Idempotent Matrix

**sparky** · February 20th 2011, 04:18 PM

I would love some guidance on how I should approach this question:

Question: If A is an idempotent matrix of order n, show that $(I+A)^n = I+(2^N-1)A$

**tonio** · February 20th 2011, 06:38 PM

Originally Posted by sparky

I would love some guidance on how I should approach this question:

Question: If A is an idempotent matrix of order n, show that $(I+A)^n = I+(2^N-1)A$

Hints:

1) Show that $\displaystyle{\sum\limits^n_{k=0}\binom{n}{k}=2^n}$ ;

2) As I, A commute with each other, $\displaystyle{(I+A)^n=\sum\limits^n_{k=0}A^k=I+A\s um\limits^n_{k=1}\binom{n}{k}=}$ ...

Tonio

**sparky** · February 23rd 2011, 04:03 PM

Thank you for your reply Tonio.

I am still trying to figure out how to show in your first point.

Here is what I understand so far:

A matrix $A$ is said to be idempotent if $A^2=A$

**tonio** · February 23rd 2011, 06:06 PM

Originally Posted by sparky

Thank you for your reply Tonio.

I am still trying to figure out how to show in your first point.

Here is what I understand so far:

A matrix $A$ is said to be idempotent if $A^2=A$

Well, understanding definitions is important but it's hardly enough for this problem...

Further hint:

Newton's Binomial Theorem: For any pair of commuting elements a,b in a ring and for

any natural n, we have that $\displaystyle{(a+b)^n=\sum\limits^n_{k=0}\binom{n} {k}a^{n-k}b^k}$

Tonio

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