3 Fields, Elementary Matrices and Calculating Inverses

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1 3 Fields, Elementary Matrices and Calculating Inverses 3. Fields So far we have worked with matrices whose entries are real numbers (and systems of equations whose coefficients and solutions are real numbers). If for a moment we call the set of real numbers K (it will become clear shortly why we re not just calling it R!), here are a few basic properties of these operations: (F) for all a,b,c K, (a+b)+c a+(b+c); (F) for all a,b K, a+b b+a; (F3) there is an element of K (which we call 0 ) with the property that a+0 0+a a for all a K; (F4) for every a K there exists an element a K such that a+( a) 0; (F5) for all a,b,c K, (ab)c a(bc); (F6) for all a,b K, ab ba; (F7) there is an element of K (which we call and which is different from 0) with the property that a a a for all a K; (F8) for every a K except 0, there exists an element a K such that aa a a ; (F9) for all a,b,c K, a(b+c) ab+ac. These nine properties are called the field axioms. Check for yourself that they are true for the real numbers. Of course lots of other things are true for the real numbers, so what s so special about these ones? It turns out these are exactly the properties which we need to make linear algebra work. This means we can replace the real numbers with any other type of scalar having these properties and things will still work in the same way! For example, the complex numbers (check!) also satisfy the field axioms, so we can do linear algebra where the scalars are complex numbers. Definition. Let K be a set equipped with operations of addition and multiplication 9. Then K is a called field if it satisfies the nine field axioms. Examples. The real numbers satisfy the axioms (that was the whole point!) so R is a field. We have already mentioned that the complex numbers satisfy the axioms, so C is a field as well. Exercise Sheet 3 gives some more examples. Non-example. The set of matrices (with real number entries) has operations of addition and multiplication, but is not a field. We saw on Exercise Sheet that it doesn t satisfy (F6). In fact it satisfies seven of the nine axioms: on Exercise Sheet 3 you can find the other one that it doesn t satisfy! Remark. For real numbers(and complex numbers) we have the extra operations of subtraction and division. The definition of a field doesn t explicitly mention these, but in fact we can use them in any field by defining: a b to be shorthand for a+( b); and a/b and a b to be shorthand for a(b ). Remark. Many other elementary facts about the real numbers can be deduced from the field axioms, which means these facts must be true in all fields. For example, we know that in the real numbers we always have (b+c)a ba+ca, which is not one of the field axioms. But in fact if a, b and c come from any field then: (b+c)a a(b+c) ab+ac ba+ca 9 Of course, the field axioms don t even make sense unless we know how to add and multiply the objects! 9

2 where the first and last equalities follow from (F6) and the middle one from (F9). Exercise Sheet 3 gives some more examples of this. Important Remark. In fact, almost everything we did in Chapters and works when the real numbers are replaced by elements of some other field (for example, by complex numbers). Matrix operations can be defined in exactly the same way, and everything we proved about them remains true. Systems of linear equations still make sense, and Gaussian and Gauss-Jordan Elimination still find the solution sets (there is an example to try on Exercise Sheet 3). Of course, wherever we referred to the real numbers 0 and (for example, in defining identity and zero matrices, or during Gaussian elimination) we use the 0 and elements of the field instead. The only thing which was really specific to the real numbers, and doesn t quite work over other fields, is the geometric interpretation of solution sets in space. Remark. It is worth taking a step back to consider the big picture of what we did in this section: We start with a particular concrete mathematical structure (in this case the real numbers). We identify the vital features which make it work, and abstract these features as axioms. We forget the original structure and study an abstract structure (a field) about which the only assumption we make is that it satisfies the axioms. Everything we learn in the abstract setting applies to every structure satisfying the axioms (for example, we can apply it back to the real numbers, but we can also apply it to the complex numbers). This process of abstraction is one of the central pillars of mathematics. Instead of studying specific individual structures in isolation, it lets us simultaneously understand lots of different structures which work in the same way. For example, instead of learning that something is true for the real numbers and then having to check all over again whether it still works for the complex numbers, we can do it for both structures at once (and for a whole load of other structures besides). This idea is the basis of abstract algebra, which you will study in detail starting with MATH00 Algebraic Structures I next semester. 3. Elementary Matrices From now on, we will mostly work with matrices whose entries come from a field K. Thus, everything we show and do will work for matrices of real numbers, complex numbers, or even more obscure fields. If you find this confusing, remember that you can always specialise back to the case of R: in other words just imagine all the entries are real numbers. Definition. An n n matrix E is an elementary matrix if it can be obtained from the identity matrix I n by a single elementary row operation. For an elementary row operation ρ we write E ρ for the corresponding matrix. There are three types of elementary matrices, corresponding to the three types of row operations. They look like: E ri r j 0

