The problem was presented like this:
[MATH]A_{2n} = \begin{bmatrix} a & 0 & \cdots & 0 & b\\ 0 & a & \cdots & b & 0\\ & & \cdots & & \\ 0 & b & \cdots & a & 0\\ b & 0 & \cdots & 0 & a \end{bmatrix} n \in N, A_{0} = 1[/MATH]
Show that [MATH]detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}[/MATH]
I tried to use induction for this case:
Base case:
[MATH]n=1: detA_{2n} = \begin{vmatrix}a & b\\ b & a\end{vmatrix} = a^{2}-b^{2} = a^{2}-b^{2} \cdot detA_{2n-2} = a^{2}-b^{2} \cdot detA_{0} = a^{2}-b^{2} \cdot 1 = a^{2}-b^{2} [/MATH]
Hypothesis:
[MATH] detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2} \Rightarrow detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} [/MATH]
[MATH]detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n+2-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n} \Leftrightarrow detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}[/MATH]
This feels circular, is this the right way to do it?
[MATH]A_{2n} = \begin{bmatrix} a & 0 & \cdots & 0 & b\\ 0 & a & \cdots & b & 0\\ & & \cdots & & \\ 0 & b & \cdots & a & 0\\ b & 0 & \cdots & 0 & a \end{bmatrix} n \in N, A_{0} = 1[/MATH]
Show that [MATH]detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}[/MATH]
I tried to use induction for this case:
Base case:
[MATH]n=1: detA_{2n} = \begin{vmatrix}a & b\\ b & a\end{vmatrix} = a^{2}-b^{2} = a^{2}-b^{2} \cdot detA_{2n-2} = a^{2}-b^{2} \cdot detA_{0} = a^{2}-b^{2} \cdot 1 = a^{2}-b^{2} [/MATH]
Hypothesis:
[MATH] detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2} \Rightarrow detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} [/MATH]
[MATH]detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n+2-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n} \Leftrightarrow detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}[/MATH]
This feels circular, is this the right way to do it?