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

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 ..

## Answers and Replies

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

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
Staff Emeritus
Science Advisor
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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?

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 $a_i$?

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

Thanks , But do you know a complete proof ?

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.

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 !!

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?

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 ?!

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

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.

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 ?!!

Hi everyone .

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

something like that :

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

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 !