3 E ri αr i E ri r i +λr j Theorem 3.. Let A and B be m n matrices over a field and suppose B is obtained from A by a row operation ρ. Then B E ρ A. Proof. Check by matrix multiplication. Corollary 3.. Suppose A can be transformed into B by applying a sequences of elementary row operations ρ,ρ,...,ρ k. Then we have B E ρk...e ρ E ρ A. Proof. By induction on k. The base case k is exactly Theorem 3.. Now let k and suppose for induction that the statement holds for smaller k. Let A be the matrix obtained from A by applying the transformation ρ. Then by Theorem 3., A E ρ A. Now B is obtained from A by the sequence of (k ) elementary row operations ρ,...,ρ k, so by the inductive hypothesis, B E ρk...e ρ A E ρk...e ρ (E ρ A) E ρk...e ρ E ρ A. Theorem 3.3. Let E be an elementary n n matrix over a field. Then E is invertible and E is also an elementary matrix. Proof. It is easy to check (exercise!) using the definition of the inverse that: the inverse of E ri r j is E ri r j ; the inverse of E ri αr i is E ri α r i ; the inverse of E ri r i +λr j is E ri r i λr j. Elementary matrices allow us to provide the promised proof that elementary row operations on an augmented matrix preserve the solution set. Theorem.. Suppose M and N are the augmented matrices of two system of linear equations. If M is row equivalent to N then the two systems have the same solution set. Proof. Let M (A B) and N (C D). Then the two systems of equations can be written in matrix form (see Section.4) as AX B and CX D where X is the column matrix whose entries are the variables.

4 Since M and N are equivalent by a sequences of row operations, A and C are equivalent by the same sequence of row operations, and so are B and D. So by Corollary 3. there are elementary matrices E,...,E k such that C E k...e A and D E k...e B. Now suppose Z is column matrix which is a solution to AX B, that is, AZ B. Then we have CZ (E k...e A)Z (E k...e )(AZ) (E k...e )B D so Z is also a solution to CX D. The converse (showing that a solution to CX D is also a solution to AX B) is very similar (exercise!). 3.3 Invertibility and Systems of Equations Theorem 3.4. For A an n n matrix over a field, the following are equivalent: (i) A is invertible; (ii) AX 0 n has only the trivial solution X 0 n ; (iii) the reduced row echelon form of A is I n ; (iv) A is row equivalent to I n ; (v) A can be written as a product of elementary matrices. Proof. We prove (i) (ii) (iii) (iv) (v) (i). (i) (ii). Suppose A is invertible. If AX 0 n then so 0 n is the only solution to AX 0 n. X I n X (A A)X A 0 n 0 n (ii) (iii). Suppose (ii) holds. Let M be the reduced row echelon form of A. Since M is row equivalent to A, it follows from Theorem. that the equation MX 0 n has only one solution. Since M is square and in reduced row echelon form, it is either the identity matrix or has a zero row at the bottom. Having a zero row at the bottom would give lots of solutions to MX 0 n (exercise: why? think about (M 0 n ) and backtracking), so it must be that M I n. (iii) (iv). By definition, A is row equivalent to its reduced row echelon form. (iv) (v). If A is row equivalent to I n then by Corollary 3. we have for some elementary matrices E,...,E k. A E k...e I n E k...e (v) (i). By Theorem 3.3 elementary matrices are invertible, and by (an inductive argument using) Lemma.3(ii) a product of invertible matrices is invertible. 3.4 Calculating Inverses Suppose A is an invertible n n matrix over a field. By Theorem 3.4 there is a sequence of row operations ρ,...,ρ k transforming A to I n. By Corollary 3. this means I n E ρk...e ρ A

5 Multiplying both sides on the right by A we get A E ρk...e ρ AA E ρk...e ρ I n. But this means (by Corollary 3. again) that applying the sequence of row operations ρ,...,ρ k to I n gives A. This observation gives us an efficient way to find the inverse A of an invertible matrix A: (i) use Gauss-Jordan elimination to transform A into reduced row echelon form (which by Theorem 3.4 must turn out to be I n if it isn t then A isn t invertible!); (ii) apply the same sequence of row operations to I n to get A We can make the process even easier by writing the matrices A and I n side-by-side in a single augmented n n matrix (A I n ), then just apply Gauss-Jordan elimination to convert the left-hand-side to I n, and let the right-hand-side follow along. Example. Let A We have r r r r 3 r 3 r r 3 r 3 r r r 3 r r +r We have now reduced the left-hand-side to its reduced row echelon form, which is I n, confirming that A is invertible. The remaining right-hand-side must be A, so A Example. Let B This time: r r r r 3 r 3 +r r 3 r 3 +r We could continue with the elimination (exercise: try it!) but the zero row which has appeared in the bottom of the left-hand-side is clearly not going away again. This means the reduced row echelon form of B is not I n, so B is not invertible! 3