If A is nilpotent , prove that the matrix ( I+ A ) is invertible

[Maths Lover]

if A is nilpotent " A^k = 0 , for some K > 0 " , prove that the matrix ( I+ A ) is invertible ..

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I found more than a topic in the website talk about this theorem biu every one of them didn't produce a complete proof !

I found the question in artin book and I tried to solve this problem and I solved it when K is even , I could calculate the inverse but I hope to find the hole proof here ..

 

[micromass]

Please show an attempt at the solution. For example, how did you solve the even case?

 

[Maths Lover]

Please show an attempt at the solution. For example, how did you solve the even case?

My attempt is wrong !

I've found that it's wrong , any way here is it

( 1 + A ) ( 1 - A ) = 1 - A^2 : 1 denote I matrix

( 1 - A^2 ) ( 1 + A^2 ) = 1 - A^4

continue in that way , if k is even , so the power of A is even to and the power of A " let us denote it by n " will take every possible even number

when you multiply time ater time n will be equal to K then A^k = 0 so
( 1 + A ) ( 1 - A ) ( 1 + A^2 ) ... ( 1 + A^ c ) = 1
where 1 + 1 + 2 + 4 + 4 + 8 + ..+ c = K

so " ( 1 - A ) ( 1 + A^2 ) ... ( 1 + A^ c ) " is the inverse

but that is not true because K may be 6 for example but you can't get 6 with this way

so this way is useful when K = 2^ m

that's my attempt

 

[Hurkyl]

but that is not true because K may be 6 for example but you can't get 6 with this way

What way can you get 6?

Alternatively, if you can't make 6, can you still do with 4 or 8 or something else?

 

[micromass]

So, if I understand you right, you want to find an inverse of the form

a_0 I+ a_1 A + a_2 A^2 + ... + a_n A^n

where A^{n+1}=0.

Now, why don't you multiply

(I+A)(a_0 I+ a_1 A + a_2 A^2 + ... + a_n A^n)

this to see what you can take as a_i?

 

[Maths Lover]

What way can you get 6?

Alternatively, if you can't make 6, can you still do with 4 or 8 or something else?

I read your reply before you edit it

you are right I can make A^ 8 = A^6 * A^2 = 0

so it's useful for even numbers

thanks a lot , I forgot that if A^k = 0 then A^(k+n) = 0 : n >0

Thanks , But do you know a complete proof ?

 

[micromass]

Thanks , But do you know a complete proof ?

Yes, we know the complete proof. But the idea is that you should figure out yourself instead of us telling you what it is.

 

[Maths Lover]

Yes, we know the complete proof. But the idea is that you should figure out yourself instead of us telling you what it is.

I understand , I tried to solve the problem when K is odd , but my mistake was that I use the inverse of A and I forgot that A is not invertible !

 

[micromass]

I understand , I tried to solve the problem when K is odd , but my mistake was that I use the inverse of A and I forgot that A is not invertible !

Did you read Hurkyl's suggestion and my suggestion? Certainly that will give you something to do?

 

[Maths Lover]

So, if I understand you right, you want to find an inverse of the form

a_0 I+ a_1 A + a_2 A^2 + ... + a_n A^n

where A^{n+1}=0.

Now, why don't you multiply

(I+A)(a_0 I+ a_1 A + a_2 A^2 + ... + a_n A^n)

this to see what you can take as a_i?

I found that : if we make

a(n) = -a(n+1)

then ,
a0 I + a1 A + a2 A^2 + .. + an A^n

the sum will be ( a0 * I )

so , let a0 = 1 then a1 = -1 then a2 = 1 and so on I will get the I matrix

so

the matrix

( 1 - A + A^2 - A^3 + ... (-+)A^n )
is the inversr of ( I + A)

Is that right ?!

 

[Maths Lover]

Did you read Hurkyl's suggestion and my suggestion? Certainly that will give you something to do?

yes , I did .. I replied both of you

 

[micromass]

I found that : if we make

a(n) = -a(n+1)

then ,
a0 I + a1 A + a2 A^2 + .. + an A^n

the sum will be ( a0 * I )

so , let a0 = 1 then a1 = -1 then a2 = 1 and so on I will get the I matrix

so

the matrix

( 1 - A + A^2 - A^3 + ... (-+)A^n )
is the inversr of ( I + A)

Is that right ?!

Yes, that is right. That is indeed the inverse of I+A.

 

[Maths Lover]

Yes, that is right. That is indeed the inverse of I+A.

ok , Thank you very much

and my greetings to Hurkyl too ..

I want to say something ,

the way that you follow is great " i mean that you made me get the inverse myself and didn't gave it to me " , I tried so much for 2 days to get this inverse but I couldn't , now I'm so happy


thank tou again ..

my last question !

Is there a proof which prove that ( 1 + A ) is invertible without getting the inverse ?

I mean , is there a way prove that it's invertible without knowing what is the inverse ?!

 

[Ebrahim3enab]

Hi everyone .

Maths Boy I advice you to use the Latex codes its very cute .

something like that :

$$I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

 

[Maths Lover]

Hi everyone .

Maths Boy I advice you to use the Latex codes its very cute .

something like that :

$$I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

I don't know how to use latex codes !