Page 296 - 35Linear Algebra

296 Kernel, Range, Nullity, Rank

Theorem 16.2.5 (Dimension Formula). Let L: V → W be a linear trans-
1
formation, with V a finite-dimensional vector space . Then:
dim V = dim ker V + dim L(V )
= null L + rank L.

Proof. Pick a basis for V :

{v 1 , . . . , v p , u 1 , . . . , u q },

where v 1 , . . . , v p is also a basis for ker L. This can always be done, for exam-
ple, by finding a basis for the kernel of L and then extending to a basis for V .
Then p = null L and p + q = dim V . Then we need to show that q = rank L.
To accomplish this, we show that {L(u 1 ), . . . , L(u q )} is a basis for L(V ).
To see that {L(u 1 ), . . . , L(u q )} spans L(V ), consider any vector w in L(V ).
j
i
Then we can find constants c , d such that:
p
1
q
1
w = L(c v 1 + · · · + c v p + d u 1 + · · · + d u q )
1
1
q
p
= c L(v 1 ) + · · · + c L(v p ) + d L(u 1 ) + · · · + d L(u q )
q
1
= d L(u 1 ) + · · · + d L(u q ) since L(v i ) = 0,
⇒ L(V ) = span{L(u 1 ), . . . , L(u q )}.
Now we show that {L(u 1 ), . . . , L(u q )} is linearly independent. We argue
j
by contradiction. Suppose there exist constants d (not all zero) such that
1
q
0 = d L(u 1 ) + · · · + d L(u q )
1
q
= L(d u 1 + · · · + d u q ).
q
1
j
But since the u are linearly independent, then d u 1 + · · · + d u q 6= 0, and
q
1
1
q
so d u 1 + · · · + d u q is in the kernel of L. But then d u 1 + · · · + d u q must
be in the span of {v 1 , . . . , v p }, since this was a basis for the kernel. This
contradicts the assumption that {v 1 , . . . , v p , u 1 , . . . , u q } was a basis for V , so
we are done.
1
The formula still makes sense for infinite dimensional vector spaces, such as the space
of all polynomials, but the notion of a basis for an infinite dimensional space is more
sticky than in the finite-dimensional case. Furthermore, the dimension formula for infinite
dimensional vector spaces isn’t useful for computing the rank of a linear transformation,
since an equation like ∞ = ∞ + x cannot be solved for x. As such, the proof presented
assumes a finite basis for V .
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