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Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 16 Matrix Multiplication MatrixVector Product Expressed as Columns. Say that we have an m × n matrix and a n × 1 matrix: c1 a11 a12 · · · a1n c2 a21 a22 · · · a2n and c = . . A= . . . .. . . . . . . . . . am1 am2 · · · amn cn We will show that the product of these two can be expressed as a linear combination of the columns of A: a1j n a2j cj colj (A) where colj (A) = . . Ac = . . j =1 amj To see why, we recall the deﬁnition of matrix multiplication: a11 a12 · · · a1n a11 c1 + a12 c2 + · · · + a1n cn c1 a21 a22 · · · a2n c2 a21 c1 + a22 c2 + · · · + a2n cn Ac = . . . = . . .. . . . . . . . . . . . am1 = = c1 col1 (A) + c2 col2 (A) + · · · + cn coln (A). cn am1 c1 + am2 c2 + · · · + amn cn am2 · · · amn a11 c1 a12 c2 a1n cn a11 a12 a2n cn a21 a22 a21 c1 a22 c2 + + ··· + = c1 . + c2 . . . . . . . . . . . . . . am1 c1 am2 c2 amn cn am1 am2 + · · · + cn a1n a2n . . . amn Using the deﬁnition of the matrix product, we have Ac = Example. We give an example of this. Consider the following matrices: 2 2 −1 −3 and c = −3 . A= 4 2 −2 4 (2) (2) + (−1) (−3) + (−3) (4) (4) (2) + (2) (−3) + (−2) (4) = −5 −6 . On the other hand, using the previous result, A c = c1 col1 (A) + c2 col2 (A) + c3 col3 (A) = (2) = −5 −6 2 4 .
1 + (−3) −1 2 + (4) −3 −2 = 4 8 + 3 −6 + −12 −8 2 MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 16 General Remarks with Linear Systems. We explain why the previous result is useful. We return to the problem of ﬁnding all solutions to a system of linear equations in the form a11 x1 a21 x1 . . . am1 x1 Let us deﬁne the matrices a11 a21 A= . . . am1 + + a12 x2 a22 x2 . . . + ··· + ··· + ··· + + a1n xn a2n xn . . . = = b1 b2 . . . + am2 x2 + amn xn = bm a12 a22 . . . am2 ... ... .. . ... a1n a2n . . . amn It is easy to see that we have deﬁned matrix products so that the system of equations can be written in the simple form A x = b. Say for the moment that b is a linear combination of the columns of A: b = c1 col1 (A) + c2 col2 (A) + · · · + cn coln (A). Then using the formulas above, we see that the nvector c1 c2 x= . . . , x= x1 x2 . . . xn , and b= b1 b2 . . . bm . cn is a solution to the system of linear equations. In fact, the system A x = b is consistent (i.e. has at least one solution) if and only if b is a linear combination of the columns of A. Algebraic Properties of Matrix Operations Review. Up to this point, we have deﬁned four operations on matrices: • Addition: Given two m × n matrices A = aij and B = bij , we deﬁne their sum as that m × n matrix A + B = aij + bij . • Transpose: Given an m × n matrix A = aij , we deﬁne its transpose as that n × m matrix AT = aji . • Scalar Multiplication: Given an m × n matrix A = aij and a number r, we deﬁne the scalar multiple of A by r as that m × n matrix r A = r aij . • Matrix Multiplication: Given an m × p matrix A = aij and a p × n matrix B = bij , we deﬁne
p their product as that m × n matrix A B =
k=1 aik bkj . We discuss more properties of these operations during today’s lecture. Matrix Addition. Let A = following properties: (a) (b) (c) (d) aij ,B= bij and C = cij all be m × n matrices. Then we have the Commutativity: A + B = B + A. Associativity: A + (B + C ) = (A + B ) + C . Identity: There is a unique m × n matrix O, called the zero matrix, such that A + O = A for all A. Inverses: For each A, there is a unique m × n matrix −A, called the negative of A, such that A + (−A) = O. MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 16 3 Examples. Before we give the proof of these statements, we give some examples. The 2 × 2 zero matrix is O= For instance, if we choose the 2 × 2 matrix A= then we ﬁnd the sum A+O = 4 2 −1 3 + 00 00 = 4+0 2+0 −1 + 0 3+0 = 4 2 −1 3 = A. 4 2 −1 3 00 00 . Similarly, the 2 × 3 zero matrix is O= For instance, if we choose the 2 × 3 matrix A= then we ﬁnd the sum A+O = 13 −2 4 −2 3 + 0 0 0 