# THE DIMENSION OF A VECTOR SPACE

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1 THE DIMENSION OF A VECTOR SPACE KEITH CONRAD This handout is a supplementary discussion leading up to the definition of dimension and some of its basic properties. Let V be a vector space over a field F. For any subset {v 1,..., v n } of V, its span is the set of all of its linear combinations: Span(v 1,..., v n ) = {c 1 v c n v n : c i F }. In F 3, Span((1, 0, 0), (0, 1, 0)) is the xy-plane in F 3. If v is a single vector in V, Span(v) = {cv : c F } = F v is the set of scalar multiples of v, which should be thought of geometrically as a line (through the origin, since it includes 0 v = 0). Since sums of linear combinations are linear combinations and the scalar multiple of a linear combination is a linear combination, Span(v 1,..., v n ) is a subspace of V. It may not be the whole space, of course. If it is, that is, if every vector in V is a linear combination from {v 1,..., v n }, we say this set spans V or it is a spanning set for V. A subset {w 1,..., w m } of V is called linearly independent when the vanishing of a linear combination only happens in the obvious way: c 1 w c m w m = 0 = all c i = 0. From this condition, one sees that a linear combination of linearly independent vectors has only one possible set of coefficients: (1) c 1 w c m w m = c 1w c mw m = all c i = c i. Indeed, subtracting gives (c i c i )w i = 0, so c i c i = 0 for all i by linear independence. Thus c i = c i for all i. If a subset {w 1,..., w m } of V is not linearly independent, it is called linearly dependent. What does this condition really mean? Well, to be not linearly independent means there is some set of coefficients c 1,..., c m in F not all zero such that (2) c 1 w c m w m = 0. We don t know which c i is not zero. If c 1 0 then we can collect all the other terms except c 1 w 1 on the other side and multiply by 1/c 1 to obtain w 1 = c 2 c 1 w 2 c m c 1 w m Thus w 1 is a linear combination of w 2,..., w m. Conversely, if w 1 = a 2 w a m w m is linear combination of w 2,..., w m then w 1 a 2 w 2 a m w m is a linear combination of all the w s which vanishes and the coefficient of w 1 is 1, which is not zero. Similarly, if c i 0 in (2) then we can express w i as a linear combination of the other w j s, and conversely if w i is a linear combination of the other w j s then we obtain a linear combination of all the w s which vanishes and the coefficient of w i is 1, which is not zero. Therefore the set 1

2 2 KEITH CONRAD {w 1,..., w m } is linearly dependent precisely when at least one of these vectors (we are not saying which, and it could be true for more than one) is a linear combination of the rest. Spanning sets for V and linearly independent subsets of V are in some sense opposite concepts. Any subset of a linearly independent subset is still linearly independent, but this need not be true of spanning sets. Any superset of a spanning set for V is still a spanning set for V, but this need not be true of linearly independent subsets. A subset of V which has both of the above properties is called a basis. That is, a basis of V is a linearly independent subset of V which also spans V. For most vector spaces, there is no God-given basis, so what matters conceptually is the size of a basis rather than a particular choice of basis (compare with the notion of a cyclic group and a choice of generator for it). The following theorem is a first result which links spanning sets in V with linearly independent subsets. Theorem 1. Suppose V {0} and it admits a finite spanning set v 1,..., v n. Some subset of this spanning set is a linearly independent spanning set. The theorem says that once there is a finite spanning set, which could have lots of linear dependence relations, there is a basis for the space. Moreover, the theorem tells us a basis can be found within any spanning set at all. Proof. While {v 1,..., v n } may not be linearly independent, it contains linearly independent subsets, such as any one single nonzero v i. Of course, such small linearly independent subsets can hardly be expected to span V. But consider linearly independent subsets of {v 1,..., v n } which are as large as possible. Reindexing, without loss of generality, we can write such a subset as {v 1,..., v k }. For i = k + 1,..., n, the set {v 1,..., v k, v i } is not linearly independent (otherwise {v 1,..., v k } is not a maximal linearly independent subset). Thus there is some linear relation c 1 v c k v k + c i v i = 0, where the c s are in F are not all of them are 0. The coefficient c i cannot be zero, since otherwise we would be left with a linear dependence relation on v 1,..., v k, which does not happen due to their linear independence. Since c i 0, we see that v i is in the span of v 1,..., v k. This holds for i = k + 1,..., n, so any linear combination of v 1,..., v n is also a linear combination of just v 1,..., v k. As every element of V is a linear combination of v 1,..., v n, we conclude that v 1,..., v k spans V. By its construction, this is a linearly independent subset of V as well. Notice the non-constructive character of the proof. If we somehow can check that a (finite) subset of V spans the whole space, Theorem 1 says a subset of this is a linearly independent spanning set, but the proof is not constructively telling us which subset of {v 1,..., v n } this might be. Theorem 1 is a top-down theorem. It says any (finite) spanning set has a linearly independent spanning set inside of it. It is natural to ask if we can go bottom-up, and show any linearly independent subset can be enlarged to a linearly independent spanning set. Something along these lines will be proved in Theorem 3. Lemma 1. Suppose {v 1,..., v n } spans V, where n 2. Pick any v V. If some v i is a linear combination of the other v j s and v, then V is spanned by the other v j s and v.

