The pictures are the answer of the question and the theorem 3, I would to ask why the answer of this question can give the result that rank(ATA)=n ? This is the question: Let A be an m×n matrix with columns c1,c2,…,cn. If rank A=n, show that {ATc1,ATc2,…,AT cn} is a basis of Rn.
The pictures are the answer of the question and the theorem 3, I would to ask why the answer of this question can give the result that rank(ATA)=n ? This is the question: Let A be an m×n matrix with columns c1,c2,…,cn. If rank A=n, show that {ATc1,ATc2,…,AT cn} is a basis of Rn.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The pictures are the answer of the question and the theorem 3, I would to ask why the answer of this question can give the result that rank(ATA)=n ?
This is the question: Let A be an m×n matrix with columns c1,c2,…,cn. If rank A=n, show that {ATc1,ATc2,…,AT cn} is a basis of Rn.
![Let A be m x n matrix with columns C₁, C₂, .
in terms of its column:
Cn ].
A = [C₁ C₂
Then A¹c, is the jth column of ATA :
Cn. We can write matrix A
O
AT A = AT [ C₁ C2
C1
Cn] =[ATC₁ ATC₂
AT cn].
=
Since we have that rank A = n, from Theorem 3 part 4 we have that ATA is
nx n and it is invertible. This means than rank(ATA). = n. Applying
Theorem 3 part 3 on ATA we have that columns of ATA are linearly
independent in R" and applying Theorem 4 part 2 we have that they also
span R". This means that {ATc₁, ATC2, ..., AT cn} is a basis of R".](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87970aa2-d45e-49d2-9569-1aab949e1122%2F8c20704c-049f-436b-af3c-075945341480%2Fzzcujx_processed.png&w=3840&q=75)
Transcribed Image Text:Let A be m x n matrix with columns C₁, C₂, .
in terms of its column:
Cn ].
A = [C₁ C₂
Then A¹c, is the jth column of ATA :
Cn. We can write matrix A
O
AT A = AT [ C₁ C2
C1
Cn] =[ATC₁ ATC₂
AT cn].
=
Since we have that rank A = n, from Theorem 3 part 4 we have that ATA is
nx n and it is invertible. This means than rank(ATA). = n. Applying
Theorem 3 part 3 on ATA we have that columns of ATA are linearly
independent in R" and applying Theorem 4 part 2 we have that they also
span R". This means that {ATc₁, ATC2, ..., AT cn} is a basis of R".
Transcribed Image Text:Theorem 5.4.3
The following are equivalent for an m × n matrix A:
1. rank A = n.
2. The rows of A span R".
3. The columns of A are linearly independent in R™.
4. The nxn matrix AT A is invertible.
5. CA In for some nxm matrix C.
= =
6. If Ax=0, x in R", then x = 0.
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