1. ## det(kA)=k^n det(A)

Prove that det(kA)= k $^n$ det(A) for any A $\in$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)

2. Originally Posted by tn11631
Prove that det(kA)= k $^n$ det(A) for any A $\in$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)
Can you explain your proof further?

You could alternatively note that if $A=[a_{i,j}]$ that $kA=[ka_{i,j}]=[a'_{i,j}]$ and thus
\displaystyle \begin{aligned}\det\left(kA\right) &=\sum_{\pi\in S_n}\text{sgn}(\pi)\prod_{j=1}^{n}a'_{j,\pi(j)}\\ &=\sum_{\pi\in S_n}\text{sgn}(\pi)\prod_{j=1}^{n}\left(k a_{j,\pi(j)\right)\ &=k^n\sum_{\pi\in S_n}\text{sgn}(\pi)\prod_{j=1}^{n}a_{j,\pi(j)}\\ &= k^n\det\left(A\right)\end{aligned}

Remark: Really the operative thing here is that the determinant of a $n\times n$ matrix over $F$ can really be thought of as a $n$-linear form
$\det:F^n\to F$
Where we've interpreted
$\det \left[\begin{array}{c|c|c}& & &\\ v_1 & v_2 & v_3\\ & & &\end{array}\right]=\det\left(v_1,v_2,v_3\right)$
and thus if $\bold{f}_1,\cdots,\bold{f}_n\in F^n$ we see that the interpretation of your question is really to show that $\det(k\bold{f}_1,\cdots,k\bold{f}_n)=k^n\det(\bold {f}_1,\cdots,\bold{f}_n)$ but this is apparent by the $n$-linearity.

3. Originally Posted by tn11631
Prove that det(kA)= k $^n$ det(A) for any A $\in$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)
That proof is not only the right idea, but quite a nice idea too.

I would, however, point out that you know that $det(kI_n) = k^n$ immediately. I am not sure what you mean when you write ...k^n det(I[n]) det(A) = k^n (I) det(A)...

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# prove det(ka)=k^n det(a)

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