0 0 0 = 1 + 0 3 + 0 −2 + 0 −2 + 0 4 + 0 3+0 = 13 −2 4 −2 3 = A. 13 −2 4 −2 3 000 000 . Proofs. We prove the statements above. For (a), we have the identity A+B = For (b), we have the identity A + (B + C ) = = aij + bij + = cij = aij + + cij bij + cij = aij = + aij + (bij + cij ) bij + cij aij + bij = aij + bij = bij + aij = B + A. (aij + bij ) + cij aij + bij = (A + B ) + C. For (c), we must show two things. First, there exists at least one m × n matrix O1 such that A + O1 = A for all A: choose O1 = 0 as that matrix with all zero entries. Now say that there is another matrix O2 such that A + O2 = A for all A. Consider the expression O1 + O2 . Since A + O1 = A for all A, we see that O2 = O1 + O2 . But since A + O2 = A for all A by assumption, we see that O1 + O2 = O1 . Hence O2 = O1 + O2 = O2 . This shows that the zero matrix O must be unique. Now we show (d). Again, for uniqueness we must show two things. First, there exists at least one m × n matrix D1 such that A + D1 = O: choose D1 = −aij . Now say that there is another matrix D2 such that A + D2 = O. Consider the expression D1 + A + D2 . We have the identity D2 = O + D2 = (D1 + A) + D2 = D1 + (A + D2 ) = D1 + O = D1 . This shows that the negative −A must be unique. Transposes. Let A = aij the following properties: (a) (b) (c) (d)
T and B = bij be m × n matrices, and r be a real number. Then we have = A. Involution: AT T Linearity: (A + B ) = AT + B T . T ( A B ) = B T AT . T (r A) = r AT . When m = n, we say that an n × n matrix A is symmetric if AT = A and skew symmetric if AT = −A. 4 MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 16 Example. As before, we give some examples before we discuss the proofs. Consider the two matrices 0 1 1 32 2 . and B= 2 A= 2 −1 3 3 −1 Then we have the matrix product AB = (1) (0) + (3) (2) + (2) (3) (2) (0) + (−1) (2) + (3) (3)
T (1) (1) + (3) (2) + (2) (−1) (2) (1) + (−1) (2) + (3) (−1) 12 7 5 −3 = 12 5 7 −3 which gives the transpose (A B ) = Similarly, we have the transposes 1 AT = 3 2 2 −1 3 and BT = 02 12 3 −1 . . Clearly AT B T = (A B ) . On the other hand, B T AT = For comparison, we compute two matrix products: The ﬁrst is (1) (0) + (2) (1) (1) (2) + (2) (2) (1) (3) + (2) (−1) 26 1 AT B T = (3) (0) + (−1) (1) (3) (2) + (−1) (2) (3) (3) + (−1) (−1) = −1 4 10 . (2) (0) + (3) (1) (2) (2) + (3) (2) (2) (3) + (3) (−1) 3 10 3
T (0) (1) + (2) (3) + (3) (2) (1) (1) + (2) (3) + (−1) (2)
T (0) (2) + (2) (−1) + (3) (3) (1) (2) + (2) (−1) + (−1) (3) = 12 7 5 −3 . This shows that B T AT = (A B ) . Proofs. We prove the statements above. For (a), we have the identity A= aij =⇒ AT = aji =⇒ AT
T = aij = A. For (b), we have the identity (A + B ) = For (c), we have the identity
p p T aij + bij T = aji + bji = aji + bji = AT + B T . AB = aij bij =
k=1 aik bkj =⇒ (A B ) =
k=1 T ajk bki . On the other hand, we have the identity
p p B T AT =
T bji aji =
k=1 bki ajk =
k=1 ajk bki . This shows that (A B ) = B T AT . Finally for (d), we have the identity (r A) = Scalar Multiplication. Let A = aij Then we have the following properties:
T r aij T = r aji bij =r aji = r AT . and B = be m × n matrices, while r and s are real numbers. (a) Associativity: r (s A) = (r s) A. (b) Distributivity: (r + s) A = r A + s A, and r (A + B ) = r A + r B . (c) Commutativity: A (r B ) = r (A B ) = (r A) B – whenever the products are deﬁned. For the sake of time, we omit the proofs. MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 16 5 Matrix Multiplication. Let A, B , and C be matrices. Then when deﬁned, we have the following properties: (a) Associativity: A (B C ) = (A B ) C . (b) Distributivity: (A + B ) C = A C + B C , and C (A + B ) = C A + C B . Again, for the sake of time, we omit the proofs. ...
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This note was uploaded on 02/25/2010 for the course MA 00265 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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