# obtaining Bernoulli numbers from determinant

I am reading a paper entitled Bernoulli Numbers Via Determinants by Hongwei Chen and I'm confused about a particular step. The author sets up a system of equations via the following: first let $B_n$ represent the $n$-th Bernoulli number. Then

$$x=(e^x-1)\sum_{n=0}^\infty B_n\frac{x^n}{n!}$$

Letting $b_n=B_n/n!$ and expanding we get

$$x=\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\left(b_0+b_1x+b_2x^2+...\right)$$

Immediately we can see that $b_0=1$. Using Cauchy Products, we then obtain an infinite sequence of equations which are coefficients of the powers of $x$. But we don't need infinite, we can look at the system for coefficients up to $x^n$. Therefore

\begin{cases} b_1=-\frac{1}{2!} \\[2ex] \frac{b_1}{2!}+b_2=-\frac{1}{3!} \\[2ex] \frac{b_1}{3!}+\frac{b_2}{2!}+b_3=-\frac{1}{4!} \\[2ex] \vdots \\[2ex] \frac{b_1}{n!}+\frac{b_2}{(n-1)!}+...+b_n=-\frac{1}{(n+1)!} \\[2ex] \end{cases}

Then the author goes on to state that applying Cramers rule produces $$b_n= \begin{vmatrix} 1 & 0 & 0 & \cdots & -\frac{1}{2!} \\ \frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\ \end{vmatrix}$$

Then

$$B_n=n! \begin{vmatrix} 1 & 0 & 0 & \cdots & -\frac{1}{2!} \\ \frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\ \end{vmatrix}$$

And finally,

$$B_n=(-1)^n n! \begin{vmatrix} \frac{1}{2!} & 1 & 0 & \cdots & 0 \\ \frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & 0 \\ \frac{1}{4!} & \frac{1}{3!} & \frac{1}{2!} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{(n+1)!} &\frac{1}{n!} & \frac{1}{(n-1)!} & \cdots & \frac{1}{2!} \\ \end{vmatrix}$$

Can someone please explain the last step? I just don't see how that last step comes from the previous. I figure it has something to do with properties of determinants, and i do know that $|cA|=c^n|A|$, but I don't see where that is coming from here.

• The last step comes from interchanging columns and $c=-1$. – Dietrich Burde Feb 15 '16 at 22:33
• And so, if you do that, you only do that $n-1$ times...then the sign of the left most column is still negative...can you also take out a $-1$ then since the column is a scalar multiple? It's been a while since I've done determinants... – Iceman Feb 15 '16 at 22:34
• @DietrichBurde, is my last statement correct? – Iceman Feb 15 '16 at 22:57

If $c$ is a constant and $A$ is an $n \times n$ matrix, then $\det c A = c^n \det A$. Note that here $c$ multiplies all columns not just the first column. Now, if $c$ multiplies only one column then $\det [A_1, A_2, \cdots, c A_i, \cdots A_n]= c \det A$.
Now for the problem. The constant $c=-1$ only multiplies the first column after being rolled from the last place ($n-1$ swaps).
Then we have the total sign is $(-1)^{n-1} (-1) = (-1)^n$.
• In Mathematica, I typed in B[n_] := (-1)^n n! Det[ Table[If[r + 1 < c, 0, 1/(r - c + 2)!], {r, 1, n}, {c, 1, n}]] followed by BernoulliB[#] - B[#] & /@ Range[10]. The result was all zeros. – heropup Mar 11 '16 at 23:20
• Yes, you could do that for rows or columns. The determinant of $A$ is equal to the determinant of $A^T$ (the transposed) so it does not matter. But there are two contributions to the sign. Swapping two columns changes the sign, and multiplying a column changes the sign by $(-1)^n$. Added together is $n+n-1 = 2n-1$ and for an odd number $(-1)^{2n-1}=-1$ – Herman Jaramillo Mar 12 '16 at 0:22