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Proof for determinant of a scalar multiplied by a vector

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Let A be an n x n matrix and [itex]\alpha[/itex] a scalar. Show that [itex]det(\alpha A) = \alpha^{n}det(A)[/itex]

    2. Relevant equations

    [itex] det(A) = a_{11}A_{11} + a_{12}A_{12} + \cdots + a_{1n}A_{1n} [/itex]

    where [itex] A_{ij} = (-1)^{i+j}det(M_{ij}) [/itex]

    3. The attempt at a solution

    [itex] det(A) = a_{11}A_{11} + a_{12}A_{12} + \cdots + a_{1n}A_{1n} [/itex]

    [itex] det(\alpha A) = \alpha a_{11}A^{\alpha}_{11} + \alpha a_{12}A^{\alpha}_{12} + \cdots + \alpha a_{1n}A^{\alpha}_{1n} [/itex]

    [itex] det(\alpha A) = \alpha (a_{11}A^{\alpha}_{11} + a_{12}A^{\alpha}_{12} + \cdots + a_{1n}A^{\alpha}_{1n}) [/itex]

    I can see that as I go through and calculate the cofactors I will continue to get an additional alpha coefficient each time, so that I will end up with [itex]det(\alpha A) = \alpha^{n}det(A)[/itex], but I am having trouble formalizing it. Thank you in advance for any help.
     
  2. jcsd
  3. Sep 20, 2011 #2
    I haven't tried it, but induction on [itex]n[/itex] might work. Alternatively, I think it might be easier to think of the determinant as a multilinear function of the columns of the matrix rather than using cofactor expansion.
     
  4. Sep 20, 2011 #3

    lanedance

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    Homework Helper

    or do you know the expression for determinant using the e-permutation symbol, see:
    http://www.math.odu.edu/~jhh/part2.PDF
    example 1.1-9
    should follow straight form there

    However, it should follow straight from your work though, note that if [itex]C_{ij} [/itex] is a cofactor of [itex]A [/itex], then [itex]\alpha C_{ij} [/itex] is a cofactor of [itex]\alpha A_{ij} [/itex]
    [tex] C_{ij} = (-1)^{i+j}M_{ij} [/tex]
     
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