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12 12.1 Basis and dimension of subspaces The concept of basis Example: Consider the set 1 2 0 , . , S= 2 −1 1 Then span(S) = R2 . (Exercise). In fact, any two of the elements of S span R2 . (Exercise). So we can throw out any one of them, for example, the second one, obtaining the set 1 2 . Sb = , 2 −1 And this smaller set Sb also spans R2 . (There are two other possibilities for subsets of S that also span R2 .) But we can’t discard an element of Sb and still span R2 with the remaining one vector. b leaving us with the set (Why not? Suppose we discard the second vector of S, 1 S̃ = . 2 Now span(S̃) consists of all scalar multiples of this single vector (a line through 0). But anything not on this line, for instance the vector 1 v= 0 is not in the span. So S̃ does not span R2 .) What’s going on here is simple: in the first instance, the three vectors in S are linearly dependent, and any one of them can be expressed as a linear combination of the remaining b we have a linearly independent set, and two. Once we’ve discarded one of these to obtain S, if we throw away one of these, the span changes.1 1 We use the word “span” in two ways: if V = spanS, then we say that S spans the subspace V . 1 This gives us a way, starting with a more general set S, to discard “redundant” vectors one by one until we’re left with a set of linearly independent vectors which still spans the original set: If S = {e1 , . . . , em } spans the subspace V but is linearly dependent, we can express one of the elements in S as a linear combination of the others. By relabeling if necessary, we suppose that em can be written as a linear combination of the others. Then span(S) = span(e1 , . . . , em−1 ). Why? If the remaining m − 1 vectors are still linearly dependent, we can repeat the process, writing one of them as a linear combination of the remaining m − 2, relabeling, and then span(S) = span(e1 , . . . , em−2 ). We can continue this until we arrive finally at a “minimal” spanning set, say {e1 , . . . , ek } which is linearly independent. Such a set will be called a basis for V : Definition: The set B = {e1 , . . . , ek } is a basis for the subspace V if • span(B) = V . • The set B is linearly independent. Remark: In definitions like that given above, we really should put ”iff” (if and only if) instead of just ”if”, and that’s the way you should read it. More precisely, if B is a basis, then B spans V and is linearly independent. Conversely, if B spans V and is linearly independent, then B is a basis. Examples: • In R3 , the set 1 0 B = 0 , 1 , 0 0 2 0 0 = {e1 , e2 , e3 } 1 is a basis.. Why? (a) Any vector a v= b c in R3 can be written as v = ae1 + be2 + ce3 , so B spans R3 . And (b): if c1 e1 + c2 e2 + c3 e3 = 0, then c1 0 c2 = 0 , c3 0 which means that c1 = c2 = c3 = 0, so the set is linearly independent. Definition: The set {e1 , e2 , e3 } is called the standard basis for R3 . • The set 1 3 −1 S= , , −2 1 1 is linearly dependent. Any two elements of S are linearly dependent and form a basis for R2 . Verify this! Exercises: 1. Any 4 vectors in R3 are linearly dependent and therefore do not form a basis. You should be able to supply the argument, which amounts to showing that a certain homogeneous system of equations has a nontrivial solution. 2. No 2 vectors can span R3 . Why not? 3. The vector 0 is never part of a basis. 4. If a set B is a basis for R3 , then it contains exactly 3 elements. This has mostly been done in the first two parts, but put it all together. 5. (**) Prove that any basis for Rn has precisely n elements. 3 Example: Find a basis for the null space of the matrix 1 0 0 3 2 A = 0 1 0 1 −1 0 0 1 2 3 . Solution: Since A is already in Gauss-Jordan form, we can just write down the general solution to the homogeneous equation. These vectors are precisely the elements of the null space of A. We have, setting x4 = s, and x5 = t, x1 = −3s − 2t −s + t x2 = x3 = −2s − 3t , x4 = s x5 = t so the general solution to Ax = 0 is given by Null(A) = {sv1 + tv2 s, t ∈ R}, where −2 −3 1 −1 v1 = −2 , and v2 = −3 . 0 1 1 0 It is obvious 2 by inspection of the last two entries in each that the set B = {v1 , v2 } is linearly independent. Furthermore, by construction, the set B spans the null space. So B is a basis. 12.2 Dimension As we’ve seen above, any basis for Rn has precisely n elements. Although we’re not going to prove it here, the same property holds for any subspace of Rn : the number of elements in any basis for the subspace is the same. Given this, we make the following 2 When we say it’s “obvious” or that something is “clear”, we mean that it can easily be proven; if you can do the proof in your head, fine. Otherwise, write it out. 4 Definition: Let V 6= {0} be a subspace of Rn for some n. The dimension of V , written dim(V ), is the number of elements in any basis of V . Examples: • dim(Rn ) = n. Why? • For the matrix A above, the dimension of the null space of A is 2. • The subspace V = {0} is a bit peculiar: it doesn’t have a basis according to our definition, since any subset of V is linearly independent. We extend the definition of dimension to this case by defining dim(V ) = 0. Exercises: 1. Show that the dimension of the null space of any matrix R in reduced echelon form is equal to the number of free variables in the echelon form. 2. Show that the dimension of the set {(x, y, z) such that 2x − 3y + z = 0} is two. 5