Homework Notes Week 6
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1 Homework Notes Week 6 Math 24 Spring 24 34#4b The sstem = = = is consistent To see this we put the matri 3 2 A b = 2 into reduced row echelon form Adding times the first row to the second and third rows we obtain the matri Now we note that the second row is twice the third so we can eliminate the third row and then multipl the second row b /4 to obtain the matri 3 2 /2 Now adding three times the second row to the first we obtain the matri /2 /2 Now since this is in reduced row echelon form eercise 3 sas that the sstem is consistent ie that it has a solution To see what s it s solution set is we just read
2 off what an arbitrar solution must be from the above matri namel that in order for to be a solution we must have = = Now finall b Theorem 35b the set /2 /2 /2 /2 is a basis for the corresponding homogeneous sstem + 34#5* Theorem A matri has onl one reduced row echelon form Proof We will prove this b induction on n the number of columns of a matri For this let A be an m n matri and suppose that B C are m n matrices which are both reduced row echelon forms of A If n = then A B C are all just column vectors and we reall onl have two options rank A = or If rank A = then A = B = C = so in particular B = C If rank A = then b Theorem 36 we know that there is a column of B that is the column vector e ie that B = e But similar C = e hence B = C Now suppose that n > and that all matrices with n columns have a unique reduced row echelon form Let A B C be the matrices obtained b deleting the nth column of A B C respectivel Since deleting a column has no effect on the row operations in the remaining columns B C are reduced row echelon forms of A and so b the induction hpothesis we have that B = C So at the ver least the first n columns of B and C are the same so the onl wa the could disagree is in column n Note that adding a column to a matri either leaves the rank the same or increases it b one we have that rank A = rank A or rank A = rank A + Throughout the rest let r = rank A 2
3 If r = rank A then b Theorem 36 part b there are columns b ji = e i of B for each i = 2 r Now since B = C we know that the columns c ji = e i as well Since column n of B is of the form d e + d r e r for some scalars d d r we know that column n of A is d a j + d r a jr this is from Theorem 36 part d But similarl column n of C is of the form d e + d re r for some scalars d d r and again column n of A is d a j + d ra jr But Theorem 36 part c sas that {a j a jr } is linearl independent and we have just shown that d a j + d r a jr = d e + d re r so b linear independence of these vectors we have d = d d 2 = d 2 d r = d r But this means eactl that the nth column of B and C are the same so finall B = C If r = rank A + then we will see that the column n of B and C is precisel e r Since B C are in reduced row echelon form and have rank r the must have all zeros in row r But since B C have rank r there must be onzero entr in row r of these matrices in particular the entr of column n and row r of B and C must be nonzero Since this is the first nonzero entr in row r it must be and this must be the onl nonzero entr in column n as B and C are reduced row echelon forms But this means precisel that the nth columns of B and C are e r hence that B = C 4#9 For an A B M 2 2 F we have detab = deta detb We know that we can write a b e f A = B = c d g h for some scalars a b h B definition deta detb = ad bceh fg 3
4 and we can compute detab directl a b e f detab = det c d g h ae + bg af + bh = det ce + dg cf + dh = ae + bgcf + dh af + bhce + dg = aecf + aedh + bgcf + bgdh afce afdg bhce bhdg = acef acef + adeh afdg + bcgf bche + bdgh bdgh = adeh afdg + bcgf bche = adeh fg bceh fg = ad bceh fg 4# Let A M 2 2 F and C the classical adjoint of A ie A22 A C = 2 A 2 A For part a we compute directl AC and CA A A AC = 2 A22 A 2 A 2 A 22 A 2 A A A = 22 A 2 A 2 A A 2 + A 2 A A 2 A 22 A 22 A 2 A 2 A 2 + A 22 A deta = deta = deta The computation for CA is essentiall identical For part b we just compute detc A22 A detc = det 2 A 2 A = A 22 A A 2 A 2 = A A 22 A 2 A 2 = deta 4
