Lecture 15: Recall defn. of row operations, row equivalence, and row space. Theorem 1: If B is row equivalent to A, then R(B) = R(A). Proof of Theorem 1: Check that, after each elementary row oper- ation, the row space does not change. 2 3 2 0 3 R1 R1 6 R 7 6 0 7 6 2 7 6 R2 7 A = 6 7 ;B = 6 7 4 ::: 5 4 ::: 5 R 0 m Rm 0 0 1. Ri = Rj; Rj = Ri. { An exchange of rows does not change the set of rows. 0 −1 0 2. Ri := aRi; Ri := a Ri. { So, any linear combination of rows of B is a linear combination of rows of A, and vice versa. 0 0 0 0 3. Rj := aRi + Rj; Rj = Rj − aRi = Rj − aRi { So, any linear combination of rows of B is a linear combination of rows of A, and vice versa. Recall defn. of REF. Theorem 2: Every matrix is row equivalent to an REF. Proof of Theorem 2: Step 1: Exchange rows so that a row with the left-most leading entry is at top. Then the leading entry in the top row is in position (1; j1) 1 Step 2: Subtract appropriate multiples of R1 from all other rows in order to \zero out" all entries of column j1 except for position (1; j1). Continue Steps 1 and 2 inductively until only zero rows remain: { Exchange remaining rows so that (among all but row 1), R2 has the left-most leading entry . Then the leading entry in R2 is in position (2; j2), with j2 > j1. { Subtract appropriate multiples of R2 from all other rows in order to \zero out" all entries of column j2 below (2; j2). { etc. Example 1 over GF (3) = Z3: Let 2 0 2 2 3 A = 4 2 1 0 5 1 1 2 Construct an REF B for A: R1 $ R2: 2 2 1 0 3 4 0 2 2 5 1 1 2 2R1 + R3 ! R3: 2 2 1 0 3 4 0 2 2 5 0 2 2 R2 + R3 ! R3: 2 2 1 0 3 4 0 2 2 5 = B 0 0 0 2 Example 2 over GF (4) = f0; 1; a; bg, with special addition/multiplication tables (on p. 36). Find REF of: 2 a b 1 3 A = 4 1 b 0 5 b 0 1 bR1 + R2 ! R3: 2 a b 1 3 4 0 1 b 5 b 0 1 aR1 + R3 ! R3: 2 a b 1 3 4 0 1 b 5 0 1 b R2 + R3 ! R3: 2 a b 1 3 4 0 1 b 5 0 0 0 Theorem 3: Let W = hSi where S is a subset of V (n; q). Let A be a matrix whose rows are the elements of S. Let B be a REF of A. Then the nonzero rows of B from a basis for W . Proof of Theorem 3: By Theorem 2, an REF B for A exists. We must show that i) the nonzero rows of B span W and ii) the nonzero rows are linearly independent. i: By Theorem 1, the row space of B is W ; thus, the nonzero rows of B span W . 3 ii: Write 2 0 3 R1 0 6 R 7 B = 6 2 7 6 ::: 7 4 0 5 Rm 0 0 Let R1;:::; Rk, k ≤ m, be the non-zero rows. Suppose k X 0 aiRi = 0: i=1 0 For i = 1; : : : ; k, let ji be the leading entry of Ri. We show that each ai = 0, 1 ≤ i ≤ k, by induction on i. For i = 1: Since all entries in column j1, except for (1; j1), are 0, k X 0 a R0 = ( a R ) = 0: 1 1;j1 i i j1 i=1 Since R0 =6 0 and F is a field, a = 0. 1;j1 1 Inductive step: Assume ai = 0 for i = 1;:::;I − 1 < k, for some I. Since R0 = 0 for all ` > j , `;jI I k X 0 a R0 = ( a R ) = 0: I I;jI i i jI i=1 Again, since R0 =6 0 and F is a field a = 0. I;jI I By Example 1 above (GF (3)), f210; 022g is a basis for h210; 112; 022i. By Example 2 above (GF (4)), fab1; 01bg is a basis for hab1; 1b0; b01i. 4 Corollary: Let S be a finite subset of V (n; q). Let A be a matrix whose rows are the elements of S. Let B be a REF of A. Then S is linearly independent iff B has no zero rows. Proof of Corollary: One can show row equivalence preserves linear independence (i.e., if A and B are row equivalent, then the rows of A are linearly independent iff the rows of B are linearly independent) { left a an exercise. Thus, S is linearly independent iff the the rows of the REF B are linearly independent. But the latter can happen iff B has no zero rows. Another proof: If B has no zero rows, then the rows of B form a basis for R(A). Thus, the dimension of R(A) is the number of rows of A. Since the rows of A span R(A) and by Theorem 4.2, some subset of a spanning set is a basis, and all bases for the same subspace have the same size, it follows that the rows of A form a basis for R(A) and thus are linearly independent. If S is linearly independent, then the rows of A form a basis for R(A). Since the nonzero rows of B also form a basis, and any two bases have the same size, it follows that B has no zero rows. Defn: A matrix is in Reduced Row Echelon form (RREF) if 1. The leading entry in each non-zero row is strictly to the right of the leading entry in the preceding row. 2. All the non-zero rows (if any) are at the bottom. 3. All leading entries are 1 4. A leading entry is the only nonzero entry of its column. Note that the difference between REF and RREF is conditions 3 and 4. 5 REF: L = a leading entry (the different appearances of L may be differ- ent nonzero elements) 2 3 0 ::: 0 L ∗ ::: ∗ a ∗ ::: ∗ b ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 L ∗ ::: ∗ c ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 L ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 RREF: 2 3 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ 0 ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 Theorem: Every matrix is row equivalent to a unique RREF. Proof: In an REF, all entries below a leading entry are 0. From RRE, multiply each leading entry a by a−1 to make the leading entry = 1. For each leading entry i, let ji denote its column; so ai;ji = 1. For each nonzero entry a = ar;ji =6 0 for some r < i (above a leading entry), replace row Rr by Rr − aRi; this \zeros out" the (r; ji) entry. After making all leading entries of REF equal to 1: 6 2 3 0 ::: 0 1 ∗ ::: ∗ a ∗ ::: ∗ b ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ c ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 Replace R1 by R1 − aR2: 2 3 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ d ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ c ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 Replace R1 by R1 − dR3: 2 3 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ 0 ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ e ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 Replace R2 by R2 − eR3: 7 2 3 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ 0 ∗ ::: ∗ 6 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 0 ∗ ::: ∗ 7 6 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 1 ∗ ::: ∗ 7 6 7 6 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 7 6 7 4 ::::::::::::::::::::::::::::::::::::::::::::: 5 0 ::: 0 0 0 ::: 0 0 0 ::: 0 0 0 ::: 0 8 Lecture 16: HW4 will be posted later today and due on Friday, March 11. Recall REF and RREF. Example 1 over GF (3) = Z3. An REF for 2 0 2 2 3 A = 4 2 1 0 5 1 1 2 is: 2 2 1 0 3 B = 4 0 2 2 5 0 0 0 To find RREF: 2R1 ! R1 : 2R2 ! R2 2 1 2 0 3 4 0 1 1 5 0 0 0 R1 + R2 ! R1: 2 1 0 1 3 RREF = 4 0 1 1 5 0 0 0 Example 2 over GF (4) = f0; 1; a; bg.