Prove that det(kA)= k det(A) for any A M_nxn(F).

Is this proof correct?

det(AB) = det(A)det(B) so

det(kA) = det(kI[n]A) = det(kI[n])det(A) = k^n det(I[n]) det(A) = k^n (I) det(A) = k^n det(A)

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- November 2nd 2010, 07:17 PM #1

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- November 2nd 2010, 08:06 PM #2
Can you explain your proof further?

You could alternatively note that if that and thus

*Remark:*Really the operative thing here is that the determinant of a matrix over can really be thought of as a -linear form

- November 3rd 2010, 01:53 AM #3