Prove induction

• November 23rd 2008, 05:56 AM
captainjapan
Prove 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.
• November 23rd 2008, 07:04 AM
CaptainBlack
Quote:

Originally Posted by captainjapan
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.

$
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 $k>1$

Then

$
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
• November 23rd 2008, 07:05 AM
Rapha
Quote:

Originally Posted by captainjapan
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.

Have you ever heard about mathematical induction?

n = 1: OK

n -> n+1

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

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

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

Edit: Too slow :(