- #1
Chris Rorres
- 4
- 0
If A is an invertible matrix and vectors (v1,v2,...,vn) is a basis for Rn, prove that (Av1,Av2,...,Avn) is also a basis for Rn.
Proving Basis of Av with Invertible A Matrix
In summary, the conversation is about proving that (Av1,Av2,...,Avn) is a basis for Rn, given that A is an invertible matrix and (v1,v2,...,vn) is a basis for Rn. The participants discuss different approaches to proving that the set spans and is linearly independent, and mention using the fact that A is invertible and its kernel to show that the coefficients must be zero.
- #2
tiny-tim
Science Advisor
Homework Helper
- 25,839
- 258
Hi Chris! 
Show us how far you get, and where you're stuck, and then we'll know how to help!
Show us how far you get, and where you're stuck, and then we'll know how to help!
- #3
ilia1987
- 9
- 0
why don't threads like this get moved to homework?
- #4
HallsofIvy
Science Advisor
Homework Helper
- 42,989
- 975
Just lazy mentors!
- #5
Chris Rorres
- 4
- 0
I need to prove that (Av1,Av2,...,Avn) spans and that it is linearly independent but this proof is so confusing to me that i don't even know where to start doing that.
- #6
Quantumpencil
- 96
- 0
You need to show that b* A(v_1) + ... b_n A(v_n) = 0 implies b_1 ... b_n equals zero, right? Well, you know since A is invertible, what is it's kernal?
1. What is the basis of A with an invertible A matrix?
The basis of A with an invertible A matrix is the set of linearly independent columns of A. This means that each column can be expressed as a linear combination of the other columns, and no column is redundant.
2. How do you prove that A has an invertible matrix?
To prove that A has an invertible matrix, we need to show that its columns are linearly independent. This can be done by setting up a system of equations using the columns of A and solving for the variables. If the only solution is the trivial solution (all variables equal to 0), then the columns are linearly independent and A has an invertible matrix.
3. Can A have an invertible matrix if its columns are not linearly independent?
No, if the columns of A are not linearly independent, then A cannot have an invertible matrix. This is because the columns of an invertible matrix must form a basis, and a basis must consist of linearly independent vectors.
4. What is the relationship between the invertibility of A and its determinant?
The determinant of A is a measure of how much the volume of a parallelepiped changes when transformed by A. A has an invertible matrix if and only if its determinant is non-zero, meaning that the transformation does not collapse the volume of the parallelepiped to 0. In other words, if A has an invertible matrix, its determinant must be non-zero.
5. How does the invertibility of A affect its solution to a system of linear equations?
If A has an invertible matrix, then the system of linear equations can be solved for a unique solution using matrix operations. This is because an invertible matrix can be used to transform the system into an equivalent system with a unique solution. However, if A does not have an invertible matrix, then the system may have either no solution or infinitely many solutions.
Similar threads
-
Calculus and Beyond Homework Help
-
Calculus and Beyond Homework Help
-
Linear and Abstract Algebra
-
Calculus and Beyond Homework Help
-
Calculus and Beyond Homework Help
-
Calculus and Beyond Homework Help
-
Linear and Abstract Algebra
-
Precalculus Mathematics Homework Help
-
Calculus and Beyond Homework Help