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Matrices/Linear Algebra Proof

  • Thread starter ahsanxr
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  • #1
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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.
 

Answers and Replies

  • #2
Dick
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What is (I-A)*(I+A)?
 
  • #3
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The Identity matrix.
 
  • #4
Dick
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The Identity matrix.
Hence? What do you conclude from that?
 
  • #5
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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 :)
 
  • #6
Dick
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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?
 
  • #7
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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).
 
  • #8
Dick
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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.
 
  • #10
Dick
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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
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Can you do that with matrices? I have read the property that (A+B)C=AC+BC but only to that extent.
 
  • #12
Dick
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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
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Yes, suppose C = D+E, hence AC+BC=(A+B)(D+E).
 
  • #14
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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
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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
Dick
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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
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A matrix multiplied by the original matrix to give Identity.
 
  • #18
Dick
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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
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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
Dick
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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
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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
Dick
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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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