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Math 221, Section 1.7 (1) Linear dependence/independence (2) Example with 1-3 vectors (3) When L.D. is automatic 1 / 13 Motivation I Remember the quiz questions: Given three non-zero vectors u, v, w in R3 , then I I I Span{u} is a line (1-dimensional), but is Span{u, v} a plane (2-dimensional) passing through the origin? is Span{u, v, w} equal to R3 (3-dimensional)? I The answers are YES sometimes, but NO in general. I The NO answers can be explained using linear combinations in Sec 1.3. I Today we will see under a special condition (linear independence) on {u, v, w}, the vectors can span a subspace with expected dimension, giving YES answers to above. 2 / 13 Linear dependence: 1 vector I A set of one vector {v} are said to be I I I Geometrically, it means that I I I {v} is linearly independent if Span{v} = {xv|x is a scalar} is a line. {0} is linearly dependent because Span{0} = 0 is a point, not a line. Arithmetically, it means that: I I I linearly independent if it is a non-zero vector, linearly dependent if it is the zero vector 0. {v} is linearly independent if the equation xv = 0 has only trivial solution x = 0. The zero vector 0 is linearly dependent because the equation x0 = 0 has many nontrivial solutions for x (any scalar is a solution). The example of one vector is not very interesting. 3 / 13 Linear dependence: 2 vectors A set of two vectors {u, v} are said to be I linearly independent if they are not multiple of each other. I linearly dependent if one is a multiple of another. For example, in R2 , I u = [ 32 ] and v = [ 62 ] are linearly independent. I u0 = [ 31 ] and v0 = [ 62 ] are linearly dependent because v0 = 2u0 . Geometrically, two vectors are linearly dependent if and only if they lie on the same line through the origin. 4 / 13 Linear dependence: 2 vectors Arithmetically, there is another way to describe this. I u, v are linearly independent if the equation xu + y v = 0 has only trivial solution [ yx ] = [ 00 ]. Check: I u0 , v0 linearly dependent if the equation xu0 + y v0 = 0 has non-trivial solutions. Check: This arithmetic definition will be used for describing 3 or more vectors to be linear independent or not. 5 / 13 Linear dependence: 3 vectors Three vectors {u, v, w} are said to be I linearly independent if the homogeneous system xu + y v + zw = 0 hx i h0i has only trivial solution y = 0 . z I 0 linearly dependent if the above equation has non-trivial solutions. 6 / 13 Example 1 in the book u = I I I h2i h4i h1i 2 , v = 5 , w = 1 . 3 0 The equation xu + y v + zw = 0 is a homogeneous system, h1 4 2i Form the coefficient matrix A = [uvw] = 2 5 1 . 360 Apply Gaussian elimination. h1 4 2i 251 → 360 I 6 → h1 4 2i 011 0 0 0 REF . Infinitely many solutions ⇒ non-trivial solution exists ⇒ {u, v, w} are linearly dependent 7 / 13 If you proceed to find the RREF h1 4 2i 011 → → 0 0 0 REF You can solve for the solution hx i y z =z h 2 −1 1 h 1 0 −2 i 01 1 00 0 RREF , i . Therefore, 2u − 1v + 1w = 0, or w = −2u + 1v, (Think: what if I write a vertical line separating the last column from the others. Which equation is it?) Geometrically, this means that w is in Span {u, v}, or that w lies in the plane generated by {u, v}. 8 / 13 Example 1 in the book u = I I I h5i h 0 , v = 0 7 2 −6 i h ,w= i . The equation xu + y v + zw = 0 is a homogeneous system, h5 7 9 i Form the coefficient matrix A = [uvw] = 0 2 4 . 0 −6 −8 Apply Gaussian elimination. h5 7 9 i 0 2 4 → 0 −6 −8 I 9 4 −8 → h 1 7/5 9/5 i 0 1 0 0 2 1 . REF unique solution ⇒ solution must be trivial ⇒ {u, v, w} are linearly independent . 9 / 13 Summary I To check whether {u, v, w} are linearly independent or not: (1) Form the matrix A = [uvw] (2) Apply Gaussian elimination on A (3) In the REF of A, if I I non-pivotal column exists ⇒ linearly dependent all columns are pivotal ⇒ linearly independent I In the linearly dependent case, to further find out the scalars s, t such that w = su + tv, read the entries from the non-pivotal column. I If a set of vectors are linearly independent, they span a subspace of dimension equal to the number of vectors in the set. 10 / 13 Some automatic situations The next two theorems describe special cases in which the linear dependence of a set is automatic. I (Theorem 8) If a set contains more vectors than there are entries in each vector, then the set is linearly dependent. I (Ex 15) [ 51 ] , [ 28 ] , [ 13 ] , 5 2 1 −1 → 183 7 −1 7 are linearly dependent. → [ 10 01 ]REF . the last two columns are non-pivotal, so there exist non-trivial solution, which means: linearly dependent. I (Reason: If A has more columns than rows, then it must have non-pivotal columns) I (In fact, if you understand the reason, then Gaussian elimination is not even necessary.) 11 / 13 Some automatic situations I (Theorem 9) If a set of vectors contains the zero vector, then the set is linearly dependent. I (Ex 17) h 5 −3 −1 i h 0 i h −7 i , 0 , 2 are linearly dependent, because 0 4 h 5 i h0i h −7 i h 0 i 0 −3 + y 0 + 0 2 = 0 −1 0 4 0 for every y , and we may take y to be non-zero. I (Again, if you understand the reason, then Gaussian elimination is not even necessary.) 12 / 13 Appendix: General definition A set of vectors {v1 , v2 . . . , vn } are said to be I linearly independent if the homogeneous system x1 v1 + x2 v2 + · · · + xn vn = 0 has only trivial solution x1 = x2 = · · · = xn = 0. I linearly dependent if the above equation has non-trivial solutions. 13 / 13