I know for sure that there is some easy way to prove what I am about to tell, but, at first, I'd like to know if I can set up a proof by induction for two "cross-referenced formulas".
I have two matrices A ($n\times n$) and B ($n\times n$). $$A_n=\left( \begin{array}{ccc} 1-a & 1 &... & 1 \\ 1 & 1-a&... &1 \\ ... & ... & ...&...\\ 1 & ... &1 & 1-a \end{array} \right) $$ $$B_n=\left( \begin{array}{ccc} 1 & 1 &... & 1 \\ 1 & 1-a&... &1 \\ ... & ... & ...&...\\ 1 & ... &1 & 1-a \end{array} \right) $$
$$\det A_n=(1-a)\det A_{n-1}-(n-1)\det B_{n-1}$$ $$\det B_n=\det A_{n-1}-(n-1) \det B_{n-1}$$
$$\det A_n=(-1)^n a^{n-1}(a-n)$$ $$\det B_n=(-1)^{n-1}a^{n-1}$$
The formulas I am really sure they are ok.I get no contradiction using them together, but how can I make a proof without killing the logic??
thank you.
The two recurrences (for $\det A_n$ and $\det B_n$) are clearly correct, so it’s just a matter of proving the two closed forms, and you can use an ordinary induction argument to prove them simultaneously.
They are both clearly correct when $n=1$. Suppose that they hold for some $n\ge 1$. Then
$$\begin{align*} \det A_{n+1}&=(1-a)\det A_n-n\deg B_n\\ &=(1-a)(-1)^na^{n-1}(a-n)-n(-1)^{n-1}a^{n-1}\\ &=(1-a)(-1)^na^n-n(1-a)(-1)^na^{n-1}-n(-1)^{n-1}a^{n-1}\\ &=(1-a)(-1)^na^n+n(-1)^na^n\\ &=(n+1-a)(-1)^na^n\\ &=(-1)^{n+1}a^n\big(a-(n+1)\big)\;, \end{align*}$$
and
$$\begin{align*} \det B_{n+1}&=\det A_n-(n-1)\det B_n\\ &=(-1)^na^{n-1}(a-n)-n(-1)^{n-1}a^{n-1}\\ &=(-1)^na^n\;, \end{align*}$$
and the result follows by induction on $n$.