18.5. THE MATRIX OF A LINEAR TRANSFORMATION 437

is just a multiple of the identity, it follows that DN = ND. Therefore, the usual binomialtheorem may be applied and this yields the following equations for k ≥ ms.

T ks = (D+N)k =

k

∑j=0

(kj

)Dk− jN j

=ms

∑j=0

(kj

)Dk− jN j, (18.5)

the third equation holding because Nms = 0. Thus T ks is of the form

T ks =

αk · · · ∗...

. . ....

0 · · · αk

 .

Lemma 18.4.5 Suppose T is of the form Ts described above in 18.4 where the constant α,on the main diagonal, is less than one in absolute value. Then

limk→∞

(T k)

i j= 0.

Proof: From 18.5, it follows that for large k, and j ≤ ms,(kj

)≤ k (k−1) · · ·(k−ms +1)

ms!.

Therefore, letting C be the largest value of∣∣∣(N j

)pq

∣∣∣ for 0≤ j ≤ ms,∣∣∣∣(T k)

pq

∣∣∣∣≤ msC(

k (k−1) · · ·(k−ms +1)ms!

)|α|k−ms

which converges to zero as k→∞. This is most easily seen by applying the ratio test to theseries

∞

∑k=ms

(k (k−1) · · ·(k−ms +1)

ms!

)|α|k−ms

and then noting that if a series converges, then the kth term converges to zero. ■

18.5 The Matrix Of A Linear TransformationTo begin with, here is an easy lemma which relates the vectors in two vector spaces havingthe same dimension.

Lemma 18.5.1 Let q : V →W be one to one, onto and linear. Then q maps any basis of Vto a basis for W. Conversely, if q is linear and maps a basis to a basis, then it is one to oneonto.

Proof: Let {v1, · · · ,vn} be a basis for V , why is {qv1, · · · ,qvn} a basis for W? Firstconsider why it is linearly independent. Suppose ∑

nk=1 ckqvk = 0. Then q(∑n

k=1 ckvk) = 0