Is I-A Non-Singular? Proving Non-Singularity and Inverse for a Matrix

  • Thread starter Thread starter ahsanxr
  • Start date Start date
  • Tags Tags
    Algebra Proof
Click For Summary
The discussion revolves around proving that if A is an n x n matrix such that A² = 0, then I - A is non-singular and its inverse is I + A. Participants explore the relationship between the matrices by multiplying (I - A) and (I + A), leading to the identity matrix, which confirms that I - A has an inverse. The connection between a matrix being non-singular and having an inverse is emphasized, clarifying that if I - A has an inverse, it is indeed non-singular. Participants express confusion about the proof process but ultimately agree that the order of proving non-singularity and finding the inverse is flexible. The conversation concludes with a consensus on the definitions and properties involved in the proof.
ahsanxr
Messages
350
Reaction score
6

Homework Statement



Let A be an n x n matrix. Show that if A2=0 then I - A is non-singular and (I - A)-1= I + A

Homework Equations





The Attempt at a Solution



Ok, so I started off with finding the general form of a 2x2 matrix which when squared gives a zero matrix, and all the properties above are satisfied. But how do I show that for an n x n matrix? Please help me out.
 
Physics news on Phys.org
What is (I-A)*(I+A)?
 
The Identity matrix.
 
ahsanxr said:
The Identity matrix.

Hence? What do you conclude from that?
 
But I only know that because the question says to show it and the "experiment" I did with my 2x2 matrix shows that. I cannot prove it. Please give a more thorough explanation :)
 
ahsanxr said:
But I only know that because the question says to show it and the "experiment" I did with my 2x2 matrix shows that. I cannot prove it. Please give a more thorough explanation :)

If A*B=I then A=B^(-1). Isn't that the definition of inverse?
 
Yes, obviously. I do not know how that is connected to the question though. It asks us to show that the inverse of I-A is I+A which at this point is not self-evident (or at least I think so).
 
ahsanxr said:
Yes, obviously. I do not know how that is connected to the question though. It asks us to show that the inverse of I-A is I+A which at this point is not self-evident (or at least I think so).

Now you are just plain confusing me. If (I-A)*(I+A)=I then as I read the definition of 'inverse' that means (I-A)^(-1)=(I+A). If you mean showing its the inverse by using the explicit calculation you probably used for 2x2 matrices, then you can't do that. You don't know what A is. It's still true though.
 
Dick said:
(I-A)*(I+A)=I

How? This is my question.
 
  • #10
ahsanxr said:
How? This is my question.

Multiply it out. (I+A)*(I-A)=? Use the distributive property! I thought you did that in post 3. I may have been wrong.
 
  • #11
Can you do that with matrices? I have read the property that (A+B)C=AC+BC but only to that extent.
 
  • #12
ahsanxr said:
Can you do that with matrices? I have read the property that (A+B)C=AC+BC but only to that extent.

Sure you can. (I+A)(I-A)=I(I-A)+A(I-A). Now distribute again.
 
  • #13
Yes, suppose C = D+E, hence AC+BC=(A+B)(D+E).
 
  • #14
Of course. That was kind of a stupid question. Thanks for clearing it up.

The question first asks us to show that I-A is non-singular. How do we show that? Or does that follow from (I-A)(I+A)=I and since I has an inverse, it must be non-singular?
 
  • #15
When two matrices are multiplied together to produce the identity, it means the two matrices are inverses of each other. What is the definition of a non-singular matrix?
 
  • #16
ahsanxr said:
Of course. That was kind of a stupid question. Thanks for clearing it up.

The question first asks us to show that I-A is non-singular. How do we show that? Or does that follow from (I-A)(I+A)=I and since I has an inverse, it must be non-singular?

You KNOW I has an inverse. It's I! It's doesn't follow from that. Please quote me the definition of what an inverse is?
 
  • #17
A matrix multiplied by the original matrix to give Identity.
 
  • #18
ahsanxr said:
A matrix multiplied by the original matrix to give Identity.

Ok, so (I+A)(I-A)=I. What's the inverse of (I-A)?
 
  • #19
I get that. That is what I'm asking. Is the proof of (I-A)'s non-singularity the fact that I+A is its inverse?
 
  • #20
ahsanxr said:
I get that. That is what I'm asking. Is the proof of (I-A)'s non-singularity the fact that I+A is its inverse?

Yes, yes, yes. If a matrix has an inverse, it's nonsingular. Now I'm going to have to ask to you look up the definition of 'nonsingular'.
 
  • #21
I know, I know. I was just verifying :P

But the question first asked us to prove the non-singularity and then the second part. Doesn't our approach do it the other way around?
 
  • #22
ahsanxr said:
I know, I know. I was just verifying :P

But the question first asked us to prove the non-singularity and then the second part. Doesn't our approach do it the other way around?

I don't think being asked to show a) and b) means you have to do them in that order. Or that you can't use the same expression to prove both.
 

Similar threads

Does there exist a 2x2 non-singular matrix with only one 1d eigenspace?
  • · Replies 2 ·
Replies
2
Views
2K
Linear Algebra: Non-Singular Matrix and Zero Matrix
  • · Replies 13 ·
Replies
13
Views
2K
Person i belongs to a clique iff ith diagonal entry of B^3 is positive
  • · Replies 9 ·
Replies
9
Views
2K
Non singular matrix M such that MAM^T=F
  • · Replies 5 ·
Replies
5
Views
2K
Proving statements about matrices | Linear Algebra
  • · Replies 25 ·
Replies
25
Views
3K
Can I Use Substitution to Prove the Residue Theorem Integral?
  • · Replies 5 ·
Replies
5
Views
2K
Linear Algebra: Question about Inverse of Diagonal Matrices
  • · Replies 6 ·
Replies
6
Views
2K
Non-Singular matrix - RREF proof
  • · Replies 2 ·
Replies
2
Views
4K
Show that a matrix has a right inverse
  • · Replies 3 ·
Replies
3
Views
2K
Show GL/O/SO(n,R) form groups under Matrix Multiplication
  • · Replies 2 ·
Replies
2
Views
2K