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

• Maths Lover
In summary: 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)

#### Maths Lover

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 ..

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

micromass said:
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

Maths Boy said:
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 $a_i$?

Hurkyl said:
What way can you get 6?

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

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 ?

Maths Boy said:
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.

micromass said:
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 !

Maths Boy said:
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?

micromass said:
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 ?!

micromass said:
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

Maths Boy said:
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.

micromass said:
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 :

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

enrahim3enab said:
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 !

## 1. What does it mean for a matrix to be nilpotent?

A matrix is said to be nilpotent if there exists a positive integer k such that A to the power of k is equal to the zero matrix.

## 2. How do you prove that a matrix is nilpotent?

To prove that a matrix A is nilpotent, you need to show that there exists a positive integer k such that A to the power of k is equal to the zero matrix. This can be done by finding the characteristic polynomial of A and showing that its highest power is equal to the zero polynomial.

## 3. What does it mean for a matrix to be invertible?

A matrix is said to be invertible if there exists another matrix B such that the product of A and B is equal to the identity matrix. In other words, A multiplied by its inverse B results in the identity matrix.

## 4. How do you prove that a matrix is invertible?

To prove that a matrix A is invertible, you need to show that there exists another matrix B such that the product of A and B is equal to the identity matrix. This can be done by using row operations to reduce A to the identity matrix, and performing the same operations on the identity matrix to obtain B.

## 5. How does the nilpotency of A relate to the invertibility of (I + A)?

If A is nilpotent, then there exists a positive integer k such that A to the power of k is equal to the zero matrix. This means that (I + A) to the power of k is equal to the identity matrix, making (I + A) invertible. This can be proven by using the binomial expansion formula on (I + A) to the power of k.