(I.D) Solving Linear Systems via Row-Reduction

Transcription

1 (I.D) Solving Linear Systems via Row-Reduction Turning to the promised algorithmic approach to Gaussian elimination, we say an m n matrix M is in reduced-row echelon form if: the first nonzero entry of each row is 1 (called a leading 1 ). We write r for the number of leading 1 s; if a column contains a leading 1, then this must be its only nonzero entry (such columns are called pivot columns ); and if a row contains a leading 1, then each row above contains a leading 1 further to the left. Note in particular that if the leading 1 of row i occurs in the k th i then k 1 <... < k r. entry, DEFINITION 1. The number r is called the rank of M. If M is m n then r min{m, n}. If r = min{m, n} we say M has maximal rank. When m n, reduced row echelon matrices of maximal rank are all of the form for m > n, I n for m = n In contrast, when m < n there are many possibilities: a 3 5 rowreduced echelon matrix of maximal rank (r = 3), for instance, can 1

2 2 (I.D) SOLVING LINEAR SYSTEMS VIA ROW-REDUCTION take any of the forms , , 0 1 0, , , , , , , or , where the stands for arbitrary value. We now show that any matrix is row-equivalent to a reduced row echelon matrix, by describing an row-reduction algorithm that always terminates on such a matrix. Begin with an (arbitrary) m n matrix A, and place your imaginary cursor at A 11, its upper lefthand entry. Now move the cursor to the right (if necessary) until it reaches a column with a nonzero entry; if the cursor entry = 0, swap the cursor row with the first row below with a nonzero entry in the cursor column; divide the cursor row by the cursor entry (to make cursor entry = 1 ); eliminate all other entries in the cursor column (by adding suitable multiples of the cursor row to all the other rows); move the cursor down and to the right, go back to the first, and repeat until we reach a reduced row echelon matrix rre f (A). Since this procedure is completely deterministic, it yields a welldefined map M mn (R) M mn (R) sending A rre f (A). This

3 (I.D) SOLVING LINEAR SYSTEMS VIA ROW-REDUCTION 3 rre f (A) is simply defined as the outcome of this particular algorithm applied to A. We have not proved that A is row-equivalent to a unique row-reduced echelon matrix (that is true, but will be proved somewhat later in these notes). EXAMPLE rre f = DEFINITION 3. For an arbitrary matrix A, we define the rank by (In the example, this is 3.) rank(a) := rank(rre f (A)). Notice that each of the bullets is accomplished by elementary row operations, which is to say via left-multiplication by an (invertible) elementary matrix. Given A as input, the above algorithm therefore spits out two well-defined matrices: namely, rre f (A) and the (invertible) product E(A) = E N... E 1 of the row-operations. These matrices are related by rre f (A) = E(A) A, which may be viewed as a decomposition of A = E(A) 1 rre f (A) into {invertible (m m)} {reduced row echelon (m n)}. Solving homogeneous equations. A x = 0 = 0 = E(A) A x = rre f (A) x ; and conversely rre f (A) x = 0 = 0 = E(A) 1 rre f (A) x = A x. So the solutions (in x ) to rre f (A) x = 0 and A x = 0 are the same. Now rre f (A) x = 0 is easily solved. Suppose rre f (A) = ; then in solving rre f (A) x = 0 we may choose the variables x 2 and x 4 freely, while the variables in the pivot columns (with leading 1 s)

