438 CHAPTER 18. LINEAR TRANSFORMATIONS

and since q is one to one, it follows that ∑nk=1 ckvk = 0 which requires each ck = 0 because

{v1, · · · ,vn} is independent. Next take w ∈W. Since q is onto, there exists v ∈V such thatqv = w. Since {v1, · · · ,vn} is a basis, there are scalars ck such that q(∑n

k=1 ckvk) = q(v) =w and so w = ∑

nk=1 ckqvk which is in the span(qv1, · · · ,qvn) . Therefore, {qv1, · · · ,qvn} is

a basis as claimed.Suppose now that qvi = wi where {v1, · · · ,vn} and {w1, · · · ,wn} are bases for V and

W . If q(∑nk=1 ckvk) = 0, then ∑

nk=1 ckqvk = ∑

nk=1 ckwk = 0 and so each ck is 0 because it

is given that {w1, · · · ,wn} is linearly independent. For ∑nk=1 ckwk an arbitrary vector of W,

this vector equals ∑nk=1 ckqvk = q(∑n

k=1 ckvk) . Therefore, q is also onto. ■Such a mapping is called an isomorphism.

Definition 18.5.2 Let V be a vector space of dimension n and W a vector space of dimen-sion m with respect to the same field of scalars F. Let

qα : Fn→V

qβ : Fm→W

be two isomorphisms as in the above lemma. Here α denotes the basis

{qα e1, · · · ,qα en}

and β denotes the basis{

qβ e1, · · · ,qβ em}

for V and W respectively. For L ∈L (V,W ) ,the matrix of L with respect to these two bases,

{qα e1, · · · ,qα en} ,{

qβ e1, · · · ,qβ em}

denoted by [L]βα

or [L] for short satisfies

Lqα = qβ [L]βα(18.6)

In terms of a diagram,L

V → Wqα ↑ ◦ ↑ qβ

Fn → Fm

[L]βα

(18.7)

Starting at Fn, you go up and then to the right using L and the result is the same if you goto the right by matrix multiplication by [L]

βαand then up using qβ .

So how can we find this matrix? Let

α ≡ {qα e1, · · · ,qα en} ≡ {v1, · · · ,vn}β ≡

{qβ e1, · · · ,qβ em

}≡ {w1, · · · ,wm}

and suppose the i jth entry of the desired matrix is ai j. Letting b ∈ Fn, the requirement 18.6is equivalent to

∑i

=([L]b)i︷ ︸︸ ︷(∑

jai jb j

)wi = L∑

jb jv j = ∑

jb jLv j.