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Lecture 3 (Row Equivalence, Row Reduction) (12) Students Should Describe How Determinant of a Square Matrix Would Change If an E

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Lecture 3 (Row Equivalence, Row Reduction) (12) Students Should Describe How Determinant of a Square Matrix Would Change If an E 3 Lecture 3 (Row equivalence, row reduction) (12) Students should describe how determinant of a square matrix would change if an elementary row operation is applied (in each of the three cases). (13) IfB is obtained fromA by an elementary row operation thenfind how to getA fromB. Students have to describe the inverse operations in each cases. Students should note that an elementary row operation is an invertible map fromM m n(R) to itself. × (14) Denote an elementary row operation byρ. IfA M m n(R) thenρ(A) =ρ(I)A (pre- multiplication ofA byρ(I)) whereI is them m identity∈ matrix.× Students should verify this fact in each case. Ifρ is an elementary× row operation thenρ(I) is referred to as an elementary matrix. (15) Use the above rule successively on a matrix and get the following two statements for anyfinitely many elementary row operationsρ 1,ρ 2,...,ρ s: (a) (ρ ρ ρ )(A) = (ρ ρ ρ )(I)A. s ◦···◦ 2 ◦ 1 s ◦···◦ 2 ◦ 1 (b) (ρs ρ 2 ρ 1)(A) =ρ s(I) ρ 2(I)ρ1(I)A. (16) A matrix◦A··· is◦ said◦ to be row equivalent··· to the matrixB if there is afinitely many elementary row operationsρ 1,ρ 2,...,ρ s such thatB=(ρ s ρ 2 ρ 1)(A). (17) Observe that row equivalence is an equivalence◦ relation.···◦ ◦ (a)A is row equivalent to itself. Recall that the identity map is composition of two elementary row operations (for instance, composition ofR R with itself). i ↔ j (b)A is row equivalent toB if and only ifB is row equivalent toA. IfB=(ρ s ρ 2 ρ1)(A), 1 1 1 ◦···◦ ◦ thenA=(ρ 1− ρ s− 1 ρ s− )(B) and inverse of an elementary row operation is an elementary row◦ operation.···◦ − ◦ (c) IfA is row equivalent toB andB is row equivalent toC thenA is row equivalent toC. If B=(ρ s ρ 2 ρ1)(A) andC=(ρ s+t ρ s+2 ρs+1)(B) thenC=(ρ s+t ρ 2 ρ1)(A). IfA is row equivalent◦···◦ ◦ toB we will often write◦···A◦ B. ◦ ◦···◦ ◦ (18) One can write an algorithm to show, “Every matrix∼ is row equivalent to a unique row reduced echelon matrix.” This statement is same as “Every equivalence class of row equivalent matrices contains a unique row reduced echelon matrix.” Step 1: Appy interchange of rows to push down the zero rows to the end of the matrix so that no zero row is before a nonzero row. Step 2: Find thefirst nonzero column (from left) (suppose it isk 1). Step3: Again apply interchange of rows to push up a row whose leading nonzero coefficient occurs infirst nonzero column (i.e. in thek 1-th column), to thefirst row. Divide thefirst row by the leading nonzero coefficient so that the leading nonzero coefficient becomes 1. Step 4. Next apply applyR i R i µR1 for suitable values ofi andµ so that thefirst nonzero column has nonzero coefficient→ only− in thefirst row. Step5: Find thefirst nonzero column (from left) when we ignore thefirst row (suppose it isk 2). Apply interchange of rows to push up a row whose leading nonzero coefficient occurs in column k2, to the second row. Divide the second the second row by the leading nonzero coefficient. Step 6: ApplyR i µR 2 for suitable values ofi andµ so thatk 2 column has nonzero coefficients only in the second− row. Step 7: Find thefirst nonzero row when we ignore thefirst two rows. continue. When there is no nonzero rows left, stop. 0 0 4 1 0 0 4 1 0 3 0 1 0 3 0 1 0 3 0 1 0 0 4 1 ApplyR 3 R 4. Get . ApplyR 1 R 2. Get .Next 0 0 0 0 ↔ 0 4 2 0 ↔ 0 4 2 0 0 4 2 0 0 0 0 0 0 0 0 0 0 1 0 1/3 0 1 0 1/3 0 0 4 1 0 0 