# Matrix little demonstration

• May 25th 2013, 02:48 AM
Matrix little demonstration
Hello,

I'm a bit bloqued on a little demontration.

A^4=0 , (I-A)^-1 = I²+A+A²+A³

Here is what i've done:

I=(I-A).(I²+AI+A²I+A³I)
=> I=I³+AI²+A²I²+A³I³-AI²-A²I-A³I-(A^4)I
=> I=I+A²+A³-A²-A³-A^4
=> I=I-A^4
=> A^4=0

But it is prove in reverse...

The right way:
If you know that matrix A^4=0, how to prove that (I-A)^-1 = I²+A+A²+A³

But in ths case (harder I supose), I don't know how to do.

Can you help me?
• May 25th 2013, 06:40 AM
zzephod
Re: Matrix little demonstration
Consider $(I-A)[I+A+A^2+A^3]$

(what are the assumptions that you have forgotten to mention? And for good measure also consider $[I+A+A^2+A^3](I-A)$)

.
• May 25th 2013, 08:30 AM
Re: Matrix little demonstration
Sorry but I have no particular assumptions.
The exercice is presented as such
A^4=0 , (I-A)^-1 = I²+A+A²+A³
It's already a problem for me at the base.

Or what do you mean?

If y consider (I-A).(I+A+A²+A³), I start in the same oposite direction that i've done no?

I still have a misanderstanding with this.
• May 25th 2013, 10:44 AM
zzephod
Re: Matrix little demonstration
Quote:

Sorry but I have no particular assumptions.
The exercice is presented as such
A^4=0 , (I-A)^-1 = I²+A+A²+A³
It's already a problem for me at the base.

Or what do you mean?

If y consider (I-A).(I+A+A²+A³), I start in the same oposite direction that i've done no?

I still have a misanderstanding with this.

$(I-A)[I+A+A^2+A^3]=[I+A+A^2+A^3] - [A+A^2+A^3+A^4]=I-A^4=I$

Hence if $(I-A)$ is invertible and $A^4=0$ then $[I+A+A^2+A^3]$ is its' inverse ..
• May 25th 2013, 11:01 AM
Re: Matrix little demonstration
Okay, I got t now!

Thanks a lot!
• May 25th 2013, 04:43 PM
Soroban
Re: Matrix little demonstration

Quote:

I'm a bit bloqued on a little demontration.

$A^4\,=\,0 \quad\Longleftrightarrow\quad (I-A)^{-1} \:=\: I+A+A^2+A^3$

Here is what I've done:

We have: . $(I-A)^{-1} \:=\:I+A+A^2+A^3$

Multiply by $(I-A):$

. . $(I-A)(I-A)^{-1} \:=\: (I-A)(I+A+A^2+A^3)$

. . . . . . . . . . . . . . $I \;=\;I^2 + IA + IA^2 + IA^3 - AI - A^2 - A^3 - A^4$

. . . . . . . . . . . . . . $I \;=\;I + A + A^2 + A^3 - A - A^2 - A^3 - A^4$

. . . . . . . . . . . . . . $I \;=\;I - A^4$

n . . . . . . . . . . . $A^4 \;=\;0$

But to prove it in reverse . . .

Just run the step in reverse . . .

We have: . $A^4 \;=\;0$

. . . . . . . . . $0 \;=\;-A^4$

Add $I\!:\qquad\; I \;=\;I-A^4$

. . . . . . . . . $I \;=\;I - A + A - A^2 + A^2 - A^3 + A^3 - A^4$

. . . . . . . . . $I \;=\;(I-A) + A(I-A) + A^2(I-A) + A^3(I-A)$

. . . . . . . . . $I \;=\;(I-A)(I+A+A^2+A^3)$

Multiply by $(I-A)^{-1}$

. . $(I-A)^{-1}I \;=\;(I-A)^{-1}(I-A)(I+A+A^2+A^3)$

. . . $(I-A)^{-1} \;=\; I+A+A^2+A^3$
• May 26th 2013, 01:56 AM