5 Finall for part c we know that if deta if and onl if A is invertible so if A is invertible then A C = and b uniqueness of inverses we have deta A = [deta] C 4#* Theorem Suppose δ : M 2 2 F F is a function with the following three properties i δ is a linear function of each row of the matri when the other row is held fied ii If the two rows of A M 2 2 F are identical then δa = iii If I is the 2 2 identit matri then δi = Then δa = deta for ever A M 2 2 F Suppose δ satisfies properties i ii and iii For the current proof given vectors F 2 let s write for the 2 2 matri whose rows are and as there will never be an danger to confuse this notation with that of a column vector We will break down the argument in three lemmas Lemma δ = δ Proof Because of propert i we have + δ = δ + δ = δ Because of propert ii we have δ = δ Therefore δ + δ + δ = and δ = or equivalentl δ + δ + = + = δ + δ 5
6 Lemma 2 We have δ ever = 2 F 2 e e = and δ e e 2 = and therefore δ e = 2 for Proof The first two facts are immediate consequences of properties ii and iii respectivel B propert i we then also have e e δ = δ e e = e + 2 e δ + 2 e 2 δ = e + 2 = 2 2 for ever vector = 2 F 2 Lemma 3 We have δ ever = 2 F 2 e2 e = and δ e2 e 2 = and therefore δ e2 = for Proof The first fact follows from propert iii and Lemma and the second fact follows from propert ii B propert i we then also have e2 e δ = δ 2 e2 e = e + 2 e δ + 2 e 2 δ = e + 2 = 2 for ever vector = 2 F 2 Now b propert i we have for an two vectors = 2 = 2 F 2 that e δ = δ + 2 e 2 e e2 = δ + 2 δ = 2 2 It then follows immediatel from the definition of 2 2 determinants that δa = deta for ever 2 2 matri A over F 42#23* Theorem The determinant of an upper triangular matri is the product of its diagonal entries 6
7 Proof This is eas to see from the alternate definition of determinants from the April 24 slides The idea is that if A is an n n matri and σ is an n-permutation then signσa σ A nσn = unless σ is the permutation 2 n where ever number is in order Indeed given an other permutation σ let i be the first number such that σ i i We must then have σ j = i for some j > i But then A jσj = A ji = since A is upper triangular Proof Suppose that A = A ij ij n is an n n upper triangular matri meaning that A ij = for all i > j We prove this b induction on n the number of columns and rows in an upper triangular matri If n = then A = a and deta = a b definition of det for matrices Now suppose that n > and that the result holds for all n n upper triangular matrices Then we compute deta b cofactor epansion along row n Theorem 44 then shows that deta = n n+j A nj detãnn = n+n A nn detãnn = A nn detãnn j= since A n = A n2 = = A nn = Now if Ãnn is an n n matri and is upper triangular with diagonal entries A A 22 A n n so b the induction hpothesis deta = A nn detãnn = A nn A A n n = A A nn which is precisel the product of the diagonal entries of A 42#24 If an n nmatri A has a row consisting entirel of zeros then deta = First write A = a i = for some i n then Theorem 43 a for some row vectors a Then suppose that 7
8 deta = det = det = det a a i a i+ a a i + a i+ a a i a i+ + det = deta + deta = 2 deta a a i a i+ So subtracting deta from both sides we get deta = 43# If M is skew-smmetric then det M = detm t = detm b Theorem 48 Since M = IM we also have det M = det I detm = n detm b Theorem 47 and Eercise 23 from Section 42 If n is odd then it follows from detm = detm that detm = Since we are working over the comple numbers + If n is even then the above sas nothing much A 2k 2k a skew-smmetric matri ma or mot be invertible The zero matri is an eample of a 2k 2k 8
9 skew-smmetric matri that is not invertible The matri B given b k I B = k I k k where I k is the k k identit matri and k is the k k zero matri then B is invertible since rankb = 2k and B is clearl skew-smmetric Indeed M = is a skew-smmetric matri with determinant 43#2 If Q M n n R is orthogonal then detq = ± Recall a few facts about determinant first we know that deti = where I is the identit matri second detab = deta detb for all A B M n n R and third that deta t = deta Using these facts we have = deti = detqq t = detq detq t = detq detq = detq 2 but the onl numbers in R whose squares are one are and Therefore detq = ± 9
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