. Hence
the columns (rows) are dependent; only in the case when every Aki
is zero, this conclusion fails. In order to obtain a general
proof, one has to go somewhat deeper into the matter. Consider:1.9 Minors of a determinant:
Again, let A be a matrix with n rows and n columns.
It has been proved in 1-8,
Proposition 3 and Theorem 2 that, if the columns of A depend on each other and
therefore do also the rows, then det A= 0. It is important
to know that the converse also is true, i.e., that, if a
determinant is zero, the columns (and rows) are interdependent.
Of course, if det A = 0, then 1.8 (11) shows that . Hence
the columns (rows) are dependent; only in the case when every Aki
is zero, this conclusion fails. In order to obtain a general
proof, one has to go somewhat deeper into the matter. Consider:
where a11 = 1, since
is an even or odd permutation according to
.
Hence
where + has to be taken for the even and - for the odd permutation of 2, .. ., n, respectively. Hence A11 is the determinant of the matrix which is generated by striking out its first row and first column. Similarly, the determinant Aki is generated by replacing aki by 1 and setting equal to zero all other elements in the k-th row, and in the i-th column. The k-th row may be interchanged by simple transposition with the (k— 1)-st, then with the (k — 2)-nd, etc. Thus, by k - 1 transpositions, the k-thTh row is shifted to the first place without altering the relative order of the remaining rows. By this operation, Aki receives the factor (— 1)k-1. Subsequently, the i-thTh column is moved to the first place and Aki is therefore replaced by (-1)i+k. Then the first element of the first row is 1, whereas the other elements of the first row and of the first column are 0. As it has been shown above, these two lines can be omitted without altering the determinant. Hence:
Proposition 1: (-1)i+kAki is equal to the determinant which is generated when the row and column, intersecting at aki are both omitted.
Definition: Let B be a matrix with n rows and n columns. If you omit m - r rows and n — r columns, the determinant of the remaining square matrix multiplied by e = ±l is called a minor of B of order r.
For example, let A be a square matrix; its
minors of highest possible order are det A and - det A. It
follows from Proposition 1 that the cofactors of the elements of
a square matrix are minors. By permutations of the rows and
columns, a minor is transformed into a minor as this permutation
means involves only multiplication by ±1. Let k1,
··· , kr be r < n
different numbers. If 
are replaced in det (a1,
··· , an)) by 
, you obtain a
determinant
r applications of Proposition 1 yield
Proposition 2: The determinant (1) is a minor of A; it is generated by omission in A of the rows k1, ··· , kr and the columns i1, ··· ,ir; the determinant of the remaining matrix multiplied by e = ±1is equal to the determinant (1). e is equal to +1 or— I according to whether i1+ ··· + ir + k1 + ··· + kr is even or odd.
As det A is a linear, homogeneous function of minors of order n - 1 [cf. 1.8, Proposition 9), det A= 0 if every minor of order n - 1 vanishes. Similarly, if every minor of order n - 2 vanishes, the same applies to every minor of order n-l and therefore to det A. By repeated application of these arguments, you arrive at
Proposition 3: If every minor of order n equals zero, then the same holds for the minors of higher order.
The Rank of a matrix is equal to zero if and only if every element is equal to zero, i.e. if every minor of order I, and therefore every minor is equal to zero. Investigate now the link between the Rank of a matrix and the maximum order of non-vanishing minors.
Proposition 4: Let B be a matrix with m £ n rows and n columns. If a minor of B of order m is different from zero, Rank B = m.
Proof: Let the minor composed of the m column vectors a, ..., r be different from zero, then these m-vectors are independent [cf. 1.8 Proposition 3]. Hence Rank B = Rank C(B) = m.
If m > n and a minor of B of order n is different from zero, the matrix can be transformed by interchange of rows and columns to the case considered in Proposition 4. Hence Rank B = n.
Theorem: If B has a minor of order r which differs from zero, but every minor of higher order (if any) is equal to zero, then Rank B = r.
Proof: Since the Rank of a matrix is not altered by permutations of rows and columns, assume without loss of generality that the determinant formed by the rows 1, ··· , r and the columns 1, ··· , r is different from zero. Thus, the matrix formed by the rows 1, ...,r
is of Rank r, and the row vectors cc are therefore independent. Consider the matrix formed by a1, ··· , ar, av, r < v £ n, and for any particular v, consider those minors of order r+1 which contain the columns 1, ··· , r:
Each of these minors is equal to zero, and the cofactors, say
have the same values for every u; they are minors cut out of the columns 1,···,r. In particular,
Hence
for k = 1, ··· , r, ··· ,n. Hence
Since A' ¹ 0, the n-vector av depends on a1, ···,a r. This is true for v=r+l,···,n. Hence a1, ···,a r form a bass of the vector space R(B) generated by the rows of B, that is, Rank B = r.
