MTH 102A - Linear Algebra II Semester

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1 MTH 02A - Linear Algebra II Semester Arbind Kumar Lal P. Vector space A set V over a field F is a vector space if x+y is defined in V for all x,y V and αx is defined in V for all x V,α F s.t. ) + is commutative, associative. 2) Each x V has a inverse w.r.t +. 3) Identity element 0 exists in V for +. 4) x = x holds x V, where is the multiplicative identity of F. 5) (αβ)x = α(βx),(α+β)x = αx+βx hold α,β F,x V. 6) α(x+y) = αx+αy holds α F, x,y V. R 3 is a vector space (VS) over R. Here (x,x 2,x 3 )+α(y,y 2,y 3 ) := (x +αy,x 2 +αy 2,x 3 +αy 3 ). M m,n := {A m n : a ij C} is a VS over C. Here [A+αB ij := a ij +αb ij. R S := {functions from S to R} is a VS over R. Here (f +αg)(s) := f(s)+αg(s). R[x= {p(x) p(x) is a real polynomial in x} is a VS over R. R[x;5= {p(x) R[x degree of p(x) 5} is a VS over R. The set {0} is the smallest VS over any field. The set of all real sequences (being R N ) is a VS over R. The set of all bounded real sequences is a VS over R. The set of all convergent real sequences is a VS over R. {(a n ) a n R, a n 0} is a VS over R. Change 0 to??? No. C((a,b),R) := {f : (a,b) R f is continuous} is a VS over R. Indian Institute of Technology Kanpur

2 C 2 ((a,b),r) := {f : (a,b) R f is continuous} is a VS over R. C ((a,b),r) := {f : (a,b) R f infinitely differentiable} is a VS. {f : (a,b) R f +3f +5f = 0} is a VS over R. {A n n a ij R, a = 0} is a VS over R. {A n n a ij R, A upper triangular} is a VS over R. {A M n n (R) A is symmetric} is a VS over R. What about the skew symmetric ones? How about A t = 2A??? Let W and V be VSs over F. We call W a subspace of V if W V. Notation: W V. Th Let V be a VS over F and W V. Then W V if and only if for all x,y W we have αx+y W for all α F.!! {0} and V are subspaces of any VS V, called the trivial subspaces. The VS R over R does not have a nontrivial subspace. Take a subspace W {0}. Then a W s.t. a 0. So αa W for each α R. So W = R. The VS R over Q has many nontrivial subspaces. Two of them are Q[ 2 = {a+b 2 : a,b Q} and Q[ 3 = {a+b 3 : a,b Q}. P. Subspaces Fix a real A m n. The set {Ax : x R n } is a subspace of R m, called the range/column space of A. Our notation: col(a). Note that col(a) is basically the set of all vectors obtained by adding and subtracting real multiples of column vectors of A. Fix a real A m n. The set {x t A : x R m } is a subspace of R n, called the row space of A. Our notation: row(a). Fix a real A m n. The set {x R n : Ax = 0} is a subspace of R n, called the null space of A. Our notation: null(a). col(a), row(a), null(a), null(a t ) are the four fundamental subspaces related to a real/complex matrix A. Let a R 2, a 0. Then, {x a t x = 0} is a nontrivial subspace of R 2. Geometrically, it represents a straight line passing through the origin. In fact, these are the only nontrivial subspaces of R 2. {x R 2 x x 2 = 0} is NOT a subspace of R 2. Important: union of axes; first time union of subspaces is not a subspace. Let x i V, α i F, i =,,k. We call k α i x i a linear combination (lin.comb) of 2

