Monday, July 15, 2019

Algebraic Manipulation for Mathematical Induction




I'm working on a mathematical induction problem. The question is as follows:



$P = \begin{pmatrix}
1-A & A \\
B & 1-B \\
\end{pmatrix}$



for A,B $\epsilon$ (0, 1). Show by induction, or otherwise that
$P^n = \frac{1}{A+B} \begin{pmatrix}

B & A \\
B & A \\
\end{pmatrix} + \frac{(1-A-B)^n}{A+B} \begin{pmatrix}
A & -A \\
-B & B \\
\end{pmatrix}$



for any n $\epsilon $ $\Bbb N$.



I understand how induction is done, however I'm lost with the algebraic manipulation. So far I've proved that P(1) is true by substituting n = 1 into the equation. When it comes to proving that p(k+1) is true however, I get lost in the algebra. I know I have to show something like




$P^k = \frac{1}{A+B} \begin{pmatrix}
B & A \\
B & A \\
\end{pmatrix} + \frac{(1-A-B)^k}{A+B} \begin{pmatrix}
A & -A \\
-B & B \\
\end{pmatrix} = \begin{pmatrix}
1-A & A \\
B & 1-B \\

\end{pmatrix}^k$



$P^{k+1} = \frac{1}{A+B} \begin{pmatrix}
B & A \\
B & A \\
\end{pmatrix} + \frac{(1-A-B)^{k+1}}{A+B} \begin{pmatrix}
A & -A \\
-B & B \\
\end{pmatrix} = \begin{pmatrix}
1-A & A \\

B & 1-B \\
\end{pmatrix}^{k+1} = \begin{pmatrix}
1-A & A \\
B & 1-B \\
\end{pmatrix}P^k = \begin{pmatrix}
1-A & A \\
B & 1-B \\
\end{pmatrix}[\frac{1}{A+B}\begin{pmatrix}
B & A \\
B & A \\

\end{pmatrix} + \frac{(1-A-B)^k}{A+B} \begin{pmatrix}
A & -A \\
-B & B \\
\end{pmatrix}]$



Any help would be greatly appreciated


Answer



Show us what you get if you carry out the computation on the right hand side of your last equation: Distribute the matrix multiplication over the matrix addition, and perform the matrix multiplication. You can take the scalar multiples outside during this, and consider them afterwards.


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