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?

Re: Matrix little demonstration

Consider $\displaystyle (I-A)[I+A+A^2+A^3]$

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

.

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.

Re: Matrix little demonstration

Quote:

Originally Posted by

**Adi0101** 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.

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

Hence if $\displaystyle (I-A)$ is invertible and $\displaystyle A^4=0$ then $\displaystyle [I+A+A^2+A^3]$ is its' inverse ..

Re: Matrix little demonstration

Okay, I got t now!

Thanks a lot!

Re: Matrix little demonstration

Hello, Adi0101!

Quote:

I'm a bit bloqued on a little demontration.

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

Here is what I've done:

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

Multiply by $\displaystyle (I-A):$

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

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

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

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

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

But to prove it in reverse . . .

Just run the step in reverse . . .

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

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

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

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

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

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

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

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

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

Re: Matrix little demonstration

Great!

It seems so simple now...

Thanks a lot for this detailed demontration!