3 THE DIMENSION OF A VECTOR SPACE 3 For example, if V is spanned by v 1, v 2, and v 3, and v 1 is a linear combination of v, v 2, and v 3, where v is another vector in V, then V is spanned by v, v 2, and v 3. Lemma 1 should be geometrically reasonable. See if you can prove it before reading the proof below. Proof. Reindexing if necessary, we can suppose it is v 1 which is a linear combination of v, v 2,..., v n. We will show every vector in V is a linear combination of v, v 2,..., v n, so these vectors span V. Pick any w V. By hypothesis, w = c 1 v 1 + c 2 v c n v n for some c i F. Since v 1 is a linear combination of v, v 2,..., v n, we feed this linear combination into the above equation to see w is a linear combination of v, v 2,..., v n. As w was arbitrary in V, we have shown V is spanned by v, v 2,..., v m. The following important technical result relates spanning sets for V and linearly independent subsets of V. It is called the exchange theorem. The name arises from the process in its proof, which relies on repeated applications of Lemma 1. Theorem 2 (Exchange Theorem). Suppose V is spanned by n vectors, where n 1. Any linearly independent subset of V has at most n vectors. If you think about linear independence as degrees of freedom, the exchange theorem makes sense. What makes the theorem somewhat subtle to prove is that the theorem bounds the size of any linearly independent subset once we know the size of one spanning set. Most linearly independent subsets of V are not directly related to the original choice of spanning set, so linking the two sets of vectors is tricky. The proof will show how to link linearly independent sets and spanning sets by an exchange process, one vector at a time. Proof. First, let s check the result when n = 1. In this case, V = F v for some v (that is, V is spanned by one vector). Two different scalar multiples of v are linearly dependent, so a linearly independent subset of V can have size at most 1. Now we take n 2. We give a proof by contradiction. If the theorem is false, then V contains a set of n + 1 linearly independent vectors, say w 1,..., w n+1. Step 1: We are told that V can be spanned by n vectors. Let s call such a spanning set v 1,..., v n. We also have the n + 1 linearly independent vectors w 1,..., w n+1 in V. Write the first vector from our linearly independent set in terms of our spanning set: w 1 = c 1 v 1 + c 2 v c n v n for some c i F. Since w 1 0 (a linearly independent set never contains the vector 0), some coefficient c j is nonzero. Without loss of generality, we can reindex the v s so that c 1 is nonzero. Then the above equation can be solved for v 1 as a linear combination of w 1, v 2,..., v n. By Lemma 1, (3) V = Span(w 1, v 2,..., v n ). Notice that we have taken one element from the initial spanning set out and inserted an element from the linearly independent set in its place, retaining the spanning property. Step 2: Let s repeat the procedure, this time using our new spanning set (3). Write w 2 in terms of this new spanning set: (4) w 2 = c 1w 1 + c 2v c nv n

