Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove that the determinant function satisfying the following properties is unique:

$\det(I)=1$ where $I$ is identity matrix, the function $\det(A)$ is linear in the rows of the matrix and if two adjacent rows of a matrix $A$ are equal, then $\det A=0$. This is how Artin has stated the properties.I find Artin's first chapter rough going and would appreciate some help on this one.

share|improve this question
1  
It would help if you told us exactly what you're having trouble with. Wikipedia has a proof of the result, can you follow that? –  EuYu Oct 12 '12 at 5:37

3 Answers 3

up vote 4 down vote accepted

An alternative way is the Gaussian elimination: for a given $n\times n$ matrix $A$ with rows $r_1,..,r_n$, the following steps are allowed to use, in order to arrive to the identity matrix or one with a zero row (by the linearity, if $A$ has a zero row, the 'Artinian determinant' has to be zero).

  1. Add a scalar multiple of a row $r_j$ to another row $r_i$, i.e.: $i\ne j$ and $$r_i':= r_i+\lambda r_j$$
  2. Multiply a row by a nonzero scalar, i.e.: $\lambda\ne 0$ and $$r_i':=\lambda\cdot r_i$$
  3. Exchange 2 rows (can also be obtained by 1. and 2.)

Let's assume, we have two 'Artinian determinants': $D$ and $D'$. Using the above mentioned fact that every matrix can be transformed to the indentity or with a zero row, we will have $D=D'$, because 1. keeps both $D$ and $D'$ (why?), 2. multiplies both $D$ and $D'$ by $\lambda$, and 3. by $-1$.

share|improve this answer

I have thought about your problem for a while now and I think there is a nice slick way to do this. Consider the space $W$ of all multilinear alternating forms $f$ in $k$ - variables

$$f : V \times \ldots \times V \to \Bbb{C}.$$

We claim that there is a canonical isomorphism between $W$ and $(\bigwedge^k V)^\ast$. Indeed, this should be clear because given any $f \in W$, the universal property of the $k$ - th exterior power tells us that there is a unique linear map $g \in (\bigwedge^k V)^\ast$ such that $f = g \circ \iota$ where $\iota : V \times \ldots \times V \longrightarrow \bigwedge^k V$ is the canonical mapping that sends the tuple $(v_1,\ldots,v_k)$ to $v_1 \wedge \ldots \wedge v_k$. Conversely given any $h \in (\bigwedge^k V)^\ast$ we can precompose it with $\iota$ to give us a mapping from $V \times \ldots \times V \to \Bbb{C}$.

In summary, we can use these facts to give us a canonical isomorphism between $W$ and $(\bigwedge^k V)^\ast$. If we put $k = n$, where $n = \dim V$ then

$$ 1= \dim_{\Bbb{C}} \bigwedge\nolimits^{\!k}V = \dim_{\Bbb{C}} \left(\bigwedge\nolimits^{\!k} V\right)^\ast $$

from which it follows that $W$ is one dimensional. In other words, any $f \in W$ is a scalar multiple of $\det$, where

$$\det : V \times V\times \ldots \times V \longrightarrow \Bbb{C}$$

is the mapping that sends the tuple $(v_1,\ldots, v_n)$ to the determinant of the matrix whose columns are the vectors $v_1, v_2, \ldots, v_n$. Now here comes the killer blow: Suppose we demand that an alternating multilinear $f$ be such that $f(e_1,\ldots,e_n) = 1$ where the $e_i$ are the standard basis vectors of $\Bbb{C}^n$. Then because

$$f(v_1,\ldots,v_n) = c\cdot \det(v_1,\ldots,v_n)$$

for some constant $c$, shoving in $(v_1,\ldots,v_n) = (e_1,\ldots,e_n)$ we must have that

$$\begin{eqnarray*} 1 &=& f(e_1,\ldots, e_n) \\ &=& c\cdot \det(e_1,\ldots,e_n) \\ &=& c \end{eqnarray*}$$

because the determinant of the identity matrix is $1$. Consequently we have shown:

Any alternating multilinear form in $\dim V$ number of variables with the value of the form on the tuple $(e_1,\ldots,e_n)$ being $1$ must be equal to the determinant.

$$\hspace{6in} \square$$

share|improve this answer
    
1 sec.Michael Artin has barely introduced linear algebra, forget about vector spaces(though I know what it is) and multinearity is not something I have seen till date.I am sorry but I do not understand anything you wrote. To be precise, can we not skip the machinery you used? I get a feeling you are correct as I can see your education in your profile.Does that mean Artin was being sloppy here? –  Richard Nash Oct 12 '12 at 9:09
1  
@RichardNash Hi, I did not realise that you were only starting artin's book. However, I believe in order to understand the fact that Artin stated either you will have to go through the barrage of sums in the wikipedia link, or there is a nice way to look at the determinant in terms of maps between a kind of vector space called the exterior power. If you don't know what it is, it is ok but I am afraid (even going through Marc's answer below) that it would be vary hard to know exactly why we have the determinant being the multilinear map with those properties. –  user38268 Oct 12 '12 at 9:18
    
Thanks mate. Is there an abstract algebra book you recommend which does not make sloppy statements which a reader cannot logically prove from the knowledge developed earlier? I was looking at Bourbaki's algebra chapters 1-3 but I do not know what background is necessary to understand it. –  Richard Nash Oct 12 '12 at 9:24
1  
It's neat that we get functoriality for free :) –  Alexei Averchenko Nov 10 '12 at 13:47
1  
@BenjaLim I forgot :D I guess I meant $\det(AB) = \det(A)\det(B)$. Happy new year, anyway! –  Alexei Averchenko Jan 1 '13 at 8:50

A basic fact about linear functions is that they are completely determined by their values on a basis of the vector space. For a multi-linear function this means (repeating this statement for each argument) that they are determined by their values where each argument independently runs through a basis of the vector space. For a function of a matrix that is linear in the rows, it means that the function is determined by the values it takes for matrices for which each row has a single entry $1$ and all other entries $0$. Concretely if such a function is written $f(v_1,\ldots,v_n)$, the arguments being rows of a matrix $A$, then by multi-linearity $$ f(A)=\sum_{j_1,j_2,\ldots,j_n=1}^n a_{1,j_1}a_{j_2,2}\ldots a_{n,j_n} \, f(e_{j_1},e_{j_2},\ldots,e_{j_n}), $$ where $e_k$ is the $k$-th standard basis vector viewed as a row.

Now we must take into account that $f$ vanishes whenever two adjacent rows are equal. This implies directly that in the above summation one can drop any terms for which $j_i=j_{i+1}$ for some $i$. But also, by a standard "polarisation" argument (namely that $g(x+y,x+y)=g(x,x)+g(x,y)+g(y,x)+g(y+y)$ for bilinear $g$, so $g(x,y)=-g(y,x)$ if in addition $g$ vanishes on equal arguments), $f$ changes sign whenever we interchange two adjacent rows. So if $j_i>j_{i+1}$ for some $i$, then we have $$ f(e_{j_1},e_{j_2},\ldots,e_{j_n}) =-f(e_{j_1},e_{j_2},\ldots,e_{j_{i+1}},e_{j_i},\ldots,e_{j_n}), $$ and the sequence of indices $j_1,j_2,\ldots,j_{i-1},j_{i+1},j_i,j_{i+2},\ldots,j_n$ on the right, in which $j_i$ and $j_{i+1}$ have been interchanged, has one less inversion than the sequence on the left (an inversion of a sequence being a pair of positions where the term in the left position is strictly larger than the one in the right position). (You may notice I am re-doing a proof that any permutation is a composition of adjacent transpositions; one could also use that fact to show that any permutation of the arguments of $f$ affects the value by the sign of that permutation.)

Now for any sequence $(j_1,j_2,\ldots,j_n)$ other than $(1,2,\ldots,n)$, we either find that $f(e_{j_1},e_{j_2},\ldots,e_{j_n})$ is zero, or that it is determined by a similar value of $f$ but at a sequence of indices with strictly less inversions. It follows (by induction on the number of inversions) that all such terms are determined by $f(e_1,\ldots,e_n)$ alone. Finally it was given that $f(e_1,\ldots,e_n)=1$, so $f$ is completely determined.

As a bonus, this argument gives the explicit Leibniz formula for the determinant, once you check that $f(e_{\pi_1},e_{\pi_2},\ldots,e_{\pi_n})=\operatorname{sg}(\pi)$ for any permutation $\pi$ and that $f(e_{j_1},e_{j_2},\ldots,e_{j_n})=0$ for any non-permutation $(j_1,j_2,\ldots,j_n)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.