MATH 304Linear AlgebraLecture 17:Basis and dimension (continued).Rank of a matrix.BasisDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Equivalently, a subset S ⊂ V is a basis for V if anyvector v ∈ V is uniquely represented as a linearcombinationv = r1v1+ r2v2+ · · · + rkvk,where v1, . . . , vkare distinct vectors from S andr1, . . . , rk∈ R.DimensionTheorem 1 Any vector space has a basis.Theorem 2 If a vector space V has a finite basis,then all bases for V are finite and have the samenumber of elements.Definition. The dimension of a vector space V ,denoted dim V , is the number of elements in any ofits bases.Examples. • dim Rn= n• Mm,n(R): the space of m×n matrices; dim Mm,n= mn• Pn: polynomials of degree less than n; dim Pn= n• P: the space of all polynomials; dim P = ∞• {0}: the trivial vector space; dim {0} = 0How to find a basis?Theorem Let V be a vector space. Then(i) any spanning set for V contains a basis;(ii) any linearly independent subset of V iscontained in a basis.Approach 1. Get a spanning set for the vectorspace, then reduce this set to a basis.Approach 2. Get a linearly independent set, thenextend it to a basis.Problem. Find a basis for the ve ctor space Vspanned by ve ctors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).To pare this spanning set, we need t o find a relationof the form r1w1+r2w2+r3w3+r4w4= 0, whereri∈ R are not all equal to zero. Equivalently,1 0 2 11 1 3 10 1 1 1r1r2r3r4=000.To solve this system of linear equations forr1, r2, r3, r4, we apply row reduction.1 0 2 11 1 3 10 1 1 1→1 0 2 10 1 1 00 1 1 1→1 0 2 101 1 00 0 01→1 0 2 001 1 00 0 01(reduced row echelon form)r1+ 2r3= 0r2+ r3= 0r4= 0⇐⇒r1= −2r3r2= −r3r4= 0General solution: (r1, r2, r3, r4)=(−2t, −t, t, 0), t ∈ R.Particular solution: (r1, r2, r3, r4) = (2, 1, −1, 0).Problem. Find a basis for the ve ctor space Vspanned by ve ctors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).We have obtained t hat 2w1+ w2− w3= 0.Hence any of vectors w1, w2, w3can be dropped.For instan ce, V = Span(w1, w2, w4).Let us check whether vectors w1, w2, w4arelinearly independent:1 0 11 1 10 1 1=1 0 11 1 00 1 0=1 10 1= 1 6= 0.They are!!! It follows that V = R3and{w1, w2, w4} is a basis for V .Row space of a matrixDefinition. The row space of an m×n matrix A isthe subspace of Rnspanned by rows of A.The dimension of the row space is called th e rankof the matrix A.Theorem 1 The rank of a matrix A is the maximal numberof linearly independent rows in A.Theorem 2 Elementary row operations do not change therow space of a matrix.Theorem 3 If a matrix A is in row echelon form, then thenonzero rows of A are linearly independent.Corollary The rank of a matrix is equal to the numb er ofnonzero rows in its row echelon form.Theorem Elementary row operations do notchange the row space of a matrix.Proof: Suppose that A and B are m×n matrices such that Bis obtained from A by an elementary row operation. Leta1, . . . , ambe the rows of A and b1, . . . , bmbe the rows of B.We have to show that Span(a1, . . . , am) = Span(b1, . . . , bm).Observe that any row biof B belongs to Span(a1, . . . , am).Indeed, either bi= ajfor some 1 ≤ j ≤ m, or bi= r aiforsome scalar r 6= 0, or bi= ai+rajfor some j 6= i and r ∈ R.It follows that Span(b1, . . . , bm) ⊂ Span(a1, . . . , am).Now the matrix A can also be obtained from B by anelementary row operation. By the ab ove,Span(a1, . . . , am) ⊂ Span(b1, . . . , bm).Problem. Find the rank of the matrixA =1 1 00 1 12 3 11 1 1.Elementary row operations do not chan ge the rowspace. Let us convert A to row echelon form:1 1 00 1 12 3 11 1 1→1 1 00 1 10 1 11 1 1→1 1 00 1 10 1 10 0 11 1 00 1 10 1 10 0 1→1 1 00 1 10 0 00 0 1→1 1 001 10 010 0 0Vectors (1, 1, 0), (0, 1, 1), and (0, 0, 1) form a basisfor the row space of A. Thus the rank of A is 3.It follows that the row space of A is the entire spaceR3.Problem. Find a basis for the ve ctor space Vspanned by ve ctors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).The vector spac e V is the row space of a matrix1 1 00 1 12 3 11 1 1.According to the solution of the previous problem,vectors (1, 1, 0), (0, 1, 1), and (0, 0, 1) form a basisfor V .Column space of a matrixDefinition. The column space of an m×n matrixA is the subspace of Rmspanned by columns of A.Theorem 1 The column space of a matrix A coincides withthe row space of the transpose matrix AT.Theorem 2 Elementary column operations do not changethe column space of a matrix.Theorem 3 Elementary row operations do not change thedimension of the column space of a matrix (although they canchange the column space).Theorem 4 For any matrix, the row space and the columnspace have the same dimension.Problem. Find a basis for the column space of thematrixA =1 1 00 1 12 3 11 1 1.The column space of A coincides with the row space of AT.To find a basis, we convert ATto row echelon form:AT=1 0 2 11 1 3 10 1 1 1→1 0 2 10 1 1 00 1 1 1→1 0 2 101 1 00 0 0 1Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis forthe column space of A.Problem. Find a basis for the column space of thematrixA =1 1 00 1 12 3 11 1 1.Alternative solution: We already know from aprevious problem t hat the rank of A is 3. It followsthat the columns of A are linearly indepe ndent .Therefore these columns form a basis for thecolumn
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