- #1
Chris Rorres
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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
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SUMMARY
In the discussion, participants explore the proof that if A is an invertible matrix and (v1, v2, ..., vn) forms a basis for Rn, then (Av1, Av2, ..., Avn) also constitutes a basis for Rn. The key steps involve demonstrating that the set (Av1, Av2, ..., Avn) spans Rn and is linearly independent. The proof hinges on the properties of invertible matrices, particularly that the kernel of an invertible matrix is trivial, which directly supports the linear independence of the transformed vectors.
PREREQUISITES- Understanding of linear algebra concepts such as basis, span, and linear independence.
- Knowledge of properties of invertible matrices, specifically their kernels.
- Familiarity with matrix-vector multiplication and its implications on vector spaces.
- Basic proof techniques in mathematics, particularly in the context of vector spaces.
- Study the properties of invertible matrices in detail, focusing on their kernels and image.
- Learn about linear transformations and how they affect bases in vector spaces.
- Explore examples of proving linear independence using matrix transformations.
- Investigate the implications of the Rank-Nullity Theorem in relation to invertible matrices.
Students of linear algebra, mathematics educators, and anyone interested in understanding the foundational concepts of vector spaces and matrix transformations.
- #2
tiny-tim
Science Advisor
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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
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why don't threads like this get moved to homework?
- #4
HallsofIvy
Science Advisor
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Just lazy mentors!
- #5
Chris Rorres
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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
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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?