Proving Linearity of Transformation T on U

[nietzsche]

Homework Statement

Suppose U is a finite dimensional vector space and A = {u1, u2, ... , un} is a basis of U. Define T : U -> R(nx1) by T(v) = [v]_A.

(In other words: U is an n-dimensional vector space, A is a basis for U, and T is the transformation that takes a vector in U and finds its coordinate vector with respect to the basis A.)

Show that T is a linear transformation. Find kerT.

Homework Equations

The Attempt at a Solution

If v and w are arbitrary vectors in U and a and b are scalars, we have

T(av+bw)
= [av+bw]_A
=[av1+bw1, ... , avn+bwn]_A
=[av1 + ... + avn]_A + [bw1 + ... + bwn]_A
=a[v1 + ... + vn]_A + b[w1 + ... + wn]_A
=aT(v) + bT(w)

So T is a linear transformation.

I just don't see if what I'm doing here is justified. Please help! Thanks.


And for finding kerT, since A is a basis, it is linearly independent, so the only way to get the zero vector is to multiply everything by 0, so kerT = {0}.

 

[mighty2000]

Your solution for showing that T is a linear transformation is correct. You have correctly shown that T satisfies the two properties of a linear transformation: T(av + bw) = aT(v) + bT(w) and T(0) = 0.

However, your reasoning for finding kerT is not entirely complete. While it is true that the only way to get the zero vector is to multiply everything by 0, you also have to consider the fact that [v]A = 0 only when v = 0. This is because A is a basis, which means that every vector in U can be uniquely represented as a linear combination of the basis vectors. Therefore, if [v]A = 0, it must be the case that v = 0.

So, the kernel of T is actually the set of all vectors v in U such that T(v) = [v]A = 0. Since this only occurs when v = 0, we can say that kerT = {0}.

I hope this helps clarify your understanding of the problem. Keep up the good work!