Saturday, June 14, 2014

Linear Algebra: #13 Determinant

  • Linear Algebra: #13 Determinant

Let M(n × n, F) be the set of all n × n matrices of elements of the field F.

A mapping det : M(n × n, F) → F is called a determinant function if it satisfies the following three conditions.

  1. det(In) = 1, where In is the identity matrix. 

  2. If A ∈ M(n×n, F) is changed to the matrix A' by multiplying all the elements in a single row with the scalar a ∈ F, then det(A') = a · det(A). (This is our row operation Si(a).) 

  3. If A' is obtained from A by adding one row to a different row, then det(A') = det(A). (This is our row operation Sij(1).) 

Simple consequences of this definition
Let A ∈ M(n × n, F) be an arbitrary n × n matrix, and let us say that A is transformed into the new matrix A' by an elementary row operation. Then we have:
  • If A' is obtained by multiplying row i by the scalar a ∈ F, then det(A') = a · det(A). This is completely obvious! It is just part of the definition of “determinants”.

  • Therefore, if A' is obtained from A by multiplying a row with −1 then we have det(A' ) = −det(A).

  • Also, it follows that a matrix containing a row consisting of zeros must have zero as its determinant. 

  • If A has two identical rows, then its determinant must also be zero. For can we multiply one of these rows with −1, then add it to the other row, obtaining a matrix with a zero row. 

  • If A ′ is obtained by exchanging rows i and j, then det(A') = −det(A). This is a bit more difficult to see. Let us say that A = (u1, . . . , ui, . . . , uj, . . . , un), where uk is the k-th row of the matrix, for each k. Then we can write
    Linear Algebra: #13 Determinant equation pic 1
    (This is the elementary row operation Sij.)

  • If A' is obtained from A by an elementary row operation of the form Sij(c), then det(A') = det(A). For we have: 
    Linear Algebra: #13 Determinant equation pic 2

Therefore we see that each elementary row operation has a well-defined effect on the determinant of the matrix. This gives us the following algorithm for calculating the determinant of an arbitrary matrix in M(n × n, F).

How to find the determinant of a matrix
Given: An arbitrary matrix A ∈ M(n × n, F).
Find: det(A).

  1. Using elementary row operations, transform A into a matrix in step form, keeping track of the changes in the determinant at each stage. 

  2. If the bottom line of the matrix we obtain only consists of zeros, then the determinant is zero, and thus the determinant of the original matrix was zero. 

  3. Otherwise, the matrix has been transformed into an upper triangular matrix, all of whose diagonal elements are 1. But now we can transform this matrix into the identity matrix In( by elementary row operations of the type Sij(c). Since we know that det(In) must be 1, we then find a unique value for the determinant of the original matrix A. In particular, in this case det(A) ≠ 0.
Note that in both this algorithm, as well as in the algorithm for finding the inverse of a regular matrix, the method of Gaussian elimination was used. Thus we can combine both ideas into a single algorithm, suitable for practical calculations in a computer, which yields both the matrix inverse (if it exists), and the determinant. This algorithm also proves the following theorem.

Theorem 34
There is only one determinant function and it is uniquely given by our algorithm. Furthermore, a matrix A ∈ M(n × n, F) is regular if and only if det(A) ≠ 0.

In particular, using these methods it is easy to see that the following theorem is true.

Theorem 35
Let A, B ∈ M(n×n, F). Then we have det(A· B) = det(A) · det(B).

If either A or B is singular, then A · B is singular. This can be seen by thinking about the linear mappings V → V which A and B represent. At least one of these mappings is singular. Thus the dimension of the image is less than n, so the dimension of the image of the composition of the two mappings must also be less than n. Therefore A · B must be singular. That means, on the one hand, that det(A · B) = 0. And on the other hand, that either det(A) = 0 or else det(B) = 0. Either way, the theorem is true in this case.

If both A and B are regular, then they are both in GL(n, F). Therefore, as we have seen, they can be written as products of elementary matrices. It suffices then to prove that det(S1)det(S2) = det(S1S2), where S1 and S2 are elementary matrices. But our arguments above show that this is, indeed, true.

Remembering that A is regular if and only if A ∈ GL(n, F), we have:

If A ∈ GL(n, F) then det(A−1) = (det(A))−1

In particular, if det(A) = 1 then we also have det(A−1) = 1. The set of all such matrices must then form a group.

Another simple corollary is the following.

Assume that the matrix A is in block form, so that the linear mapping which it represents splits into a direct sum of invariant subspaces (see theorem 29). Then det(A) is the product of the determinants of the blocks.

Linear Algebra: #13 Determinant equation pic 3

That is, for the matrix Ai*, all the blocks except the i-th block are replaced with identity-matrix blocks. Then A = A1* · · · Ap*, and it is easy to see that det(Ai*) = det(Ai) for each i.

The special linear group of order n is defined to be the set
 SL(n, F) = {A ∈ GL(n, F) : det(A) = 1}. 

Theorem 36
Let A' = C−1AC. Then det(A ′ ) = det(A).

This follows, since det(C−1) = (det(C))−1.

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