MCV4U1 – UNIT SIX LESSON SEVEN Lesson Seven: Linear Combinations and Spanning Sets Recall that the vectors iˆ, ˆj , and kˆ are called basis vectors. In two dimensions, iˆ and ˆj are called a basis for the plane, and in three dimensions, iˆ, ˆj , and kˆ are called a basis for space. Or we just use the general term “spanning sets.” However, these unit vectors are not the only ones that can make a spanning set. Any two non-collinear/parallel vectors can be a spanning set in two dimensions. Ex. Write u 11,2 as a linear combination of v 1,2 and w 2,1 , thereby showing that 1,2, 2,1 is a spanning set. Solution: Let u kv lw 11,2 k 1,2 l 2,1 If LS=RS, then the individual components must equal too. 11 k 2l 2 2k l 11,2 31,2 42,1 or u 3v 4w Ex. Is Solve to get k 3 and l 4 . 4,2, 6,3 a spanning set? Solution: Since 3 4,2 6,3, the vectors are parallel and therefore cannot be a spanning set. (i.e., they don’t define a two 2 dimensional plane, or surface, they only define a line.) Defines a line, not a surface. Defines a plane Now it gets a bit abstract. In three dimensions, any three non-coplanar vectors can be a spanning set. Ex. Write u 24,30,47 as a linear combination of v 1,3,2 and w 2,4,5, and x 5,3,7 , thereby showing that 1,3,2, 2 4,5, 5,3,7 is a spanning set. Solution: Let u kv lw mx 24,30,47 k 1,3,2 l 2,4,5 m 5,3,7 MCV4U1 – UNIT SIX LESSON SEVEN If LS=RS, then the individual components must equal too. Set up a system of three equation in three unknowns, and solve. 24 k 2l 5m 30 3k 4l 3m 47 2k 4l 7m u 2v 3w 4x and Spanning set in three dimensions. k 2, l 3, m 4 1,3,2, 2 4,5, 5,3,7 is a spanning set. Not a spanning set in three dimensions.