110 CHAPTER 5. LINEAR TRANSFORMATIONS

Theorem 5.2.12 Let A ∈L (X ,Y ). Then rank(A) = rank(M) where M is the matrix of Ataken with respect to a pair of bases for the vector spaces X , and Y. Here M is consideredas a linear transformation by matrix multiplication.

Proof: Recall the diagram which describes what is meant by the matrix of A. Here thetwo bases are as indicated.

β = {v1, · · · ,vn} X A−→ Y {w1, · · · ,wm}= γ

qβ ↑ ◦ ↑ qγ

Fn M−→ Fm

Let {Ax1, · · · ,Axr} be a basis for AX . Thus{qγ Mq−1

βx1, · · · ,qγ Mq−1

βxr

}is a basis for AX . It follows that {

Mq−1X x1, · · · ,Mq−1

X xr}

is linearly independent and so rank(A) ≤ rank(M) . However, one could interchange theroles of M and A in the above argument and thereby turn the inequality around. ■

The following result is a summary of many concepts.

Theorem 5.2.13 Let L ∈L (V,V ) where V is a finite dimensional vector space. Then thefollowing are equivalent.

1. L is one to one.

2. L maps a basis to a basis.

3. L is onto.

4. If Lv = 0 then v = 0.

Proof: Suppose first L is one to one and let β = {vi}ni=1 be a basis. Then if ∑

ni=1 ciLvi =

0 it follows L(∑ni=1 civi) = 0 which means that since L(0) = 0, and L is one to one, it must

be the case that ∑ni=1 civi = 0. Since {vi} is a basis, each ci = 0 which shows {Lvi} is a

linearly independent set. Since there are n of these, it must be that this is a basis.Now suppose 2.). Then letting {vi} be a basis, and y ∈ V, it follows from part 2.) that

there are constants, {ci} such that y = ∑ni=1 ciLvi = L(∑n

i=1 civi) . Thus L is onto. It hasbeen shown that 2.) implies 3.).

Now suppose 3.). Then L(V ) =V . If {v1, · · · ,vn} is a basis of V, then

V = span(Lv1, · · · ,Lvn) .

It follows that {Lv1, · · · ,Lvn} must be linearly independent because if not, one of the vec-tors could be deleted and you would then have a spanning set with fewer vectors thandim(V ). If Lv = 0,

v = ∑i

xivi