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

  • #1
if A is nilpotent " A^k = 0 , for some K > 0 " , prove that the matrix ( I+ A ) is invertible ..

****
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

  • #2
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Please show an attempt at the solution. For example, how did you solve the even case?
 
  • #3
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
 
  • #4
Hurkyl
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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?
 
  • #5
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3,278
So, if I understand you right, you want to find an inverse of the form

[tex]a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n[/tex]

where [itex]A^{n+1}=0[/itex].

Now, why don't you multiply

[tex](I+A)(a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n)[/tex]

this to see what you can take as [itex]a_i[/itex]?
 
  • #6
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 ?
 
  • #7
22,097
3,278
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.
 
  • #8
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 !!
 
  • #9
22,097
3,278
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?
 
  • #10
So, if I understand you right, you want to find an inverse of the form

[tex]a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n[/tex]

where [itex]A^{n+1}=0[/itex].

Now, why don't you multiply

[tex](I+A)(a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n)[/tex]

this to see what you can take as [itex]a_i[/itex]?
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 ?!
 
  • #11
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
 
  • #12
22,097
3,278
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.
 
  • #13
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 ?!!
 
  • #14
Hi everyone .

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

something like that :

$$I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
 
  • #15
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 !
 

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