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Math 221, Section 2.5 and 2.6, Part 1 (1) Subspaces (2) Basis and dimension 1 / 16 Subspace Recall from Sec 1.3: for example in R3 , h1i I Take u = 2 then 3 h1i Span{u} = {s 2 |where s is a scalar} 3 I is a 1-dimensional subspace (a line). h1i h1i Take u = 2 and v = 0 , then u, v are linearly independent 2 3 Span{u, v} = {s I h1i 2 3 +t h1i 0 2 |where s, t are scalars} is a 2-dimensional subspace (a plane). h3i Take w = 2 , then 7 [uvw] = h1 1 3i 202 327 → ··· → h1 0 1i 012 0 0 0 RREF , implies that w = 1u + 2v, and Span{u, v, w} = Span{u, v}. 2 / 16 Defining properties of subspace I A subspace is a subset H of Rn that has three properties. I I I The zero vector 0 is in H. If u, v are in H, then the sum u + v is in H. If u is in H and c is a scalar, then the scalar multiple cu is in H. In words, a subspace is closed under addition and scalar multiplication 3 / 16 I Verify that: H = Span{v1 , v2 } is a subspace of Rn . I I I 0 is in H, because 0 = 0v1 + 0v2 is a linear combination of v1 , v2 . If u = av1 + bv2 and v = cv1 + dv2 , then u + v = (a + c)v1 + (b + d)v2 which shows that u + v is a linear combination of v1 , v2 . If u = av1 + bv2 , then cu = c(av1 + bv2 ) = cav1 + cbv2 which shows that cu is a linear combination of v1 , v2 . 4 / 16 Examples I Rn itself is a subspace because it has the three properties required for a subspace. I The zero subspace {0}, consisting of only the zero vector in Rn , is a subspace. I The subspace spanned by v1 , . . . , vr , Span{v1 , . . . , vr } = {all linear combinations of v1 , . . . , vr } is a subspace of Rn . (The verification of this statement is similar to the argument given on the previous page.) 5 / 16 Non-examples I A line L not passing through the origin is not a subspace, because it does not contain the origin, as required. I Fig. 2 shows that L is not closed under addition or scalar multiplication. 6 / 16 Non-examples I In general, a translation of a subspace away from the origin by a non-zero vector p, which is p + Span{v1 , . . . , vr } is not a subspace of Rn . 7 / 16 Basis: the smallest spanning subset I Because a subspace contains an infinite number of vectors, some problems involving a subspace are handled best by working with a small finite subset of vectors that span the subspace. I The fewer the spanning vectors, the better. I The smallest possible spanning set must be linearly independent. I (Recall from Sec 1.7, if we put the vector together to form a matrix A, then all columns of REF (A) contains a pivot.) I h1i h1i h3i Example: again take u = 2 , v = 0 , w = 2 . 3 2 7 We checked that w = u + 2v, so H = Span{u, v, w} = Span{u, v} 8 / 16 Basis and dimension I A basis for a subspace H of Rn is a linearly independent subset in H that spans H. I I Note: Given a subspace, there are more than one choices of bases. The dimension of a subspace H is the number of vectors in any basis of H. I Although the choices of bases are different, but they all have the same number, which must be the dimension. 9 / 16 Basis and dimension Example: again take the vectors in Sec 1.3. h3i h1i h1i I Take u = 2 , v = 0 , w = 2 . 3 I 2 7 We checked that w = u + 2v, so H = Span{u, v, w} = Span{u, v} I The subset {u, v, w} is linearly dependent, so this subset is not a basis of H. I The subset {u, v} is linearly independent, so this subset is a basis of H, and dim(H) = 2. 10 / 16 Basis and dimension Other examples I The columns of an n × n identity matrix are the standard vectors denoted by e1 , . . . , en . h1i h0i h0i I E.g. in R3 , e1 = I e1 , e2 , e3 are linearly independent. Hence they form a basis for the subspace h1i h0i h0i Span{e1 , e2 , e3 } = {x 0 + y 1 + z 0 |where x, y , z are scalars} 0 0 1 hx i = { y |where x, y , z are scalars} I 0 0 , e2 = 1 0 , e3 = 0 1 . z It is the whole R3 . I The set e1 , . . . , en is called the standard basis for Rn . 11 / 16 Basis and dimension Other subset of n vectors in Rn form a basis, as long as they are linearly independent. h1i h1i h1i I Take u = 2 , v = 0 , w = 2 . 3 I 2 5 We checked that u, v, w are linearly independent. So H = Span{u, v, w} is a subspace in R3 . I dim(H) is equal to the number of vectors which is 3, so it must be the whole R3 . 12 / 16 Invertible matrices and basis I Suppose you have r linearly independent vectors {v1 , . . . , vr } in Rn . I If r ≤ n, then they form a basis of Span{v1 , . . . , vr }, which is an r -dimensional subspace in Rn If r = n, then I I I Span{v1 , . . . , vn } = Rn . Combining with the Invertible Matrices Theorem (a.⇔ e.), if A = [v1 . . . vn ], then for {v1 , . . . , vn } forming a basis of Rn , A must be invertible. 13 / 16 Invertible matrices and basis Examples I Previous examples I I I 1 2 3 1 2 3 form 1 1 0 2 is invertible, so the columns form a basis in R3 . 2 5 1 3 0 2 is non-invertible/singular, so the columns do not 2 7 a basis in R3 . (Sec 1.8, Ex 15) Are the columns of The REF has 4 −2 16 −3 form a basis in R2 ? pivots, so the columns a basis for R2 . 14 / 16 Invertible matrices and basis Examples I (Sec 1.8, Ex 17) 0 0 Are the columns of −2 5 6 0 3 form a basis in R3 ? 4 2 The REF has pivots, so the I (Sec 1.8, Ex 18) 1 3 Are the columns of 1 −1 −3 2 columns a basis for R3 . 5 1 form a basis in R3 ? −4 The REF has pivots, so the columns a basis for R3 . I (Sec 1.8, Ex 19) 3 6 Are the columns of −8 2 form a basis in R3 ? 1 −5 No, because 15 / 16 Quiz 4 The bold numbers indicate similar types of questions that may appear in Quiz 4. I Sec 2.2: Practice problem 2, Exercises 1,3,5,8,9a-d,10,13, 17, 29, 31 (use the algorithm I showed in class), 35 I Sec 2.3: Practice problems 1,2,3, Exercises 1,3,5, 11, 12, 13, 14, The following questions require the big Invertible Matrix Theorem and some thinking. They will not be in the quiz, but some of them could be in the next midterm and the final. You are strongly suggested to think about them. I Sec 2.2: 21, 22, 23, 24, I Sec 2.3: 15, 17, 18, 19, 20, 21, 24, 35, 36, 39 You may check the solution at: http://www.slader.com/textbook/9780321385178-linear-algebraand-its-applications-4th-edition/ 16 / 16