Vector Spaces & Bases

The two conditions pull in opposite directions. Spanning needs enough vectors to reach everything; independence forbids any redundant ones. In $\mathbb R^2$, the pair $(2, 1), (4, 2)$ fails to span (both lie on one line), the triple $(1, 0), (0, 1), (1, 1)$ spans but is dependent (the third is the sum of the first two), and $(1, 0), (1, 1)$ is just right. A basis is a spanning set with nothing wasted, which is why its size measures the space.

Why each condition costs one dimension: the solution set of $k$ homogeneous equations in $n$ unknowns is a null space, and rank–nullity gives its dimension as $n - \text{rank}$. If the $k$ conditions are independent, the rank is $k$ and the dimension is $n - k$. For the hyperplane $x_1 + x_2 + x_3 + x_4 = 0$, three variables are free and the fourth is determined, so a basis is $(1, 0, 0, -1)$, $(0, 1, 0, -1)$, $(0, 0, 1, -1)$.

The coordinates $(5, 2)$ are directions for a walk: five steps along $(1, 1)$, then two along $(1, -1)$. On the grid built from the new basis, $(7, 3)$ is the lattice point five lines over and two lines across. Check: $5(1, 1) + 2(1, -1) = (7, 3)$ ✓. The point never moved; only the ruler changed.

The conditions are exactly what linear combinations need: if sums and multiples stay inside, every combination stays inside, so $W$ is a vector space in its own right and span, basis and dimension all make sense for it. Solution sets of $A\mathbf x = \mathbf 0$ pass automatically, because $A(\mathbf u + \mathbf w) = A\mathbf u + A\mathbf w = \mathbf 0$ and $A(c\mathbf u) = cA\mathbf u = \mathbf 0$. A nonzero right-hand side breaks the very first condition: $A\mathbf 0 = \mathbf 0 \ne \mathbf b$.

Every element is $(x, y, z) = (-2s + t,\ s,\ t) = s(-2, 1, 0) + t(1, 0, 1)$. So $(-2, 1, 0)$ and $(1, 0, 1)$ span $W$. They are independent — look at the second and third coordinates: $s$ and $t$ can be read off directly, so the only combination giving $\mathbf 0$ has $s = t = 0$. Hence they form a basis and $\dim W = 2$, as "three dimensions minus one condition" predicts. Check that both lie in $W$: $-2 + 2 - 0 = 0$ ✓ and $1 + 0 - 1 = 0$ ✓.

Take $\mathbf u = (1, 2)$ and $\mathbf w = (-2, -1)$, both in $Q$ ($2 \ge 0$ and $2 \ge 0$). Their sum $(-1, 1)$ has $xy = -1 < 0$, so $Q$ is not closed under addition and is not a subspace — one counterexample settles it. The line $L$ fails even sooner: $(0, 0)$ is not on it, since $0 \ne 2\cdot0 + 1$. A line is a subspace of $\mathbb R^2$ only if it passes through the origin.

The full space $\{a + bx + cx^2\}$ has dimension $3$ and we impose one condition, so expect dimension $2$. Solve for $a = -b - c$: then $p = b(x - 1) + c(x^2 - 1)$. So $\{x - 1,\ x^2 - 1\}$ spans $W$; the two are independent since neither is a multiple of the other (they have different degrees). So $\dim W = 2$. The factor theorem says the same thing another way: $p(1) = 0$ exactly when $p = (x - 1)(\text{linear})$, and the linear factor has two free coefficients.

The first formula is just the column picture of $P[\mathbf x]_B$: a matrix times a vector is the combination of the columns weighted by the vector's entries. $P$ is invertible precisely because its columns form a basis — independent columns mean no nonzero combination vanishes, so $P$ has no null space. Going from standard coordinates to $B$-coordinates means solving $P\mathbf c = \mathbf x$, which is the same as multiplying by $P^{-1}$.

Multiply: $[\mathbf x]_B = P^{-1}\mathbf x = \begin{pmatrix}0 + 2\\-1 + 4\end{pmatrix} = \begin{pmatrix}2\\3\end{pmatrix}$. So $\mathbf x = 2\,\mathbf b_1 + 3\,\mathbf b_2$. Check: $2(2, 1) + 3(-1, 0) = (1, 2)$ ✓. On the grid drawn from $\mathbf b_1$ and $\mathbf b_2$, $\mathbf x$ is the lattice point two lines over and three across. Unlike the orthogonal basis earlier, $\mathbf b_1 \cdot \mathbf b_2 = -2 \ne 0$ here, so the one-dot-product shortcut does not apply; the inverse matrix does the work instead.

Differentiate both sides: $p'(x) = c_1 + 2c_2(x - 1)$, and at $x = 1$, $c_1 = p'(1) = 2 + 2\cdot1 = 4$. Differentiate again: $p''(x) = 2c_2$, so $c_2 = \tfrac{p''(1)}{2} = \tfrac22 = 1$. So $p = 6 + 4(x - 1) + (x - 1)^2$ and the coordinate vector is $(6, 4, 1)$. Check by expanding: $6 + 4x - 4 + x^2 - 2x + 1 = 3 + 2x + x^2$ ✓. This is the Taylor expansion at $1$ — Taylor coefficients are coordinates in a shifted basis.

Row reduce: if every column earns a pivot, only the zero combination works — independent; a pivotless column means a free variable, hence a nonzero combination summing to zero — dependent, and back-substitution hands you the actual dependence. The same computation delivers dimension: the pivot columns form a basis of the span, so $\dim(\text{span}) = \text{rank}$. This is the great economy of the subject — span, independence, basis, dimension, and solvability are all read off one echelon form.

Step 2 — read: pivots in columns 1, 2 only — Dependent, dimension of the span is $2$.

Step 3 — recover the dependence: the free column says $c_3$ is free; rows give $c_1 = -c_3$ and $c_2 = -c_3$; take $c_3 = 1$: $-(1,0,1) - (0,1,1) + (1,1,2) = (0,0,0)$ ✓ — i.e. $\mathbf v_3 = \mathbf v_1 + \mathbf v_2$, the hidden relation, now explicit.

Step 4 — basis: keep the pivot columns $\{(1,0,1), (0,1,1)\}$; they span the same plane with nothing wasted.

The orthogonal shortcut: because $\mathbf b_1\cdot\mathbf b_2 = 1 - 1 = 0$, each coefficient is an independent projection: $a = \dfrac{(7,3)\cdot(1,1)}{(1,1)\cdot(1,1)} = \dfrac{10}{2} = 5$ and $b = \dfrac{(7,3)\cdot(1,-1)}{2} = \dfrac{4}{2} = 2$ — no system solved, no interaction between coordinates. Check: $5(1,1) + 2(1,-1) = (7, 3)$ ✓. This is precisely why orthogonal bases (and eventually Fourier series) are prized: coordinates decouple.