Page 54 - Linear Algebra Done Right

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Tv = a 1 Tv 1 + ··· + a n Tv n .
In particular, the values of Tv 1 ,...,Tv n determine the values of T on
arbitrary vectors in V. Chapter 3. Linear Maps
Linear maps can be constructed that take on arbitrary values on a
basis. Specifically, given a basis (v 1 ,...,v n ) of V and any choice of
vectors w 1 ,...,w n ∈ W, we can construct a linear map T : V → W such
that Tv j = w j for j = 1,...,n. There is no choice of how to do this—we
must define T by

T(a 1 v 1 +· · ·+ a n v n ) = a 1 w 1 +· · ·+ a n w n ,
where a 1 ,...,a n are arbitrary elements of F. Because (v 1 ,...,v n ) is a
basis of V, the equation above does indeed define a function T from V
to W. You should verify that the function T defined above is linear and
that Tv j = w j for j = 1,...,n.
Now we will make L(V, W) into a vector space by defining addition
and scalar multiplication on it. For S, T ∈L(V, W), define a function
S + T ∈L(V, W) in the usual manner of adding functions:

(S + T)v = Sv + Tv

for v ∈ V. You should verify that S + T is indeed a linear map from V
to W whenever S, T ∈L(V, W). For a ∈ F and T ∈L(V, W), define a
function aT ∈L(V, W) in the usual manner of multiplying a function
by a scalar:
(aT)v = a(Tv)

for v ∈ V. You should verify that aT is indeed a linear map from V to W
whenever a ∈ F and T ∈L(V, W). With the operations we have just
defined, L(V, W) becomes a vector space (as you should verify). Note
that the additive identity of L(V, W) is the zero linear map defined
earlier in this section.
Usually it makes no sense to multiply together two elements of a
vector space, but for some pairs of linear maps a useful product exists.
We will need a third vector space, so suppose U is a vector space over F.
If T ∈L(U, V) and S ∈L(V, W), then we define ST ∈L(U, W) by

(ST)(v) = S(Tv)

for v ∈ U. In other words, ST is just the usual composition S ◦T of two
functions, but when both functions are linear, most mathematicians