4 4 (I.D) SOLVING LINEAR SYSTEMS VIA ROW-REDUCTION are determined by those choices: x 1 = 2x 2 3x 4, x 3 = x 4, x 5 = 0. The upshot is that for (rre f (A) x = 0 )A x = 0 to have nontrivial solutions (i.e. solutions other than x = 0 ), we ve got to have columns without leading 1 s. There are n columns and r leading 1 s, so r < n is what we need. Therefore m < n = r min{m, n} = m < n = of nontrivial solutions. If m = n then the only possible rre f (A) with all columns filled up by leading 1 s is the identity I n, and so the condition for an interesting solution is for rre f (A) not to be I n : that is, {rre f (A) = I n } {A x = 0 has only the trivial solution}. Now {rre f (A) = I n } = A = E(A) 1 rre f (A) = E(A) 1 = A is invertible = {A x = 0 has only the trivial solution} since we can multiply both sides by A 1. So the following four items are equivalent for a square matrix A : A x = 0 has only the trivial solution rre f (A) = I n E(A) = A 1 A is invertible. Now we could use these equivalences to see quickly that A leftinvertible = A invertible, but then it s difficult to see what is going on. So let s be more deliberate: Let A be any n n matrix with a left-inverse L, so that L A = I n. Suppose x solves A x = 0 ; then x = I n x = (L A) x = L(A x) = 0. Therefore A x = 0 has only the trivial solution, and so does rre f (A) x = 0 [since rre f (A) = E(A) A where E(A) is invertible (this is the first key)]. But if x = 0 is the only solution to rre f (A) x = 0, then all columns of rre f (A) must contain a leading 1. Therefore rre f (A) = I n, and A = E(A) 1 rre f (A) = E(A) 1 is invertible. This also establishes the claim from I.C that for a square matrix, left-invertibility implies right-invertibility and vice-versa. Solving inhomogeneous equations and computing inverses. Consider an augmented matrix (A B); this is just a big m (n 1 + n 2 ) matrix made up of a m n 1 block A and an m n 2 block B. We

5 (I.D) SOLVING LINEAR SYSTEMS VIA ROW-REDUCTION 5 shall define rre f (A B) by performing the row-reduction algorithm on A (as if to compute rre f (A) ) and carrying the row operations across to B. 1 This yields (rre f (A) E(A) B), since E(A) operates on both blocks; we stop here rather than further reducing the righthand part of the augmented matrix. If B is just the vector y, then this gives a way of solving A x = y : invertibility of E(A) = solutions of A x = y and rre f (A) x = E(A) y coincide, and solutions to the latter are easily obtained. Note that regardless of the number of equations (= m ) an inhomogeneous system may have no solutions. For A an m n matrix, the columns of E(A) are E(A) ê i, i = 1,..., m. So rre f (A ê i ) = ( rre f (A) {ith column of E(A)} ) and putting all the columns together, rre f (A I m ) = (rre f (A) E(A)). For A invertible (m m) [ rre f (A) = I m ], notice that E(A) = A 1 and rre f (A : I m ) = (I m : A 1 ). To make sure you understand this process, try using it to rederive Example I.C.1. EXAMPLE 4. Given the inhomogeneous linear system 3x 1 6x 2 + 2x 3 x 4 = 1 2x 1 + 4x 2 + x 3 + 3x 4 = 4 x 3 + x 4 = 2 x 1 2x 2 + x 3 = 1, we write the augmented matrix , Warning: this is different from taking rref of the whole m (n1 + n 2 ) matrix!

6 6 (I.D) SOLVING LINEAR SYSTEMS VIA ROW-REDUCTION then apply the rref algorithm to the 4 4 block (carrying them over to the last column) to obtain Now (interpreting this as a linear system) write the pivot variables x 1 = 2x 2 + x 4 1 x 3 = 2 x 4 in terms of the free variables x 2, x 4, which then parametrize all the solutions. Note that if we were to replace the original third equation by x 3 + x 4 = 3, the system would be inconsistent. REMARK 5. If you know how to do rre f, then you also know how to do rce f (reduced column echelon form). Flip the matrix, take rre f, and flip it back: rce f (A) := t (rre f ( t A)) = t (E( t A) ta) = A te( t A) where the column-operations t E( t A) are invertible, and occur on the right. Exercises (1) Use the rref algorithm to prove that a 1 0 c b 1 is invertible, and to compute this inverse, for arbitrary a, b, c. (2) Find all solutions of 2x 1 3x 2 7x 3 + 5x 4 + 2x 5 = 2 x 1 2x 2 4x 3 + 3x 4 + x 5 = 2 2x 1 4x 3 + 2x 4 + x 5 = 3 x 1 5x 2 7x 3 + 6x 4 + 2x 5 = 7.

7 EXERCISES 7 What is the rank of the matrix A in this case? (3) Prove that if A is an m n matrix, B is an n m matrix, and n < m, then AB is not invertible.