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 C₂Then A¹c, is the jth column of ATA :Cn. We can write matrix AOAT A = AT [ C₁ C2C1Cn] =[ATC₁ ATC₂AT cn].=Since we have that rank A = n, from Theorem 3 part 4 we have that ATA isnx n and it is invertible. This means than rank(ATA). = n. ApplyingTheorem 3 part 3 on ATA we have that columns of ATA are linearlyindependent in R" and applying Theorem 4 part 2 we have that they alsospan R". This means that {ATc₁, ATC2, ..., AT cn} is a basis of R". Theorem 5.4.3The 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.

Since determinant of ATA=determinant of AT*determinant of A…

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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Question

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,…,c_n. If rank A=n, show that {A^Tc₁,A^Tc₂,…,A^T c_n} 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".

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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