Math Is Fun Forum

basmah

Member

Registered: 2006-10-02

Posts: 18

Math help required!

Q). Let T: V --> W be a linear transformation, and let U be a subspace of W. Prove that the set 

                            T-¹ (U) = {v Є V: T(v) Є U}
   
is a subspace of V?

What is T-¹ (U) if U = {0}

I'm normally good with proofs but this one just does not make sense to me ....i'm sooo confused can anyone help me? (also if you can show all your steps, that would be great, its sometimes hard for me to understand when there are alot of steps skipped).

Last edited by basmah (2006-11-19 13:38:35)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Math help required!

What is T-¹ (U) if U = {0}

T-¹ (U) is all the elements of V which map into 0.  This is also known as the kernel of T.

Theorem: Let V be a vector space over the field K, and let W be a subset of V. Then W is a subspace if and only if it satisfies the following 3 conditions:

   1. If the zero vector, θ, is in W.
   2. If u and v are elements of W, then the sum u + v is an element of W;
   3. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;

So lets go one by one.  We know that T-¹ (U) is a subset of V because T-¹ (U) consists of only elements in V.  Thus, all we have to do is meet the above 3 and we're done.

1. The zero vector.  For U to be a subspace of W, it must contain the zero vector.  Thus, 0 Є U.  But since T is a linear transformation, T(x + y) = T(x) + T(y) for any x, y in V.  We know that V is a vector space, so 0 Є V.  So T(0 + 0) = T(0) + T(0).  So T(0) - T(0) = 0 = T(0).  Thus, T(0) = 0, but we already know that 0 Є U.  Thus, 0 Є T-¹ (U).

2. We must show that if x Є T-¹ (U) and y Є T-¹ (U), then x + y Є T-¹ (U).  Since x Є T-¹ (U) and y Є T-¹ (U), we know that T(x) and T(y) are both in U.  We know that U is a vector space, thus T(x) + T(y) is also in U.  Final nail in the coffin is T(x + y) = T(x) + T(y).  x + y is mapped to T(x) + T(y), and T(x) + T(y) is in U.  Thus, x + y is in T-¹ (U).

3. Try to use similar methods that I have to prove this one.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

basmah

Member

Registered: 2006-10-02

Posts: 18

Re: Math help required!

okay now i don't understand this part:
x + y is mapped to T(x) + T(y),

i know we have to prove: T(x+y) = T(x) + T(y) and then  T(cx)=cT(x) but i don't understand how you got there.....can you explain please...

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Math help required!

Sure thing.  As I wrote it, it sounded weird to me as well, but I didn't have time to refine it.

We know that T(x + y) = T(x) + T(y), this is the definition of a linear map.

This means that applying the function to x + y gives us T(x) + T(y).  In other words:

Function T:

Domain   | Range
---------------------
x            | T(x)
y            | T(y)
x + y      | T(x + y) = T(x) + T(y)

x is mapped to T(x), y is mapped to T(y), and x + y is mapped to T(x) + T(y)

Is that a little more clear?

Since T(x) + T(y) is in U, T(x + y) is in U, which means that x + y is in T-¹ (U):

T-¹ (U) = {v Є V: T(v) Є U}

x Є V and y Є V, so it must be that x + y Є V.  And as stated above, T(x + y) = T(x) + T(y), and T(x) Є U and T(y) Є U, thus, T(x) + T(y) Є U and so T(x + y) Є U.  This satisfies all needed requirements for x + y is in T-¹ (U).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."