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

[413]

why is det(kA)=k^ndet(A)?
[show that it is true for all n and all k]

 

[Hurkyl]

Well, what do you know about determinants? What thoughts have you already had on this problem?

 

[radou]

As Hurkly suggested, use the definition of the determinant, and the answer should arise easily.

 

[HallsofIvy]

In particular, how may "k"s are there in the matrix kA? What is the determinant of kI where I is the identity matrix?