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Proving that determinants aren't linear transformations?

  1. Jul 12, 2017 #1
    1. The problem statement, all variables and given/known data
    "Determine whether the function ##T:M_{2×2}(ℝ)→ℝ## defined by ##T(A)=det(A)## is a linear transformation.

    2. Relevant equations
    ##det(A)=\sum_{i=1}^n a_{ij}C_{ij}##

    3. The attempt at a solution
    I'm assuming that it isn't a linear transformation because ##det(A+B)≠det(A)+det(B)##. So I have here:

    ##det(A)=\sum_{i=1}^n a_{ij}C_{ij}##
    ##det(B)=\sum_{i=1}^n b_{ij}C_{ij}##

    ##det(A)+det(B)=\sum_{i=1}^n (a_{ij}C_{ij}+b_{ij}C_{ij})##

    Meanwhile, I have...

    ##det(A+B)=\sum_{i=1}^n (a_ij+bij)(C_{ij}+D_{ij})##

    I'm fairly certain that the equation directly above is incorrect; perhaps that of the one above that one, too.
     
  2. jcsd
  3. Jul 12, 2017 #2

    fresh_42

    Staff: Mentor

    What do you get for simple examples like ##A=\lambda I## and ##B=-A## before you try to prove the general case?
     
  4. Jul 12, 2017 #3

    Math_QED

    User Avatar
    Homework Helper

    You are correct. It is not a linear transformation because ##det(A + B) \neq det(A) + det(B)## (a counterexample is easy to find, so if you do this this is sufficient to prove the statement, there is really no need to use these definitions). Moreover, ##det(\lambda A) = \lambda^n det(A)## if ##A## is a ##n \times n## matrix, which means ##det(\lambda A) \neq \lambda det(A)## in general.
     
  5. Jul 13, 2017 #4
    I see. In any case, I'm just going to do the problem like this and hope that my teacher accepts it:

    "Let ##A=[I_n]## and ##B## be an ##n×n## matrix such that ##ent_{ij}(B)=β≠0## iff ##i=j=1## and ##ent_{ij}(B)=0## otherwise. Then ##det(A)=1## and ##det(B)=0##. However, ##det(A+B)=1+β≠1=det(A)+det(B)##. Therefore, the determinant is not a linear transformation from ##M_{n×n}(ℝ)→ℝ##."
     
  6. Jul 13, 2017 #5

    Math_QED

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    Homework Helper

    Counterexamples are more elegant imo, and you are sure there is no flaw in the reasoning, but your approach is fine.
     
  7. Jul 13, 2017 #6
    Thanks...
     
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