Results 1 to 3 of 3

Math Help - det(kA)=k^n det(A)

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    146

    det(kA)=k^n det(A)

    Prove that det(kA)= k ^n det(A) for any A \inM_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)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by tn11631 View Post
    Prove that det(kA)= k ^n det(A) for any A \inM_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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by tn11631 View Post
    Prove that det(kA)= k ^n det(A) for any A \inM_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)...
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum