5.21. Existence of the Determinant Function

Apostle treated Theorem 5.20 as an existence theorem for the determinant function.  However, this existence was already demonstrated by the expansion by row formulae (5.20) and (5.34).  Therefore, it seems more meaningful to treat Theorem 5.20 as a formula for an expansion by column.

 

Theorem 5.20.

Assume determinants of order n-1 exist.  Let f be a function of the n rows of an n ´ n matrix A=( a ij )  such that

f( A 1 ,, A n )= j=1 n ( ) j+1 a j1 det A j1          (5.37)

Then f satisfies all 3 axioms for a determinant function of order n.  Therefore, by induction, determinants of arbitrary order n exists.

Proof

By definition, the minor A j1  is a matrix of order n-1.  Hence, det A j1  satisfies all 3 axioms for a determinant function of order n-1.  Now,

        f( ,t A k , )= ( ) k+1 ( t a k1 )det A k1 + j( k )=1 n ( ) j+1 a j1 ( tdet A j1 )   =tf( A 1 ,, A n )

so that f satisfies axiom 1.  Next, consider the matrix B obtained from A by adding row i to row k.  Thus,

        B k = A i + A k   and  B j = A j   for all jk

so that

        b k1 = a i1 + a k1   and  b j1 = a j1   for all jk

Furthermore, since det B j1  are of order n-1, they obey axiom 2.  Denoting the rows with the 1st element truncated by a ~, we have

        det B j1 =f ( , A ˜ i i ,, A ˜ i + A ˜ k k , ) no   A ˜ j   =det A j1                for all jik

        det B k1 =f ( , A ˜ i i ,, ) no   A ˜ k =det A k1

        det B i1 =f ( , A ˜ i + A ˜ k k , ) no   A ˜ i   =f ( , A ˜ k k , ) no   A ˜ i    +f ( , A ˜ i k , ) no   A ˜ k

                        =det A i1 + ( ) ik det A k1

where the factor ( ) ik  comes from moving A ˜ i  from the kth to the ith row.  Hence,

        f( B 1 ,, B n )= j=1 n ( ) j+1 b j1 det B j1

        = ( ) k+1 b k1 det B k1 + ( ) i+1 b i1 det B i1 + j( ik )=1 n ( ) j+1 b j1 det B j1

        = ( ) k+1 ( a i1 + a k1 )det A k1 + ( ) i+1 a i1 ( det A i1 + ( ) ik det A k1 )

                + j( ik )=1 n ( ) j+1 a j1 det A j1

        = ( ) k+1 a k1 det A k1 + ( ) i+1 a i1 det A i1 + j( ik )=1 n ( ) j+1 a j1 det A j1

        =f( A 1 ,, A n )

so that f satisfies axiom 2.  Finally, setting A= I ( n )  gives

        f( I 1 ,, I n )= a 11 det A 11 =1×det I ( n1 ) =1

so that axiom 3 is also satisfied.  QED.

Comment

The foregoing proof is easily adapted to show that

        f( A 1 ,, A n )= j=1 n ( ) j+k a jk det A jk                  (5.38)

for any k=1,,n .  Furthermore (5.38) is a column expansion formula.  This is to be contrasted with the row expansion formula (5.34).

Theorem 5.21.

det A t =detA

Proof

The proof is by induction.  Skipping the trivial case of n=1 , we see that the case n=2  is easily verified using eq(5.1).  Next, we assume the theorem is true for case n-1 and try to prove case n.

To begin, let A=( a ij )  and B= A t =( b ij ) , where b ij = a ji .  Expanding detA and detB by the 1st column and row expansions, respectively, we have

        detA= j=1 n ( ) j+1 a j1 det A j1        

detB= j=1 n ( ) j+1 b 1j det B 1j   = j=1 n ( ) j+1 a j1 det B 1j

Since Aj1 and B1j are order n-1 matrices, we have

det B 1j =det A j1 t =det A j1

Þ            detA=detB                     QED.