Basis and dimension

Result 1. A basis spans V, so weights exist; it is independent, so they are unique (Result 6 of Linear independence).

Result 2. Let V = \operatorname{span}\{\mathbf{w}_1, \dots, \mathbf{w}_m\}, and let \mathbf{u}_1, \dots, \mathbf{u}_k in V have k > m. Write each \mathbf{u}_j = a_{1j}\mathbf{w}_1 + \dots + a_{mj}\mathbf{w}_m. The homogeneous system a_{i1}c_1 + \dots + a_{ik}c_k = 0, for i = 1, \dots, m, has more unknowns than equations: its k coefficient columns lie in \mathbb{R}^m, so they are dependent (Result 4 of Linear independence), and it has a solution with c_1, \dots, c_k not all zero. Collecting the terms in each \mathbf{w}_i,

So the \mathbf{u}_j are dependent, and an independent list in V has k \leq m.

Result 3. Take two bases of V, with p and q vectors. The first spans V and the second is independent in V, so q \leq p by Result 2; exchanging their roles gives p \leq q. This There exists, and there is exactly one shows the two numbers equal by bounding each by the other. The standard basis gives \dim \mathbb{R}^n = n.

A lemma for growing a list. If \mathbf{u}_1, \dots, \mathbf{u}_j is independent and \mathbf{w} is not in its span, then \mathbf{u}_1, \dots, \mathbf{u}_j, \mathbf{w} is independent. In a relation with weight d on \mathbf{w}, d \neq 0 would let us solve for \mathbf{w} as a combination of the \mathbf{u}_i; so d = 0, and then every other weight is 0.

Result 4. If V = \{\mathbf{0}\}, the empty list is a basis. Otherwise start with any \mathbf{v}_1 \neq \mathbf{0} in V and, while the span of the list is not V, add a vector of V outside it. The lemma keeps the list independent, and the span stays inside V, which contains the span of any of its vectors (Subspaces). Independent lists in \mathbb{R}^n have at most n vectors, so this stops within n steps, and only when the span is V. This proves the claim left open in Subspaces: every subspace of \mathbb{R}^n is the span of at most n vectors.

Result 5. To shrink, go left to right and delete each vector that is \mathbf{0} or a combination of the kept vectors before it. The span does not change, and what remains is independent by Result 5 of Linear independence. To extend, add vectors of V outside the span as in Result 4; Result 2, with a basis of V as the spanning list, caps the length at \dim V.

Result 6. If k independent vectors in V did not span V, extending would give k + 1 independent vectors in a subspace spanned by k, against Result 2. If k vectors span V, shrinking gives a basis, which has k vectors by Result 3, so nothing was deleted. The last sentence is Result 2 read both ways.

Result 7. A basis of U is independent in V, so it has at most \dim V vectors. If it has exactly \dim V, it spans V by Result 6, so V \subseteq U.

Result 8. A subspace S of \mathbb{R}^3 has dimension 0 to 3 by Result 4. Dimension 1 means S = \operatorname{span}\{\mathbf{u}\} with \mathbf{u} \neq \mathbf{0}, a line; dimension 2 means the span of two independent vectors, a plane; dimension 3 gives S = \mathbb{R}^3 by Result 7. This finishes the argument in Subspaces: three vectors, each outside the span of those before, are independent by the lemma, so they span \mathbb{R}^3 by Result 6.

Result 9. (a) A relation among the columns is a solution of a homogeneous system, which row operations do not change, so the original vectors satisfy exactly the relations of the echelon columns. Setting one non-pivot weight to 1 and the other free weights to 0 writes that vector as a combination of pivot vectors, so it can be deleted. The pivot columns alone have a pivot in every column, so they are independent (Result 3 of Linear independence). (b) The general solution \mathbf{x} = t_1\mathbf{s}_1 + \dots + t_r\mathbf{s}_r shows the \mathbf{s}_i span. Each \mathbf{s}_i has 1 at its own free variable’s position and 0 at the other free positions, so that entry of a combination is t_i, and a zero combination has every t_i = 0.

