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Matrices/Linear Algebra Proof

by ahsanxr
Tags: algebra, matrices or linear, proof
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ahsanxr
#1
Sep8-10, 10:29 PM
P: 341
1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. 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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Dick
#2
Sep8-10, 10:35 PM
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What is (I-A)*(I+A)?
ahsanxr
#3
Sep8-10, 10:36 PM
P: 341
The Identity matrix.

Dick
#4
Sep8-10, 10:39 PM
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Matrices/Linear Algebra Proof

Quote Quote by ahsanxr View Post
The Identity matrix.
Hence? What do you conclude from that?
ahsanxr
#5
Sep8-10, 10:41 PM
P: 341
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 :)
Dick
#6
Sep8-10, 10:42 PM
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Quote Quote by ahsanxr View Post
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?
ahsanxr
#7
Sep8-10, 10:44 PM
P: 341
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).
Dick
#8
Sep8-10, 10:49 PM
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Quote Quote by ahsanxr View Post
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.
ahsanxr
#9
Sep8-10, 10:52 PM
P: 341
Quote Quote by Dick View Post
(I-A)*(I+A)=I
How? This is my question.
Dick
#10
Sep8-10, 10:55 PM
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Quote Quote by ahsanxr View Post
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.
ahsanxr
#11
Sep8-10, 11:02 PM
P: 341
Can you do that with matrices? I have read the property that (A+B)C=AC+BC but only to that extent.
Dick
#12
Sep8-10, 11:04 PM
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Quote Quote by ahsanxr View Post
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.
Samuelb88
#13
Sep8-10, 11:05 PM
P: 162
Yes, suppose C = D+E, hence AC+BC=(A+B)(D+E).
ahsanxr
#14
Sep8-10, 11:08 PM
P: 341
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?
Samuelb88
#15
Sep8-10, 11:10 PM
P: 162
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?
Dick
#16
Sep8-10, 11:11 PM
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Quote Quote by ahsanxr View Post
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?
ahsanxr
#17
Sep8-10, 11:13 PM
P: 341
A matrix multiplied by the original matrix to give Identity.
Dick
#18
Sep8-10, 11:18 PM
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Quote Quote by ahsanxr View Post
A matrix multiplied by the original matrix to give Identity.
Ok, so (I+A)(I-A)=I. What's the inverse of (I-A)?


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