4 1 applyR 1 1/3R 1 . Get . ApplyR 3 R 3 4R 1, get . → 0 4 2 0 → − 0 0 2 4/3 − 0 0 0 0 0 0 0 0 0 1 0 1/3 0 1 0 1 0 0 1 1/4 0 0 1 1/4 AppyR 2 1/4R 2, get . ApplyR 3 R 3 2R 2, get . → 0 0 2 4/3 → − 0 0 0 11/6 − − 0 0 0 0 0 0 0 0 4 0 1 0 1 0 0 1 1/4 Apply,R 3 6/11R3, get . ApplyR 1 R 1 R 3 andR 2 R 2 1/4R 3 get → − 0 0 0 1 → − → − 0 0 0 0 0 1 0 0 0 0 1 0 , which is RRE. 0 0 0 1 0 0 0 0 Remark: There are more than one ways of getting the row reduced echelon form of a matrix. Lecture 4 (Rank and invertibility of Matrices) (19) Following applications of row equivalence of matrices are to be discussed: (a) Finding rank of a matrix. IfA is row equivalent to the row reduced echelon matrixR then the number of nonzero rows ofR is called the rank ofA. In particular, the rank of a row reduced echelon matrix is the number of nonzero rows. We will see later that the row rank of a matrix (i.e. dimension of the row space) is the rank of the matrix. 0 0 4 1 1 1 1 1 1 0 3 0 1 1 2 1 3 1 The rank of is 3. The Rank of is two. Student must 0 0 0 0 0 1 0 2 0 0 4 2 0 1 0 1 1 1 − 1 0 1 1 1 0 1 0− 2 0 show that this matrix is row equivalent to which is row reduced echelon 0 0 0 0 0 0 0 0 0 0 (so that the rank of the given matrix is 2). Remark: We will see use of rank of a matrix. (b) Determining whether a square matrix is invertible. In case it is invertiblefinding the inverse. Theorem: SupposeR is row reduced echelon and row equivalent toA. ThenA is invertible if and only ifR is the identity matrix. In fact, Algorithm: Suppose (A I) is row equivalent to (R B) andR is row reduced echelon. | | 1 ThenA is invertible if and only ifR=I and in this caseB=A − . 1 If (ρ ρ ρ )(A) =I, then (ρ ρ ρ )(I) A=I. ThereforeA − = k ◦···◦ 2 ◦ 1 k ◦···◦ 2 ◦ 1 · (ρk ρ 2 ρ 1)(I). ◦···◦ ◦ 1 1 1 Example: SupposeA= 1 2 1 . Apply elementary row operations on the matrix 1 2 3 (A I) tofind (R B) whereR is the RRE matrix row equivalent toA. | | 1 1 1 100 | ApplyR 2 R 2 R 1 and thenR 3 R 1 on 1 2 1 010 and get → − → 1 2 3|001 | 1 1 1 100 | 0 1 0 1 1 0 . Then applyR 1 R 1 R 2 andR 3 R 3 R 1 and get 0 1 2|−1 0 1 → − → − |− 1 0 1 2 1 0 | − 0 1 0 1 1 0 . Then applyR 3 1/3R 3 and thenR 1 R 1 R 3 and get 0 0 2|0− 1 1 → → − | − 1 0 0 2 1/2 1/2 2 1/2 1/2 | − − 1 − − 0 1 0 1 1 0 . Conclude thatA − = 1 1 0 . 0 0 1|0− 1/2 1/2 −0 1/2 1/2 | − − Remark: Algorithm tofind row reduced echelon form of a matrix, also determines whether a square matrix is invertible and if it is invertible, it can be used tofind the inverse. (c) Solving a system of linear equations. Obviously,AX = 0 always has a solution (for instance,x i = 0 i). Suppose we have the following system of linear equations:AX=B. Observe the following∀ statements: 5 i) If (A B) is row equivalent to (A � B �), then the two systemsAX=B andA �X=B � have the| same solutions. | Reasoning: Sinceρ(PQ) =ρ(P)Q we haveAX=B if and onlyρ(A)X=ρ(B) for any compositionρ offinitely many elementary row operations. ii) IfA is row reduced echelon matrix then one can determine whetherAX=B has a solution and what are all solutions just by inspection. Reasoning: SupposeA is a row reduced echelon matrix. If there is a rowi of (A B) such thati-th row ofA a zero row buti-th row ofB nonzero thenAX=B has no solution.| SupposeAX=B has a solution. Assume thatA hasr nonzero rows and the leading nonzero coefficient ofi-th row is in thek i-th coefficients. Callx k1 , . xkr dependent un- knowns. Assign arbitrary values to allx i fori k 1, k2, . , kr then writex k1 , . , xkr as a linear combination as the independent unknowns. �∈{ } Discuss enough examples. (We will discuss later, if the system is homogenous then the dimension of the solution space isn r.) −.
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