Prove induction

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November 23rd 2008, 05:56 AM
captainjapan

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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.

$<br /> A^1=A=\left[\begin{array}{cc}<br /> a&0\\0&b<br /> \end{array}\right]<br />$

which will give us a base case.

Suppose true for some $k>1$

Then

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

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 :(

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