Two linear transformations agree, subspace

  • Thread starter 1MileCrash
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  • #1
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Main Question or Discussion Point

I've been up way too long, so pardon me if this doesn't make sense, but..

Let V and W be vector spaces.
Let T and U be linear transformations from V to W.

Consider the set of all x in V such that T(x) = U(x)

1.) I think that this is a subspace of V.
2.) Can I say anything about its dimension?

The dimension is at most V, when the two transformations are the same. What is the minimum dimension in terms of dimensions for V, W, null spaces, or ranges?

Is it AT LEAST the smaller of the two nullities?
 

Answers and Replies

  • #2
PeroK
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There are lots of examples for T and U where the only element they map equally is 0. If U = aT (a not = 1) for example.
 
  • #3
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There are lots of examples for T and U where the only element they map equally is 0. If U = aT (a not = 1) for example.
Surely..

But if T and U only map from 0 equally, then the nullspace of either T or U is just {0}.

That's why I think the minimum dimension of this space can be related to the nullities of T and U.
 
  • #4
PeroK
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It's only if the nullities of T & U exceed the dimension of V that they must have more than 0 in common! And, then you could have anything from NullT + NullU - DimV to min{NullT, Null U}
 
  • #5
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It's only if the nullities of T & U exceed the dimension of V that they must have more than 0 in common!
OK, I understand. But how could we have nullities that exceed the dimension of V? Isn't the nullspace a subspace of V?

EDIT or do you mean the sum of the two nullities?
 
  • #6
PeroK
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Yes the sum of the nullities. And, what goes on outside the null spaces is equally independent. They may have no null vectors in common (except 0) but be equivalent outside of NullT U NullU
 
  • #7
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I think I am confusing myself. Could you maybe see something wrong with the following?

Let T and R be arbitrary linear transformations.

rank(T+R) <= rank(T) + rank(R)

(dim-null)(T+R) <= (dim-null)(T) + (dim-null)(R)
null(T+R) >= null(T) + null(R)
null(T+R) >= max{null(T), null(R)}

Let (T+R)(x0) = 0
Then T(x0) + R(x0) = 0
So T(x0) = -R(x0)
So nullspace(T+R) = {k | T(k) = -R(k)}

Thus

dim{k | T(k) = -R(k)} >= max{null(T), null(U)}

Now let U(x) = -R(x)

dim{k | T(k) = U(k)} >= max{null(T), null(-U)}

And, null(-U) = null(U)

dim{k | T(k) = U(k)} >= max{null(T), null(U)}


Thanks again.
 
  • #8
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You can think of your subspace as the kernel of the linear map [itex]T-U[/itex].
 
  • #9
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Whoops. I see you were already basically doing that. My bad.
 
  • #10
PeroK
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I think I am confusing myself. Could you maybe see something wrong with the following?

Let T and R be arbitrary linear transformations.

rank(T+R) <= rank(T) + rank(R)

(dim-null)(T+R) <= (dim-null)(T) + (dim-null)(R)

PK: this doesn't hold. Consider R = -T.

null(T+R) >= null(T) + null(R)
null(T+R) >= max{null(T), null(R)}

Let (T+R)(x0) = 0
Then T(x0) + R(x0) = 0
So T(x0) = -R(x0)
So nullspace(T+R) = {k | T(k) = -R(k)}

Thus

dim{k | T(k) = -R(k)} >= max{null(T), null(U)}

Now let U(x) = -R(x)

dim{k | T(k) = U(k)} >= max{null(T), null(-U)}

And, null(-U) = null(U)

dim{k | T(k) = U(k)} >= max{null(T), null(U)}

PK: This is not true. Consider T to be unity matrix with 0 in the (1,1) place. And U to be twice the unity matrix with 0 in the (2, 2) place. Then:

T(k) = U(k) only for 0. But null(T) = null(U) = 1.


Thanks again.
T & U could each be null on half the basis vectors (different halves), so be different on every vector except 0.

There is no relationship between what the LT's do outside their null spaces and how big their null spaces are.
 
  • #11
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I greatly appreciate the insight, PeroK. I will think about these things.
 

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