I was given this proof dealing primarily with the information found in 4.1-4.3 o

Question

I was given this proof dealing primarily with the information found in 4.1-4.3 of "Linear Algebra and its applications" by david lay. I understand the individual aspects of it but can't correlate them into a unified proof. please help.

Let T : V ? W and U : V ? X be linear transformations of vector spaces, and suppose

that ker(T) ? ker(U). Show that for v ? V , the value U(v) depends only on T(v) ? im(T)

and use this to show that there is a linear transformation U : im(T) ? X such that

U ? T = U.

Explanation / Answer

Basically they want to say that if we have some vector in im(T), say w, then there could be many vectors in V which are mapped to w, but the value of U at any of these vectors is same and depends only on w.

This is because if v1 and v2 in v both get mapped to w, we have T(v1-v2) = T(v1)-T(v2) = 0

=> v1-v2 is in ker(T)

=> v1-v2 is in ker(U)

=> U(v1-v2) = 0

=> U(v1)-U(v2) = 0, i.e U(v1) = U(v2).

The foregoing shows that the following is a well defined linear transformation.

S : im(T) -> X, S(w) = U(v), where v is such that T(v) = w.

Note that im(T) being a subspace of W is a vector space.

That this is well defined is shown above.

Suppose w1,w2 are in im(T)

Also S(w1)+S(w2) = U(v1)+U(v2), where T(v1)=w1, T(v2) = w2.

Now T(v2+v2) = T(v1)+T(v2)=w1+w2.

Thus S(w1+w2) = U(v1+v2) (definition of S, noting v1+v2 is mapped to w1+w2 by T)

Now since U is linear transformation U(V1+v2) = U(v1)+U(v2).

This shows S(w1+w2) = U(v1+v2) = U(v1)+U(v2) = S(w1)+S(w2).

That is S(w1+w2) = S(w1)+S(w2).

Now w in im(T), k scalar

kS(w) = k.U(v) , where v is such that T(v)=w.

Now T(kv) = k.T(v) = kw

=> S(kw) = U(kv) (again definition of S, kv is mapped to kw by T)

Now since U(kv) = kU(v), we have S(kw) = k S(w).

Thus S is a linear transformation.

Now T: V->W, S : Im(T)->X,

so we can take ST

ST(v) = S(w), where T(v)=w

= U(v) (since by definition of S(w) = U(v), where v is any element so that T(v)=w).

Thus ST(v) = U(v). This is true for any v in V

Hence ST = U.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.