Proving Matrix Equality with Inverse Properties

In summary, the conversation discusses the question of whether X and Z are equal if XY = I and YZ = I, and presents a proof showing that this is indeed the case. The proof involves using the fact that if XY = I, then YX = I as well, and incorporating the left and right inverses of Y into the equation. It is also noted that the existence of inverses may need to be proven in order for this proof to hold.
  • #1
ConeOfIce
13
0
The question is: Suppose one has n×n square matrices X, Y and Z such that
XY = I and Y Z = I. Show that it follows that X = Z.

Now if this were XY and ZY, I would just say that:

XY=ZY -> XY-ZY=0 ->Y(X-Z)=0 -> X-Z=0 -> X=Z.

I am wondering that since the Y is on different sides of the Z and X this still holds? Thanks!
 
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  • #2
I don't think you can just pull the Y out like that.

If I recall correctly, if XY = I, then doesn't YX = I aswell? If this is true perhaps you could encorporate it into your proof somehow.
 
  • #3
Actually I think I just figured it out.
X^-1 Y =X^-1 I -> Y=X^-1

then sub that in for Y in the other equation
X^-1 Z = I
XX^-1 Z = X
Z=X

Thanks for your help!
 
  • #4
it goes like this: XY = I so XYZ = IZ = Z.

but YZ = I too, so XYZ = XI = X. so X = XYZ = Z.

the problem with your argument is that there is no X^-1 unless yoiu prove this problem.

but if yu are careful, yu can use the fact that X = the left inverse of Y and Z = the right inverse of Y.
 
  • #5
Oh, I forgot the inverses don't always exist...it has been a bit since I have done and linear algebra work. Your solution makes perfect sense, thanks!
 

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