Suppose you define as follows :

  • for $(a,b,c,d)\in \mathbb{R}^4$, $\det \begin{pmatrix} a & b \\ c & d\end{pmatrix} = ad-bc$.
  • for $A$ a square matrix of size $n$, you define $\det A$ recursively using the Laplace expansion with respect to the first row.

Clearly, the determinant of a matrix is uniquely defined (since you impose the expansion with respect to the first row).

The question is how do you prove with this definition the basic following properties :

  • $\det(AB)=\det A \det B$.
  • the Laplace expansion with respect to the others rows or the columns.
  • $\det ^tA = \det A$
  • or any other basic property that is not totally clear from this definition.
  • 1
    $\begingroup$ Why start there; just define $det(a)=a$ for a $1\times 1$ matrix for your base case. $\endgroup$ – vadim123 Feb 25 '14 at 18:20
  • $\begingroup$ Of course, there is no reason to start with $2\times 2$ matrices. You can also assume that with start with $1\times 1$ matrices. $\endgroup$ – user37238 Feb 25 '14 at 18:23
  • $\begingroup$ Prove that your definition gives a determinant in classic sense (i.e. defined with the help of permutations). Should be doable (but tiresome) by recurrence. Then you can immediately obtain all the standard properties. $\endgroup$ – TZakrevskiy Feb 25 '14 at 18:29
  • $\begingroup$ @TZakrevskiy Of course, if it is doable, it is a valid proof. But I'm more looking for a "direct" proof from the definition if you understand what I mean. $\endgroup$ – user37238 Feb 25 '14 at 18:37
  • $\begingroup$ Induction on the size of the matrix $\endgroup$ – TBrendle Feb 25 '14 at 18:50

First prove that this definition gives a multilinear alternating $n$-form of unit norm on the columns, that is

  • Multilinear (linear in each component): \begin{align} \det (v_1,\dots,v_{i-1},\lambda v_i,v_{i+1},\dots,v_n) &= \lambda\det(v_1,\dots,v_n), \\ \det (v_1,\dots,v_{i-1},v_i+v_i',v_{i+1},\dots,v_n) &= \det(v_1,\dots,v_n) + \det(v_1,\dots,v_{i-1},v_i',v_{i+1},\dots,v_n). \end{align}
  • Alternating: $$\det(v_1,\dots,v_n) = 0\quad\text{if $v_i=v_j$ for some $i\neq j$.}$$
  • Unit norm: $$\det(e_1,\dots,e_n) = 1.$$

If that is proven, you know your definition is equivalent to the axiomatic approach, from which all the properties follow. The subsequent proofs can be found in most textbooks on linear algebra.

I don't see any point in struggling with direct proofs from the unhandy definition you gave.

  • 1
    $\begingroup$ This definition may appear unhandy as you say but the goal is to present the determinant as a tool for people that are not doing math and the basic properties of the determinant (without proof). But one may wonder how we can actually prove all of this properties and the answer "well I gave you the very bad definition, you just have to go from this definition to the real one and then it is easy" isn't satisfying. $\endgroup$ – user37238 Feb 25 '14 at 18:50
  • $\begingroup$ Don't teach the determinant with a single definition. Give multiple definitions, prove or just sketch their equivalence and pick the definition best suited to prove each individual property. The definition as "oriented volume" of a parallelotope can come in very handy! $\endgroup$ – Christoph Feb 25 '14 at 18:53
  • $\begingroup$ I don't have the time nor the authority to do such a thing sadly. $\endgroup$ – user37238 Feb 25 '14 at 19:30
  • $\begingroup$ Then you also don't have time to prove $\det(AB)=\det(A)\det(B)$ from your definition. You can always show some examples though. $\endgroup$ – Christoph Feb 25 '14 at 19:31
  • $\begingroup$ Indeed, we're not going to prove this. I'm just wondering if it is possible from the definition without speaking about all the different definitions. $\endgroup$ – user37238 Feb 25 '14 at 19:35

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