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I know the definition of Determinant function that it is a mapping $D: \mathbb{K}^{n \times n} \rightarrow \mathbb{K}$ such that (i) $D$ is n-linear (ii) $D(A) = 0$, if two rows are equal (iii) $D(I) = 1$ for the identity matrix I. where $\mathbb{K}$ is a commutative ring and $\mathbb{K}^{n \times n}$ is the set of all $n \times n$ matrices.

Can anyone tell me by using the above definition of determinant function how to prove $det(AB) = det(A) det(B)$ and $det(A) = det(A^T)$?

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  • $\begingroup$ I greatly doubt you can prove this important theorem only using the general properties of it you mention and without going through way more stuff... $\endgroup$ – DonAntonio Sep 19 '16 at 15:37
  • $\begingroup$ For this I have a definition of alternating n-linear function also, I tried to prove this with the help of this definition also but couldn't succeed. $\endgroup$ – Amanda Sep 19 '16 at 15:48
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    $\begingroup$ @Amanda: Too many "this" in your comment. Do you want to prove that $D$ is an alternating function from your definition? It's easy, just expand $D(\ldots,a+b,\ldots,a+b,\ldots)$. $\endgroup$ – Canis Lupus Sep 19 '16 at 16:00
  • $\begingroup$ I have defined $D(A) = D(A_1 A_2 .... A_n) = det(AB)$ and also proved that $D(A_1 A_2 .... A_n)$ is an alternating n-linear function. Whether it is correct to define $D(A) = det(AB)$? $\endgroup$ – Amanda Sep 19 '16 at 16:10
  • $\begingroup$ Now I understood. Thanks to all for the help :) $\endgroup$ – Amanda Sep 19 '16 at 16:11
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Here is a sketch: fix $A$ and define $D(B) = detAB$. It is easy to show that $D$ preserves addition and scalar multiplication in the columns of $B$, and that D is alternating in the columns of $B$. Therefore, by uniqueness of the alternating, multilinear function, we must have $D(B)=D(I)\ detB=\ detAI\ detB=\ detA\ detB$.

Now use the $QR$ factorization of $A^T$ and what we have just proved, to show the second claim.

Yet another way would be to use your definition to show that the $det$ function $must$ have the form

$\displaystyle \sum_\lambda \operatorname{sgn} \left({\lambda}\right) b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)}$ where the sum is taken over all permutations $\lambda:\left \{ 1,\cdots, n \right \}\to \left \{ 1,\cdots, n \right \}$

and from here use the arguments you are probably used to.

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