CHAPTER II DIMENSION In the present chapter we investigate ...

So, for each v, there is a vector u in V such that T u = v. Here u is uniquely
determined by v. To see this, assume there is another vector w such that T w = v.
Then T (u − w) = T u − T w = v − v = 0. This shows that u − w is in ker T . The
assumption, ker T = {0} tells us that u − w = 0 and hence u = w. Now we can define
the inverse T −1 of T by putting T −1 v = u if T u = v holds. We have proved
Corollary 2.4.2. If T is a linear operator on a finite dimensional space V with
ker T = {0}, then T is invertible.
The argument for proving the above corollary also shows
Corollary 2.4.3. If T is a linear mapping on a finite dimensional space V into
itself and if the homogeneous equation T x = 0 has no nontrivial solution, then, for any
vector b in V , the equation T x = b has a unique solution.
Corollary 2.4.4. If M is an m × n matrix with m < n, then the homogeneous
equation Ax = 0 has nontrivial solutions. (Notice that m is the number of equations and
n is the number of unknowns in the system of equations in vector form Ax = 0.)
Proof. Define the linear map T : C n → C m by putting T x = Ax (in other
words, T = MA). Then dim ker T + dim T (C n ) = dim C n = n. On the other hand,
since T (C n ) is a subspace of C m and dim C m = m, we have dim T (C n ) ≤ m. Thus
dim ker T = dim C n − dim T (C n ) ≥ n − m > 0. Hence ker T = {0}. Any nonzero vector
in ker T gives a nontrivial solution to Ax = 0.
Let A = [A1 A2 · · · An] be an m × n matrix. As usual, we study the homogeneous
equation Ax = 0 by introducing the linear map T : C n → C m . A solution of this equation
is just a vector in the kernel of T . So the dimension of the solution space is the nullity of
T , which is also the number of parameters needed in writing down the general solution.
The rank of A is defined to be the rank of T , which is the dimension of the range T (V )
of V . From then identity
⎢
T x = Ax = [A1 A2 · · · An] ⎢
⎣
⎡
x1
x2
.
xn
⎤
⎥
⎦ = A1x1 + A2x2 + · · · + Anxn
we see that the range of T is just the column space of A, that is, the linear span of
the column vectors A1, A2, . . . , An of the matrix A. The leading vectors of this list of
column vectors is a basis of the column space of A. The number of the leading vectors is
the rank of A. Recall that leading vectors can be identified by the echelon form of A.
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