4 4 KEITH CONRAD for some c i in F. We want to use this equation to show w 2 can be inserted into (3) and one of the original vectors can be taken out, without destroying the spanning property. Some care is needed, because we want to keep w 1 in the spanning set rather than accidentally swap it out. (This is an issue that we did not meet in the first step, where no new vectors had yet been placed in the spanning set.) Certainly one of c 1, c 2,..., c n is nonzero, since w 2 is nonzero. But in fact we can say something a bit sharper: regardless of the value of c 1, one of c 2,..., c n is nonzero. Indeed, if c 2,..., c m are all zero, then w 2 = c 1 w 1 is a scalar multiple of w 1, and that violates linear independence (as {w 1,..., w m } is linearly independent, so is the subset {w 1, w 2 }). Without loss of generality, we can reindex v 2,..., v n so it is c 2 which is nonzero. Then we can use (4) to express v 2 as a linear combination of w 1, w 2, v 3,..., v n. By another application of Lemma 1, using our new spanning set in (3) and the auxiliary vector w 2, it follows that V = Span(w 1, w 2, v 3,..., v n ). Step 3: Now that we see how things work, we argue inductively. Suppose for some k between 1 and n 1 that we have shown V = Span(w 1,..., w k, v k+1,..., v n ). (This has already been checked for k = 1 in Step 1, and k = 2 in Step 2, although Step 2 is not logically necessary for what we do; it was just included to see concretely the inductive step we now carry out for any k.) Using this spanning set for V, write (5) w k+1 = a 1 w a k w k + a k+1 v k a n v n with a i F. One of a k+1,..., a n is nonzero, since otherwise this equation expresses w k+1 as a linear combination of w 1,..., w k, and that violates linear independence of the w s. Reindexing v k+1,..., v n if necessary, we can suppose it is a k+1 which is nonzero. Then (5) can be solved for v k+1 as a linear combination of w 1,..., w k, w k+1, v k+2,..., v n. By Lemma 1, using the spanning set {w 1,..., w k, v k+1,..., v n } and the auxiliary vector w k+1, we can swap w k+1 into the spanning set in exchange for v k+1 without losing the spanning property: V = Span(w 1,..., w k, w k+1, v k+2,..., v n ). We have added an new vector to the spanning set and taken one of the original vectors out. Now by induction (or, more loosely, repeating this step n k 1 more times ), we arrive at the conclusion that (6) V = Span(w 1,..., w n ). However, we were starting with n + 1 linearly independent vectors w 1,..., w n+1, so w n+1 is not in the span of w 1,..., w n. That contradicts the meaning of (6), which says every vector in V is a linear combination of w 1,..., w n. We have reached a contradiction, so no linearly independent subset of V contains more than n vectors, where n is the size of a spanning set for V. Example 1. Consider M 3 (R), the 3 3 real matrices. It is a vector space over R under matrix addition and the usual multiplication of a matrix by a real number. This vector space has a 9-element spanning set, namely the 9 matrices with a 1 in one component and 0 elsewhere. Therefore any linearly independent subset of M 3 (R) has at most 9 elements in it.

