Find a Linear Transformation such that T squared = T

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To find a linear transformation T such that T² = T, one can define T as a projection operator on an arbitrary subspace, avoiding the zero and identity operators. A suggested approach involves selecting a basis and defining T(xi) = ai*xi, where ai can be either 0 or 1, ensuring at least one ai is 0 and one is 1. This construction guarantees that T is neither the zero operator nor the identity operator while satisfying the condition T² = T. The discussion also touches on the proof that the direct sum of im(T) and ker(T) forms a vector space. Understanding these concepts is crucial for successfully defining the desired transformation.
mmmboh
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Hi, I had to prove that if T2=T then the direct sum of im(T) and ker(T) is a vector space, and I did that, but now I am suppose to find such a T that isn't the zero operator (I'm not even sure what that is, just a transformation that makes any vector zero?) or the identity operator. Problem is I can't think of any :confused:. I am just trying to think of a transformation T2=T, but none of what I have thought of works, any help please?
 
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mmmboh said:
Hi, I had to prove that if T2=T then the direct sum of im(T) and ker(T) is a vector space, and I did that, but now I am suppose to find such a T that isn't the zero operator (I'm not even sure what that is, just a transformation that makes any vector zero?) or the identity operator. Problem is I can't think of any :confused:. I am just trying to think of a transformation T2=T, but none of what I have thought of works, any help please?

T is a projection operator on arbitrary subspace.
 
mmmboh said:
the zero operator (I'm not even sure what that is, just a transformation that makes any vector zero?)
That's right.

Problem is I can't think of any :confused:.
Then let's solve for one. If we can't figure things out with matrix algebra, then break things into components, and turn the problem into one of number algebra!

I'll get you started: let's suppose T is a 0x0 matrix. Oh, this can't work -- the only 0x0 matrix is the zero matrix. (Which is also an identity matrix), so nevermind.

Well, let's try 1x1, so that T = [a]. So in terms of a, what do we want?
 
What if I write Tv= a 2x2 matrix where the top row entries are v and the bottom row entries are zero.
Sorry that is suppose to be a matrix. Anyway does that make sense as an answer?
 
I don't think what I wrote even makes sense :S
 
mmmboh said:
I don't think what I wrote even makes sense :S
It makes sense if the top row is v. (It doesn't make sense for the entries to be v -- at least not in the arithmetic framework you're using)

But then what does T2 work out to be?
 
Sorry I didn't see the end of your post haha, well we want an a that preserves addition and scalar multiplication for one, and when you transform it again gives you a still.
 
T2 ends up being a matrix where the top row is a matrix of V? I'm not sure...
 
Why don't you pick a basis, any basis. {x1...xn}. Now define T(xi)=ai*xi where ai is equal to zero or one as you wish. That defines a T. Is T^2=T?
 
  • #10
Yes but isn't that T the zero or identity operator in that case?
 
  • #11
mmmboh said:
Yes but isn't that T the zero or identity operator in that case?

It's zero if you pick ALL of the ai's=0. It's the identity if you pick ALL of the ai's=1. If you don't want either of those then pick some of them to be 0 and some to be 1.
 
  • #12
Ah so can I write Tx1,...xn=a1x1,...,aixi where there is at least one a=0 and one a=1 (but a can only be 0 or 1)?
 
  • #13
mmmboh said:
Ah so can I write Tx1,...xn=a1x1,...,aixi where there is at least one a=0 and one a=1 (but a can only be 0 or 1)?

That's the idea.
 
  • #14
Thanks!
 
  • #15
How were you able to prove that the direct sum of im(T) and ker(T) is a vector space?
 

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