3 x,,x k. The set { k α ix i α i F} is a subspace of V. Let S V. By span of S we mean { m α i x i α i F,x i S}. is R n. Notation: LS(S). This is a subspace of V. Convention: LS( ) = {0}. Lete i = (0,...,0,,0,...,0) t R n withatthei-thcoordinate. Then, LS(e, e 2,...,e n ) P. Linear Span Let x t = [ 2 and y t = [ 2 { [ } 3. Then, ls(x,y) = a[ 2 +b 2 : a,b R 3 {[ } {[ } a+b x = 2a 2b : a,b R = y : 4x y 2z = 0 as 4(a + b) (2a 2b) a+3b z 2(a+3b) = 0. Note that we have obtained conditions on x,y and zsuch that the system a + b = x,2a 2b = y and a+3b = z has a solution for all choices of a,b and c. If U,W are subspaces of V then, U+W := {u+w u U,w W} is a subspace of V, called the sum of U and W. When U W = {0}, it is called the internal direct sum of U and W. Notation: U W. Internal: they are inside V; external: will come later; x-axis + y-axis in R 2, R 3 ; x-axis + xy-plane in R 3 ; xy-plane + xz-plane in R 3. Th Let U,W V. Then, U + W is the smallest subspace of V which contains both U and W.!! Important Is U U+W? Is W U+W? Let X V s.t. U,W X. Is U+W X? If U W and W V, then U V. Let {U t U t V} be a nonempty class. Then t U t is a subspace.!! Let U,W V. Then U W V if and only if either U W or W U.!! Q If U W V, is it necessary that U V? Let V,W be VSs over F. Then V W is a VS over F with (v,w )+(v 2,w 2 ) := (v +v 2,w +w 2 ) and α(v,w) = (αv,αw). Notation: V W. Called external direct sum of V and W. Ex Give a bijection f : R 3 R[x;2 s.t. f(αu+v) = αf(u)+f(v) for each u,v R 3. Take 3 subspaces of R 3. Observe their image in R[x;2. Ex Give vector spaces such that U U 2 U 3. Ex Let U,W V. Then V = U W (int/ext?) if and only if for each v V, unique u U and unique w W s.t. v = u+w. 3

4 {[ x Put V=M 2 (R), W= x 2 x 3 0 Are W,U subspaces of V? Yes. Is V = W+U? Yes. } {[ } x : x i R, U= 0 : x 0 x i R. 2 Is V = W U? No. {[ } x 0 What is the subspace W U? : x R. 0 0 Th Let V be a VS and S V. Let L = {U U V, S U}. Then ls(s) = U.!! U L P. Linear dependency Ex Let A be row-equivalent to B. Then row(a) = row(b). Converse? Let x,...,x k V. If a nonzero lin.comb of x i s becomes 0, then we say x,...,x k are linearly dependent (lin.dep). We say x,...,x k are linearly [ independent, [ if they [ are not lin.dep. In R 2 0, the vectors x =,e 2 = and e 0 2 = are lin.dep, as x e 2e 2 = 0. [ [ [ 3 4 v = 0,v 2 =,v 3 = as 2v +2v 2 v 3 +0v 4 = 0.,v 4 = [ 2 are lin.dep in R 3, [ [ v =,v 2 = are lin.ind in R 3. How? Answering the previous question: the vectors will be linearly dependent (lin.dep) if we can [ find α 0 s.t. α[ β [ [ α β [ 0 +β[ = 0. That is, if = 0. 0 [ [ α 0 Apply GJE, if there are no free variables, then only solution is =. In that case β 0 we conclude lin.ind. [ α If there are free variables, then we have nonzero solutions for. In this case we β conclude[ lin.dep. [ [ 0 v =,v 2 =,v 3 = 0 are lin.dep. Why? As the variable for v 3 will be free. 0 Or 0 v +0 v 2 +αv 3 = 0 for all α R. Th Vectors v,w,z in R 2 are linearly dependent. [ [ α [ v Po. Solving αv+βw+γz = 0 means solving w z 0 β =. v 2 w 2 z 2 0 γ [ v That is, we have to find the RREF of w z 0. v 2 w 2 z 2 0 Applying GJE, as there are only two rows, the number of pivots is at most 2. As the number variables is 3, we will have a free variable. Hence we will have a nonzero solution. 4