1. Coordinates (3, 2). (-2, 4) = k(3, 1) would need \displaystyle k = -\frac{2}{3} and k = 4, so the two are independent, and so they span \mathbb{R}^2 by Result 6. 3(3, 1) + 2(-2, 4) = (5, 11).
2. Coordinates (1, 1, 1). With the vectors as columns, R_3 - R_1 gives (0, -2, 1) and R_3 + 2R_2 gives (0, 0, 7): a pivot in every row and column. (1, 0, 1) + (2, 1, 0) + (0, 3, 1) = (3, 4, 2).
3. Basis (-1, 1, 0), (2, 0, 1); dimension 2; coordinates (2, 1). Solving x = -y + 2z gives (x, y, z) = y(-1, 1, 0) + z(2, 0, 1), independent since entries 2 and 3 read y and z. 2(-1, 1, 0) + (2, 0, 1) = (0, 2, 1).
4. Basis (0, -1, 1, 0), (0, -1, 0, 1); dimension 2. With x_1 = 0, x_2 = -x_3 - x_4, so each such change is x_3(0, -1, 1, 0) + x_4(0, -1, 0, 1); entries 3 and 4 show independence.
5. If three vectors spanned \mathbb{R}^4, the four independent \mathbf{e}_1, \dots, \mathbf{e}_4 would lie in a space spanned by three, against Result 2. (These three are dependent anyway: (2, 5, 6, 9) = 2(1, 2, 3, 4) + (0, 1, 0, 1).) Five vectors in \mathbb{R}^4 are dependent by Result 4 of Linear independence.
6. A list containing \mathbf{0} is dependent (Result 2 of Linear independence). If the list without \mathbf{v}_j still spanned V, then \mathbf{v}_j would be a combination of the others, and the basis would be dependent by Result 1 there (with no others, \mathbf{v}_j = \mathbf{0}).
7. Every vector of V is a combination, so the list spans V. The trivial weights give \mathbf{0}, and by hypothesis no other weights do, so the list is independent.
8. Basis (1, 2, -1), (0, 1, 2); dimension 2; a plane through the origin. With the vectors as columns, R_2 - 2R_1 and R_3 + R_1 give rows (0, 0, 1, 1) and (0, 0, 2, 2), and R_3 - 2R_2 clears the last: pivots in columns 1 and 3.
9. Basis (-3, 1, 0, 0), (1, 0, 1, 0), (-2, 0, 0, 1); dimension 3. Solve x_1 = -3x_2 + x_3 - 2x_4 with x_2, x_3, x_4 free.
10. \mathbf{f}_1, \mathbf{f}_2, \mathbf{f}_5; dimension 3. \mathbf{f}_3 = \mathbf{f}_1 + \mathbf{f}_2 and \mathbf{f}_4 = \mathbf{f}_1 + 2\mathbf{f}_2 can be deleted. \mathbf{f}_1, \mathbf{f}_2 are not multiples, and a\mathbf{f}_1 + b\mathbf{f}_2 = (3a + b, a + b, 2a, 2b) equals \mathbf{f}_5 only if \displaystyle a = b = \frac{1}{2}, but then 3a + b = 2 \neq 1. So \mathbf{f}_5 is outside their span, and the lemma makes the three independent.
11. Each entry sum is 0. If a(1, 2, -3, 0) + b(0, 1, 1, -2) + c(3, 0, -1, -2) ={} \mathbf{0}, the fourth entry gives b = -c, the first a = -3c, and the second 2a + b = -7c = 0, so all weights are 0. Three independent vectors in W, of dimension 3, are a basis by Result 6.
12. Add (1, 0, 0), for example. a(1, -1, 2) + b(2, 0, 1) = (1, 0, 0) forces a = 0 (second entry) and b = 0 (third), but then the first entry is 0 \neq 1. So the lemma makes the three independent, and they are a basis by Result 6.
13. An echelon form has at most one pivot per row, so at most 3 pivots among 5 columns, leaving at least 2 free variables, and by Result 9(b) the dimension is the number of free variables.
14. Let the planes be \operatorname{span}\{\mathbf{a}, \mathbf{b}\} and \operatorname{span}\{\mathbf{c}, \mathbf{d}\}, each pair independent. Four vectors in \mathbb{R}^3 are dependent: x\mathbf{a} + y\mathbf{b} + z\mathbf{c} + w\mathbf{d} = \mathbf{0}, weights not all 0. Then \mathbf{p} = x\mathbf{a} + y\mathbf{b} = -(z\mathbf{c} + w\mathbf{d}) lies in both planes. If \mathbf{p} = \mathbf{0}, independence of each pair forces all four weights to be 0, which they are not. So both planes contain the line \operatorname{span}\{\mathbf{p}\}.
15. A nonzero multiple keeps its zeros in the same places, and these three have zeros in different places (the last has none), so no two are multiples. Yet (2, 1, 1) = (1, 1, 0) + (1, 0, 1), so the list is dependent, and so it is not a basis.
16. \dim \mathbb{R}^n = n, so by Result 6 a list of n vectors is a basis exactly when it is independent, which by Result 3 of Linear independence happens exactly when every column of an echelon form has a pivot.