If therefore A has n rows and n columns and det A = 0, then Rank A < n, i.e., its rows (columns) are interdependent.
Generalized
cofactors* The formula 
expresses the determinant is as the scalar product of the n-vector
(ai1,
···,ain) and the n-vector (Ai1,
···,Ain); thus, det A is represented as a function which is
linear in two different sets of variables (bilinear function), one set comprising minors of order 1, the other set
minors of order n - 1. This representation can be
generalized to a representation as a bilinear function by one set
of minors of order m and one set of minors of order n
— m. For this purpose, the indices 1,···,n
are subdivided into two parts 1, ··· , m and t, ··· ,n,
where 1 < m, and m+1 = t < n.
Every term of det A = S ± aij···anp can be represented as the product of two terms
where e + d is even or odd depending on whether r, ··· , s, r', ··· , s' is an even or an odd permutation of 1, ··· , n. Every term generates a partition of the lower indices into two classes, one class r, ··· , s of m elements and another class r', ···, s' of n—m elements.
* This section may be omitted at a first reading.
There exist 
such partitions; each
partition corresponds to m!(n-m)! terms of the
determinant, since the elements of the first class admit m!
permutations. Naturally, m!(n-m)! is of
the number of the terms of the determinant. Firstly, consider the
partition, where
Set e = 0 or = 1 if the permutation of r, ··· , s is even or odd, respectively. Since e +d is even or odd when r, ··· , s, r', ··· , s' is an even or an odd permutation, d is even or odd when r', ··· , s' is an even or odd permutation. The sum of these terms equals
An arbitrary even permutation of the subscripts of the product on the left hand side of (1) transforms this term into another term of the determinant without changing the sign + or — . Now let
be an even permutation of the superscripts 1, ··· , n, when the terms
which one obtains by multiplication of the two
sums on the left hand side of (3) the terms of the determinant
with the correct signs + or -. If by the permutation 
of the superscripts dashed
indices are exchanged with those without dashes, none of these
terms will be a term in (3). Hence one obtains n! terms
of the determinant with correct signs one and only once in the
following manner: Perform all the different partitions of
the indices 1, ··· , n into m indices
without dashes and n - m indices with dashes;
arrange the dashed and undashed indices in such a manner that (4)
is an even permutation and execute the product (3). This product
contains the m!(n - m)! terms of the
determinant corresponding to the partition. The sum of all these
products equals the determinant, whence
where the sum must be taken as explained above. This expansion of a determinant is named after Laplace.
Since an even permutation of rows does not alter a determinant, the superscripts 1, ··· , m, t, ··· , n (where t = m + 1) can be replaced by any particular permutation of these terms. This permutation must be the same for all the terms of (5). If the permutation is odd, the sum (5) is equal to - det A, if not all superscripts differ, the sum equals 0.
1.10 Solution of systems of linear equations by means of determinants: Consider the m linear equations in n variables
Multiply the equations by the cofactors of the k-t
column and add them; then 
, where in the determinant on the right
hand side a0 occurs at the k-th place.
If det A ¹ 0,
This condition is necessary, but as there exists a solution and x1, ··· , xn are uniquely determined by (2), the condition is also sufficient. Verify this result by substituting (2) into (1):
As this result holds for j = I, ..., m., Equations (1) are satisfied by (2). For the case when the number of the equations differs from the number of unknowns, the method demands a modification. If the Rank of the full matrix is greater than the Rank of the homogeneous portion, the system has no solution. Let the two Ranks be equal, and omit the equations depending on the others. The Rank of the matrices is therefore supposed to be equal to the number m of the equations
There exists therefore a determinant formed by m columns of the coefficients on the left hand side which is different from zero. Without loss of generality, assume that it is the determinant
Multiplying the equations by the cofactors of the k-th column (k = 1, ··· , m) and dividing by d, you obtain
where
In the determinant on the right hand side, a0 and am are assumed to be put in the k-th place.
l.10.l. Comparison of different methods for solving systems of linear equations: Three methods for the solution of linear equations have been discussed : Sweep out, orthogonalization and determinants; moreover, the method of substitution has been referred to in the introduction. Common features of these methods are:
(1) A given system of linear equations, say
1.10, (1) or (3) is replaced by another one which either yields
the solution, if there exist only one, [cf. 1.10, (2)] or shows a
method of finding any number of solutions [cf. 1-10, (4)], if
there exist more than one solution
(2) The derived linear equations are homogeneous linear
combinations of the initial equations such that, if the original
equations are satisfied, the derived equations apply, i.e., the
derived equations are necessary
conditions.
