Proof of det(kA)=k^n det(A) for all n and k

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Homework Help Overview

The discussion revolves around the proof of the property of determinants, specifically that det(kA) = k^n det(A) for any scalar k and n-dimensional matrix A. Participants are exploring the underlying principles of determinants in linear algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are prompted to consider their existing knowledge about determinants and how they relate to the scalar multiplication of matrices. Questions about the definition of determinants and the specific structure of the matrix kA are raised.

Discussion Status

The conversation is ongoing, with participants offering guidance on using the definition of the determinant as a potential pathway to understanding the proof. There is an emphasis on exploring the implications of the scalar k within the context of the determinant.

Contextual Notes

Participants are encouraged to clarify their thoughts and assumptions about determinants, particularly regarding the identity matrix and the number of scalars involved in the matrix kA.

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why is det(kA)=k^ndet(A)?
[show that it is true for all n and all k]
 
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Well, what do you know about determinants? What thoughts have you already had on this problem?
 
As Hurkly suggested, use the definition of the determinant, and the answer should arise easily.
 
In particular, how may "k"s are there in the matrix kA? What is the determinant of kI where I is the identity matrix?
 

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