5 THE DIMENSION OF A VECTOR SPACE 5 Corollary 1. Suppose V {0} and V admits a finite basis. Any two bases for V have the same size. Proof. Let {v 1,..., v n } and {v 1,..., v m} be bases for V. Treating the first set as a spanning set for V and the second set as a linearly independent subset of V, the exchange theorem tells us that m n. Reversing these roles (which we can do since bases are both linearly independent and span the whole space), we get n m. Thus m = n. The (common) size of any basis of V is called the dimension of V (over F ). Example 2. From the known bases of R n and M n (R), these spaces have dimension n and n 2 over R, respectively. Example 3. The dimension of C as a real vector space is 2, with one basis being {1, i}. Example 4. The vector space {0} has no basis, or you might want to say its basis is the empty set. In any event, it is natural to declare the zero vector space to have dimension 0. Theorem 3. Let V be a vector space with dimension n 1. Any spanning set has at least n elements, and contains a basis inside of it. Any linearly independent subset has at most n elements, and can be extended to a basis of V. Finally, an n-element subset of V is a spanning set if and only if it is a linearly independent set. Proof. Since V has a basis of n vectors, let s pick such a basis, say v 1,..., v n. We will compare this basis to the spanning sets and the linearly independent sets in V to draw our conclusions, taking advantage of the dual nature of a basis as both a linearly independent subset of V and as a spanning set for V. If {u 1,..., u k } is a spanning set for V, then a comparison with {v 1,..., v n } (interpreted as a linearly independent subset of V ) shows n k by the exchange theorem. Equivalently, k n. Moreover, Theorem 1 says that {u 1,..., u k } contains a basis for V. This settles the first part of the theorem. For the next part, suppose {w 1,..., w m } is a linearly independent subset of V. A comparison with {v 1,..., v n } (interpreted as a spanning set for V ) shows m n by the exchange theorem. To see that the w s can be extended to a basis of V, apply the exchange process from the proof of the exchange theorem, but only m times since we have only m linearly independent w s. We find at the end that V = Span(w 1,..., w m, v m+1,..., v n ), which shows the w s can be extended to a spanning set for V. This spanning set contains a basis for V, by Theorem 1. Since all bases of V have n elements, this n-element spanning set must be a basis itself. Taking m = n in the previous paragraph shows any n-element linearly independent subset is a basis (and thus spans V ). Conversely, any n-element spanning set is linearly independent, since any linear dependence relation would let us cut down to a spanning set of fewer than n elements, but that violates the first result in this proof: a spanning set for an n-dimensional vector space has at least n elements. Theorem 4. If V is an n-dimensional vector space, any subspace W is finite-dimensional, with dimension at most n. Proof. This theorem is trivial for the zero vector space, so we may assume V {0}, i.e., n 1.

6 6 KEITH CONRAD Any linearly independent subset of W is also a linearly independent subset of V, and thus has size at most n by Theorem 3. Choose a linearly independent subset {w 1,..., w m } of W where m is maximal. Then m n. We will show Span(w 1,..., w m ) = W. For any w W, the set {w, w 1,..., w m } has more than m elements, so it can t be linearly independent. Therefore there is some vanishing linear combination aw + a 1 w a m w m = 0 where a, a 1,..., a m are in F and are not all 0. If a = 0 then the a i s all vanish since w 1,..., w m are linearly independent. Therefore a 0, so we can solve for w: w = a 1 a w 1 a m a w m. Thus w is a linear combination of w 1,..., w m. Since w was arbitrary in W, this shows the w i s span W. So {w 1,..., w m } is a spanning set for W which is linearly independent by construction. This proves W is finite-dimensional with dimension m n. Theorem 5. If V has dimension n and W is a subspace with dimension n, then W = V. Proof. When W has dimension n, any basis for W is a linearly independent subset of V with n elements, so it spans V by Theorem 3. The span is also W (by definition of a basis for W ), so W = V. It is really important that throughout our calculations (expressing one vector as a linear combination of others when we have a nontrivial linear combination of vectors equal to 0) we can scale a nonzero coefficient of a vector to make the coefficient equal to 1. For example, suppose we tried to do linear algebra over the integers Z instead of over a field. Then we can t scale a coefficient in Z to be 1 without possibly needing rational coefficients for other vectors in a linear combination. That suggests results like the ones we have established for vector spaces over fields might not hold for vector spaces over Z. And it s true: linear algebra over Z is more subtle than over fields. For example, Theorem 5 is false if we work with vector spaces over Z. Consider the integers Z and the even integers 2Z. By any reasonable definitions, both Z and 2Z should be considered one-dimensional over the integers, where Z has basis {1} and 2Z has basis {2} (since every integer is a unique integral multiple of 2). But 2Z Z, so a one-dimensional vector space over Z can lie inside another without them being equal. This is a pretty clear failure of Theorem 5 when we use scalars from Z instead of from a field.

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