-l} \\ \frac{c}{s-k} - \frac{c}{s-l} & \frac{k-a}{s-k} + \frac{a-l}{s-l} \end{bmatrix} \right \}$

$\large \displaystyle e^{At} = \frac{1}{k-l} \begin{bmatrix} (k-d)e^{kt} + (d-l)e^{lt} & be^{kt} - be^{lt} \\ ce^{kt} - ce^{lt} & (k-a)e^{kt} + (a-l)e^{lt} \end{bmatrix}$

Making $t = \ln(n)$ , where $n$ is a real number:

$\large \displaystyle n^A = \frac{1}{k-l} \begin{bmatrix} (k-d)n^k + (d-l)n^l & bn^k - bn^l \\ cn^k - cn^l & (k-a)n^k + (a-l)n^l \end{bmatrix}$

$\large \displaystyle \text{tr}(n^A) = \frac{1}{k-l} [ n^k (2k-a-d) + n^l (a+d-2l) ]$

By Vieta $k+l = a+d$ (sum of roots). Then:

$\large \displaystyle \text{tr}(n^A) = \frac{1}{k-l} [ n^k (k-l) + n^l (k-l) ]$

$\color{#20A900} \boxed{ \large \displaystyle \text{tr}(n^A) = n^k + n^l }$

To calculate $A!$ one can use the definition from the Gamma function :

$\large \displaystyle A! = \int_0^{\infty} t^A e^{-t} dt$

So:

$\large \displaystyle \text{tr}(A!) = \int_0^{\infty} \text{tr}(t^A) e^{-t} dt$

$\large \displaystyle \text{tr}(A!) = \int_0^{\infty} (t^k + t^l) e^{-t} dt$

$\large \displaystyle \text{tr}(A!) = \int_0^{\infty} t^k e^{-t} dt + \int_0^{\infty} t^l e^{-t} dt$

$\color{#20A900} \boxed{ \large \displaystyle \text{tr}(A!) = k! + l! }$

Again, where $k$ and $l$ are the roots of the characteristic polynomial of $A$ , $p(s)$ . In the given case $p(s) = s^2 - 5s$ , so its roots are $5$ and $0$ . Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \text{tr}(A!) = 5! + 0! = 121 }$