Homework Notes Week 6

Homework Notes Week 6 Math 24 Spring 24 34#4b The sstem + 2 3 3 + 4 = 2 + 2 + 3 4 = 2 + 2 3 = 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 3 2 4 2 4 2 2 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

off what an arbitrar solution must be from the above matri namel that in order for 2 3 4 to be a solution we must have 2 + 2 3 = 2 4 + 2 2 4 + 4 = 2 4 + 4 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

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

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

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

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

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

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

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