5 We call a set S V linearly dependent if contains a finite lin.dep set. Thus is lin.ind and any set containing 0 is lin.dep. So S V is lin.ind if and only if each finite subset of S is lin.ind. Take A m n, m > n. Then R = rref(a) has last row 0. That is, rows of A are lin.dep. How? GJE: R = FA. Then n f mia(i,:) = 0 t. P. Linear dependency: crucial result Th Let S = {v,, } V and T = {w,,w m } ls(s) s.t. m = T > S. Then T is linearly dependent. Po. Write w i = a i v +a i2 v 2 + +a in. So w i = A(i,:). and w a v + +a n a a n v. =. =.... w m a m v + +a mn a m a mn As m > n, we see that rows of A = (a ij ) are lin.dep. That is, there exists α i, not all m zero, s.t. α i A(i,:) = 0 t. Thus m α i w i = m α i v ( A(i,:)) m v v A(i,:).. = α i.. = 0 t.. = 0. v Cor Any n+ vectors in R n is lin.dep. Po. Follows as R n = linear span(e,...,e n ). Th A n n is invertible if and only if columns of A are linearly independent. Po. It is a restatement of A is invertible if and only if Ax = 0 has a unique solution. Th A n n is invertible if and only if rows of A are linearly independent. Po. Follows as A is invertible if and only if A t is invertible. Let u,,u m V be lin.ind. Choose α,...,α m F. Put u = m α iu i. Can you find another set of β i F s.t. u = m β iu i? No. If so, we have (α i β i )u i = 0. Then u i s are lin.dep., A contradiction. Ex u,,u k R n are lin.ind if and only if Au,,Au k are lin.ind for any invertible A n. Ex Let u,v V. Then u,v are lin.ind if and only if u+v,u v are lin.ind. P. Fundamental results: Linear dependence and linear span 5

6 Th Let S V be lin.ind, x V\S. Then S {x} is lin.dep if and only if x ls(s). Po. Let S {x} be lin.dep. Then some m α i s i + α m+ x = 0, where α (the vector of α i s) is not 0. Note: α m+ cannot be 0, otherwise S is lin.dep. So x = m ( α i α m+ )s i. So x ls(s). Conversely, let x ls(s). So x = k α i s i, for some α i F,s i S, not all zero. So x k α i s i = 0. So S {x} is lin.dep. Cor Let S V be lin.ind, x V\S.Then S {x} is lin.ind if and only if x / ls(s).!! Cor Let S V be lin.ind. Then ls(s) = V if and only if each proper superset of S is lin.dep.!! Th Let A = ER, where R = rrefa and E is invertible. Then rows of A corresp. to the pivotal rows of R are lin.ind. Also columns of A corresp. to the pivotal columns of R are lin.ind. Po. Pivotal rows of R are lin.ind due to the pivotal s. Let the pivotal rows of R be R(,:),...,R(k,:) and R be the submatrix formed by these rows. Let A be the submatrix of A which gives R. So R = rref(a ). That is, A = E R, for some invertible E. Suppose there exists x 0 0 s.t. x t 0 A = 0. So x t 0 E R = 0. Note that x t 0 E = y t 0 0. So y t 0R = 0, a contradiction. Other part: note that pivotal columns in R are lin.ind, due to the pivotal s. Let R(:,j ),...,R(:,j k ) be the pivotal columns of R. Since E is invertible, A(:,j ) = ER(:,j ),...,A(:,j k ) = ER(:,j k ) are lin.ind. Th Let x,,x n be lin.dep, x 0. Then, there exists k > s.t. x k is a lin.comb of x,,x k. Po. Consider {x },{x,x 2 },,{x,x 2,,x n } one by one. Take the smallest k > s.t. S k = {x,,x k } is lin.dep. So S k is lin.ind. In that case x k ls(s k ). Let S = {v,...,v k } R n be lin.dep, v 0. One may take help of GJE to find out the first v i which is a lin.comb of the earlier ones and to find a lin.ind subset T S s.t. lst = lss. [ [ [ [ 0 v =, v 2 = 0, v 3 =, v 4 =. A = 2 [ 0 0 rrefa = So v 3 = 2v 2 v ; T = {v,v 2,v 4 } [