Page 196 - Linear Algebra Done Right

The terms on the first line on the right are all in null N ∩ range N; the
terms on the second line are all in W. Thus the last equation and 8.41
imply that Jordan Form 185
0 = a 1,m(v 1 ) N m(v 1 ) v 1 +· · ·+ a j,m(v j ) N m(v j ) v j
8.43 = a 1,m(v 1 ) N m(u 1 ) u 1 + ··· + a j,m(v j ) N m(u j ) u j
and

8.44 0 = a j+1,0 v j+1 + ··· + a k,0 v k .

Now 8.43 and (ii) imply that a 1,m(v 1 ) =· · ·= a j,m(v j ) = 0. Because
(v j+1 ,...,v k ) is a basis of W, 8.44 implies that a j+1,0 =· · ·= a k,0 = 0.
Thus all the a’s equal 0, and hence the list of vectors in (a) is linearly
independent.
Clearly (ii) implies that dim(null N ∩ range N) = j. Along with 8.41,
this implies that

8.45 dim null N = k.
Clearly (i) implies that

j

dim range N = (m(u r ) + 1)
r=0
j

8.46 = m(v r ).
r=0
The list of vectors in (a) has length

k j

(m(v r ) + 1) = k + m(v r )
r=0 r=0
= dim null N + dim range N
= dim V,

where the second equality comes from 8.45 and 8.46, and the third
equality comes from 3.4. The last equation shows that the list of vectors
in (a) has length dim V; because this list is linearly independent, it is a
basis of V (see 2.17), completing the proof of (a).
Finally, note that

(N m(v 1 ) v 1 ,...,N m(v k ) v k ) = (N m(u 1 ) u 1 ,...,N m(u j ) u j ,v j+1 ,...,v k ).