(3) The initial linear equations are homogeneous linear
combinations of the derived ones. Hence the derived equations are
sufficient conditions. By the method of sweep out as well
as of substitution, this reduction of the given linear equations
occurs step by step. Consider the method of substitution.
'The symbols on the left hand side are only abbreviations for the linear functions in the centre section.
If a1 ¹ 0, x1 = [ a0 - a2x2, ··· , - an xn]/a1, substituting this value into the other equations, we get equations of the type
This equation is identical to b(x) - b1 a(x)/a1 = 0. Thus this substitution is the same as the sweep out of the first column. With the aid of one of these new equations, x2 is represented as a function of x3, ··· , xn; this value is substituted into the remaining n - 2 equations, so that the second column is swept out in n-2 rows. Substitution leads after n — 1 steps to a system of the type (provided the Rank of the matrix of the homogeneous portion is equal to n):
The matrix is swept out below the diagonal. By setting xn = t0/tn in the other equations, the n-th column is swept out. The (n - 1_-th equation is transformed into s xn-1 - s0 = 0; setting xn-1 = s/s into the other equation, the (n- 1)-th column is swept out, etc. Finally, the matrix is swept out completely and the values of x1, ··· , xn are determined.
The method of substitution does not differ essentially from the method of sweep out. The method of determinants does not use a step by step procedure. One determines numbers A, B,..., K such that
is independent of x2, ··· , xn and is therefore of the type
This condition is necessary, hut it may not he sufficient. In the preceding sections, it has been shown that one obtains by the method of determinants necessary and sufficient conditions for the unknown quantities x1, ··· , xn in a suitable form, provided the fundamental condition for the Ranks of matrices holds [cf. 1.6, Theorem 2]. For numerical calculations, is at times useful to combine the methods of elimination and of sweep out. When one proceeds in this way, it is advisable to confirm very carefully whether the necessary conditions stated in this manner are also sufficient.
In general, the methods of determinants and orthogonalization are not very suitable for numerical calculations. In order to evaluate a determinant, it is in general not advisable to determine the n! terms of 1.8, (7), but to simplify the determinant by sweeping out the matrix below or above the diagonal. The value of the determinant is then equal to the product of diagonal elements. A determinant can be swept out by row-addition as well as by column-addition.. In certain cases, it is useful to calculate a determinant with the help of 1.8, (5).
l-11 Linear transformations: In the preceding section, the n-vector
has been an unknown quantity, whereas the coefficients were assumed to be given. For many applications of the theory (e.g., applications to Geometry), it is necessary to investigate the link between the numbers, the n-vectors and matrices in these formulae.
Consider
there corresponds to every n-vector x an n-vector
This correspondence will be denoted by an arrow:
This formula may now be read: A transforms x into h. Formula (2) is called a linear transformation. It follows from (2) (using the same notations as previously) that
If
then
Hence, if x assumes all the n -vectors of a vector-space V, the corresponding vectors form a vector space V', as it follows from x ® h, x '® h', that x + x' ® h + h' and cx ® ch [cf. 1.3, Proposition 10]. A system of dependent n-vectors is transformed into a system of dependent n-vectors, but the converse may not be true.
Let V be a complete vector space of Rank n, then x = (x1, ··· , xn ) = x1e1+···+xnen , whence x ® x1a1+···+xnan. Every vector of V' can be represented in this manner and x1, ··· , xn assume independently values; therefore V' is generated by a1, ··· , an and Rank V" = Rank A. This yields:
Theorem 1: A linear transformation (2) transforms a vector space into a vector space and a vector space of Rank n is transformed into a vector space of Rank equal to Rank A.
The notion of linear transformation can be fully characterized by the manner in which sums of vectors as well as products of numbers and vectors are transformed. This fact is discussed
Theorem 2: If n-vectors (1) are represented by n-vectors (3) in such a manner that there corresponds to the sum of two vectors the sum of the corresponding vectors and to the product of a number c and an n-vector the product of c and the corresponding n-vector, then the representation is realized by a linear transformation.
Proof: Let 
be the n-vectors which represent the unit
vectors ek (k = 1, ··· , n).
Then xkek
is represented by xkak
and x = (x1,···,xn)=Sxkek
is represented by Sxkak,
i.e., by the n-vector h, determined
by (2) and (3). Hence follows the theorem.
