If A = a 0

0 b

then

A^n = a^n 0

0 b^n

[IMG]file:///C:/Users/Owner/AppData/Local/Temp/moz-screenshot.jpg[/IMG]

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- Nov 23rd 2008, 05:56 AMcaptainjapanProve induction
If A = a 0

0 b

then

A^n = a^n 0

0 b^n

[IMG]file:///C:/Users/Owner/AppData/Local/Temp/moz-screenshot.jpg[/IMG]

Please see attachment its hard to display the question in text. Thanks. - Nov 23rd 2008, 07:04 AMCaptainBlack
$\displaystyle

A^1=A=\left[\begin{array}{cc}

a&0\\0&b

\end{array}\right]

$

which will give us a base case.

Suppose true for some $\displaystyle k>1$

Then

$\displaystyle

A^{k+1}=AA^k=\left[\begin{array}{cc}a&0\\0&b\end{array}\right]\left[\begin{array}{cc}a^k&0\\0&b^k\end{array}\right]

$

Now do the multiplication abd you should be there.

CB - Nov 23rd 2008, 07:05 AMRapha
Have you ever heard about mathematical induction?

n = 1: OK

n -> n+1

$\displaystyle A^{n+1} = A^n \cdot A = \begin{pmatrix} a^n & 0 \\ 0 & b^n \end{pmatrix}\cdot \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} $

$\displaystyle = \begin{pmatrix} a^n \cdot a + 0 & 0 \\ 0 & b^n \cdot b \end{pmatrix}$

$\displaystyle = \begin{pmatrix} a^{n+1} & 0 \\ 0 & b^{n+1} \end{pmatrix}$

Edit: Too slow :(