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

[ahsanxr]

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.

 

[Dick]

What is (I-A)*(I+A)?

 

[ahsanxr]

The Identity matrix.

 

[Dick]

The Identity matrix.

Hence? What do you conclude from that?

 

[ahsanxr]

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]

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]

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]

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]

(I-A)*(I+A)=I

How? This is my question.

 

[Dick]

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]

Can you do that with matrices? I have read the property that (A+B)C=AC+BC but only to that extent.

 

[Dick]

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]

Yes, suppose C = D+E, hence AC+BC=(A+B)(D+E).

 

[ahsanxr]

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]

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]

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]

A matrix multiplied by the original matrix to give Identity.

 

[Dick]

A matrix multiplied by the original matrix to give Identity.

Ok, so (I+A)(I-A)=I. What's the inverse of (I-A)?

 

[ahsanxr]

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?

 

[Dick]

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'.

 

[ahsanxr]

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?

 

[Dick]

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.