1.11.1 Composition of transformations. Product of matrices: Let x be transformed into h by 1.11 (2) and h be transformed into z by another linear transformation
Denote the matrix ((bkj)) by B. Then
Thus x is transformed into z by a linear transformation which is said to consist of the transformations x ® h, and h ® z. The matrix ((gks) = G is said to be the product
G = BA.
In order to obtain the elements gk of G, one must multiply the elements of the k-th row of B by the corresponding elements of the s-th column of A and add the products. In terms of the scalar product [cf. 1.7]
where bk denotes the k-th row vector of B.
In general, AB and BA are different matrices, i.e., the commutative law does not apply to matrix multiplication.
Let
whence P = Q, that is,
(CB)A = C(BA). (3)
This formula is expressed by the
Theorem: The associative law applies to multiplication of matrices.
l.11.11. n-vectors considered as matrices: It is often useful to consider an n-vector
as a matrix, e.g., as
(1)
where the first column is formed by the coordinates of x, whereas the other elements are equal to zero. If we multiply such a matrix from the left by an arbitrary matrix with n rows and columns, the result is again a matrix of Type (1). The linear transformation 1.11 (2) can therefore be expressed by a matrix equation
A(x) = (y).
Similarly, 1.11.1 (1) is equivalent to
B(y) = (z),
whence
BA(x) = z.
which expresses 1.11.1 (2) in matrix form.
1.11.12 Special matrices: A diagonal matrix is a square matrix, all elements of which are equal to zero except for those in the diagonal:
An elementary matrix is the square matrix
where
Multiplication of A from the left by Ers(l) is row addition in A, by which the row ar of A is replaced by ar +\ las. Multiplication of A from the right by Ers(l) is column-addition in A, by which the column as of A is replaced by as+\ lar.
1.11.2 Decomposition of matrices: It has been shown by sweep out that every matrix can be transformed into a matrix of a special type by row addition, row deletion and row multiplication [cf. 1.4 Theorem 1]. In a similar manner, it will now be shown that a square matrix can be transformed into a diagonal matrix by row and column additions. In terms of matrix-multiplication, this proposition is stated by the
Theorem: Every square matrix A can be represented as a product
A = P1DP2, (1)
where P1 and P2 are products of elementary matrices and D is a diagonal matrix.
Proof: If the matrix is the 0-matrix, it is already a diagonal matrix, otherwise (if necessary) one can arrange by column addition that at least one element in the first column differs from 0 and by row addition that a11 ¹ 0. Thus, one can sweep out the first column by row addition and subsequently the first row by column addition without altering the first column. If the matrix is not already diagonal, one can arrange now by row and column addition without altering the first row and the first column that a22 ¹ 0; then the second column and the second row are swept out. This procedure can be continued until the matrix is the diagonal matrix D. As every row addition implies multiplication by an elementary matrix from the left and, similarly, a column addition corresponds to an elementary matrix as a right hand side factor, Equation (1) applies.
A representation of A as a product of diagonal and elementary matrices is also referred to as decomposition of A.
1.11.3. Determinant of a matrix product: Let D be a diagonal matrix so that
As a multiplication by an elementary matrix from the left (right) signifies only a row (column) addition, multiplication by elementary matrices or by products of them does not alter the determinant. Multiplication of any matrix by D means that the determinant is multiplied by det D (cf. 1.11.12 (1) and (2)). Hence, if
A = P1DP2,
then
det A = d1 ··· dn.
Consider AB.
det AB = det DP2B,
det P2B = det B,
det AB = det D(P2B) = d1 ··· dn
detP2B = det A P2B = det A det B,
whence
det AB = det A det B. (2)
1.11.4 Inverse of a linear transformation:
The transformation
A(x) = (y) (1)
transforms the n-vector (x)
into the vectors (y). If det A = 0, then the Rank of the vector
space generated by the n-vector (y) is smaller
than n, whence the (x) is not generated by a
linear transformation of the n-vector (y).
Let det A ¹ 0. Then 
, whence
A'(y) = (x),
where
and bik = Aki/det A.
Hence AA' transforms every n-vector into itself. Let C be any matrix. C (ei) is an n-vector which is equal to the i-th column of C, hence, if C(ei) = (ei), then C is the diagonal matrix
Hence
A A' = E.
Moreover,
A' A(x)= (x),
whence
A' A = E.
The matrix A' is said to be the inverse matrix of A and is mostly denoted by A-1. The necessary and sufficient condition for the existence of an inverse matrix is det A ¹ 0.
E is called the unit matrix. For every matrix A,
AE = A = EA (4)
and E = E-1.
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