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

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Homework Help Overview

The discussion revolves around proving that the matrix \( I - A \) is non-singular given that \( A^2 = 0 \). Participants are exploring the implications of this condition and the relationship between \( I - A \) and its proposed inverse \( I + A \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of matrices, particularly focusing on the identity matrix and the concept of inverses. There are attempts to relate the multiplication of \( (I - A)(I + A) \) to the identity matrix and questions about how to generalize findings from a 2x2 matrix to an n x n matrix.

Discussion Status

The discussion is active, with participants questioning the connection between the definitions of non-singularity and inverses. Some have suggested that the proof of non-singularity follows from demonstrating that \( I + A \) is the inverse of \( I - A \), while others are seeking clarification on the definitions and properties involved.

Contextual Notes

Participants are navigating through the definitions of non-singular matrices and inverses, with some expressing uncertainty about the order of proving the statements in the problem. There is an emphasis on understanding the implications of the matrix properties rather than simply applying formulas.

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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.
 
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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.
 

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