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Mathematics LibreTexts

9.6: Linear Transformations

( \newcommand{\kernel}{\mathrm{null}\,}\)

Outcomes

  1. Understand the definition of a linear transformation in the context of vector spaces.

Recall that a function is simply a transformation of a vector to result in a new vector. Consider the following definition.

Definition 9.6.1: Linear Transformation

Let V and W be vector spaces. Suppose T:VW is a function, where for each xV,T(x)W. Then T is a linear transformation if whenever k,p are scalars and v1 and v2 are vectors in V T(kv1+pv2)=kT(v1)+pT(v2)

Several important examples of linear transformations include the zero transformation, the identity transformation, and the scalar transformation.

Example 9.6.1: Linear Transformations

Let V and W be vector spaces.

  1. The zero transformation
    0:VW is defined by 0(v)=0 for all vV.
  2. The identity transformation
    1V:VV is defined by 1V(v)=v for all vV.
  3. The scalar transformation Let aR.
    sa:VV is defined by sa(v)=av for all vV.
Solution

We will show that the scalar transformation sa is linear, the rest are left as an exercise.

By Definition 9.6.1 we must show that for all scalars k,p and vectors v1 and v2 in V, sa(kv1+pv2)=ksa(v1)+psa(v2). Assume that a is also a scalar. sa(kv1+pv2)=a(kv1+pv2)=akv1+apv2=k(av1)+p(av2)=ksa(v1)+psa(v2)

Therefore sa is a linear transformation.

Consider the following important theorem.

Theorem 9.6.1: Properties of Linear Transformations

Let V and W be vector spaces, and T:VW a linear transformation. Then

  1. T preserves the zero vector. T(0)=0
  2. T preserves additive inverses. For all vV, T(v)=T(v)
  3. T preserves linear combinations. For all v1,v2,,vmV and all k1,k2,,kmR, T(k1v1+k2v2++kmvm)=k1T(v1)+k2T(v2)++kmT(vm).
Proof
  1. Let 0V denote the zero vector of V and let 0W denote the zero vector of W. We want to prove that T(0V)=0W. Let vV. Then 0v=0V and T(0V)=T(0v)=0T(v)=0W.
  2. Let vV; then vV is the additive inverse of v, so v+(v)=0V. Thus T(v+(v))=T(0V)T(v)+T(v))=0WT(v)=0WT(v)=T(v).
  3. This result follows from preservation of addition and preservation of scalar multiplication. A formal proof would be by induction on m.

Consider the following example using the above theorem.

Example 9.6.2: Linear Combination

Let T:P2R be a linear transformation such that T(x2+x)=1;T(x2x)=1;T(x2+1)=3.

Find T(4x2+5x3).

We provide two solutions to this problem.

Solution 1:

Suppose a(x2+x)+b(x2x)+c(x2+1)=4x2+5x3. Then (a+b+c)x2+(ab)x+c=4x2+5x3.

Solving for a, b, and c results in the unique solution a=6, b=1, c=3. Thus T(4x2+5x3)=T(6(x2+x)+(x2x)3(x2+1))=6T(x2+x)+T(x2x)3T(x2+1)=6(1)+13(3)=14.

Solution 2:

Notice that S={x2+x,x2x,x2+1} is a basis of P2, and thus x2, x, and 1 can each be written as a linear combination of elements of S.

x2=12(x2+x)+12(x2x)x=12(x2+x)12(x2x)1=(x2+1)12(x2+x)12(x2x).

Then T(x2)=T(12(x2+x)+12(x2x))=12T(x2+x)+12T(x2x)=12(1)+12(1)=0.T(x)=T(12(x2+x)12(x2x))=12T(x2+x)12T(x2x)=12(1)12(1)=1.T(1)=T((x2+1)12(x2+x)12(x2x))=T(x2+1)12T(x2+x)12T(x2x)=312(1)12(1)=3.

Therefore, T(4x2+5x3)=4T(x2)+5T(x)3T(1)=4(0)+5(1)3(3)=14.

The advantage of Solution 2 over Solution 1 is that if you were now asked to find T(6x213x+9), it is easy to use T(x2)=0, T(x)=1 and T(1)=3: T(6x213x+9)=6T(x2)13T(x)+9T(1)=6(0)13(1)+9(3)=13+27=40.
More generally, T(ax2+bx+c)=aT(x2)+bT(x)+cT(1)=a(0)+b(1)+c(3)=b+3c.

Suppose two linear transformations act in the same way on v for all vectors. Then we say that these transformations are equal.

Definition 9.6.2: Equal Transformations

Let S and T be linear transformations from V to W. Then S=T if and only if for every vV, S(v)=T(v)

The definition above requires that two transformations have the same action on every vector in order for them to be equal. The next theorem argues that it is only necessary to check the action of the transformations on basis vectors.

Theorem 9.6.2: Transformation of a Spanning Set

Let V and W be vector spaces and suppose that S and T are linear transformations from V to W. Then in order for S and T to be equal, it suffices that S(vi)=T(vi) where V=span{v1,v2,,vn}.

This theorem tells us that a linear transformation is completely determined by its actions on a spanning set. We can also examine the effect of a linear transformation on a basis.

Theorem 9.6.3: Transformation of a Basis

Suppose V and W are vector spaces and let {w1,w2,,wn} be any given vectors in W that may not be distinct. Then there exists a basis {v1,v2,,vn} of V and a unique linear transformation T:VW with T(vi)=wi.

Furthermore, if v=k1v1+k2v2++knvn

is a vector of V, then T(v)=k1w1+k2w2++knwn.


This page titled 9.6: Linear Transformations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform.

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