Advanced Linear Combinations & Systems of Linear Equations 3

1.4 Linear Combinations Reduction Span Determining Subspaces 3 Spanning Sets in R (cont.) Example   4 Let v1 =  2  2   6 and v2 =  3 . 3 Is Span{v1 , v2 } a line or a plane? v2 is not a multiple of v1 Span{v1 , v2 } =plane through the origin Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 21 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Spanning Sets Example  1 Let A =  3 0 spanned by the Solution:    2 8 1  and b =  3 . Is b in the plane 5 17 columns of A? ? Do x1 and x2 exist so that       1 2 8 x1  3  + x2  1  =  3  0 5 17 Corresponding augmented matrix:       1 2 8 1 2 8 1 2 8  3 1 3  ∼  0 −5 −21  ∼  0 −5 −21  0 5 17 0 5 17 0 0 −4 So b is not in the plane spanned by the columns of A Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 22 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces A Shortcut for Determining Subspaces Theorem (1) If v1 , . . . , vp are in a vector space V , then Span{v1 , . . . , vp } is a subspace of V . Proof: In order to verify this, check properties a, b and c of definition of a subspace. a. 0 is in Span{v1 , . . . , vp } since 0= v1 + v2 + · · · + vp b. To show that Span{v1 , . . . , vp } closed under vector addition, we choose two arbitrary vectors in Span{v1 , . . . , vp } : u =a1 v1 + a2 v2 + · · · + ap vp and v =b1 v1 + b2 v2 + · · · + bp vp . Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 23 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces A Shortcut for Determining Subspaces (cont.) Then u + v = (a1 v1 + a2 v2 + · · · + ap vp ) + (b1 v1 + b2 v2 + · · · + bp vp ) =( v1 + v1 ) + ( v2 + v2 ) + · · · + ( vp + vp ) = (a1 + b1 ) v1 + (a2 + b2 ) v2 + · · · + (ap + bp ) vp . So u + v is in Span{v1 , . . . , vp } . c. To show that Span{v1 , . . . , vp } closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span{v1 , . . . , vp } : v =b1 v1 + b2 v2 + · · · + bp vp . Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 24 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces A Shortcut for Determining Subspaces (cont.) Then cv =c (b1 v1 + b2 v2 + · · · + bp vp ) = v1 + v2 + · · · + vp So cv is in Span{v1 , . . . , vp } . Since properties a, b and c hold, Span{v1 , . . . , vp } is a subspace of V. Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 25 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Determining Subspaces: Recap Recap 1 To show that H is a subspace of a vector space, use Theorem 1. 2 To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 26 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Determining Subspaces: Example Example Is V = {(a + 2b, 2a − 3b) : a and b are real} a subspace of R2 ? Why or why not? Solution: Write vectors in V in column form:       a + 2b a 2b = + 2a − 3b 2a −3b  = 1 2   + 2 −3  So V =Span{v1 , v2 } and therefore V is a subspace of Theorem 1. Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra by Spring, 2015 27 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Determining Subspaces: Example Example     a + 2b  Is H =  a + 1  : a and b are real a subspace of R3 ?   a Why or why not? Solution: 0 is not in H since a = b = 0 or any other combination of values for a and b does not produce the zero vector. So property fails to hold and therefore H is not a subspace of R3 . Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 28 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Determining Subspaces: Example Example  Is the set H of all matrices of the form 2a b 3a + b 3b  a subspace of M2×2 ? Explain. Solution: Since       2a b 2a 0 0 b = + 3a + b 3b 3a 0 b 3b   =a  Therefore H =Span  +b 2 0 3 0  .    0 1 , and so H is a 1 3 subspace of M2×2 . Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 29 / 30 1.4 Linear Combinations Reduction Span Determining Subspaces Spanning Sets Theorem (1.5) The span of any subset S of a vector space V is a subspace of V . Moreover, any subspace of V that contains S must also contain the span of S. Definition The subspace spanned (or subspace generated) by a nonempty set S of vectors in V is the set of all linear combinations of vectors from S: < S >= span(S) = {c1 v1 + · · · + cn vn | ci ∈ F , vi ∈ S} When S = {v1 , · · · , vn } is a finite set, we use the notation < v1 , · · · , vn > or span(v1 , · · · , vn ). A set S of vectors in V is said to span V , or generate V , if V = span(S). Jiwen He, University of Houston Math 4377/6308, Advanced Linear Algebra Spring, 2015 30 / 30