Let the rank of vector group A1, A2. Am be r, then any r linearly independent vectors in A1, A2,. Am constitute its maximal linearly independent group

Let the rank of vector group A1, A2. Am be r, then any r linearly independent vectors in A1, A2,. Am constitute its maximal linearly independent group

Counter proof: if any r linearly independent vectors in A1, A2,. Am are not its maximal linearly independent group
Let's remember that B1, B2,... Br are the R linearly independent vectors
Because it is not the maximal linearly independent group of the original vector group
Then at least one of the remaining vectors can be added to B1, B2,... Br
Then B1, B2,... BR, Br + 1 are linearly independent groups
Then the rank of vector group A1, A2. Am must be greater than or equal to R + 1
The contradiction between the topic and the design

Vector group B: B1, B2 And BM can be represented by vector group A: A1, A2 If and only if ()

R(A)=R(A,B)..

If vector group A1, A2 It is proved that: (1) vector group A1, a1 + A2 ,a1+a2+… +(2) B1 = a1 + k1am, B2 = A2 + k2am,..., BM-1 + km-1am are linearly independent

It can be proved by definition
Here's another proof for you, which uses two conclusions
(1)
(a1,a1+a2,… ,a1+a2+… +am)=(a1,a2,… am)K
Where k=
1 1 ...1 1
0 1 ...1 1
......
0 0 ...0 1
Because | K | = 1 ≠ 0, K is reversible
So r (A1, a1 + A2 ,a1+a2+… +am)=r(a1,a2,… am) = m
So A1, a1 + A2 ,a1+a2+… +Am is linearly independent
(2)
(b1,b2,...,bm-1) = (a1,a2,… am)K
Where k=
1 0 ...0
0 1 ...0
......
0 0 ...1
k1 k2 .km-1
Because A1, A2 Am is linearly independent
So r (B1, B2,..., BM-1) = R (k) = M-1
So B1, B2,..., BM-1 are linearly independent