Basis — definition, uniqueness of representation: Exercises

Reminders in one place (everything below is earned here, never assumed):

A basis is a set that is both linearly independent (only the all-zero combination gives $0$ ) and spanning (builds every vector).
The coordinates $[w]_{B} = (c_{1}, \dots, c_{n})$ are the unique scalars with $w = c_{1} v_{1} + \dots + c_{n} v_{n}$ .
$dim V$ is the size of any basis — a fixed number for the space (see Dimension).

Level 1 — Recognition

L1.1

Is $B = {(2, 0), (0, 3)}$ a basis of $R^{2}$ ? State which two properties you check.

Recall Solution

We check independence and spanning.

Independence: $c_{1} (2, 0) + c_{2} (0, 3) = (2 c_{1}, 3 c_{2}) = (0, 0)$ forces $2 c_{1} = 0$ and $3 c_{2} = 0$ , i.e. $c_{1} = c_{2} = 0$ . Only the trivial combination gives $0$ . ✔
Spanning: any $(x, y)$ is $\frac{x}{2} (2, 0) + \frac{y}{3} (0, 3)$ . ✔ Both hold, so yes, $B$ is a basis. (Two independent vectors in a 2-dimensional space always form a basis.)

L1.2

Which of these is a basis of $R^{2}$ ? (a) ${(1, 2)}$ , (b) ${(1, 0), (0, 1), (1, 1)}$ , (c) ${(1, 2), (3, 5)}$ .

Recall Solution

(a) One vector cannot span $R^{2}$ (its span is a single line). Not a basis.
(b) Three vectors in $R^{2}$ are dependent ( $(1, 1) = (1, 0) + (0, 1)$ ). Spans but not independent. Not a basis.
(c) Two vectors, and they're independent since $det (1235) = 1 \cdot 5 - 3 \cdot 2 = - 1 \neq = 0$ . This is the basis.

L1.3

True or false: the coordinates of $w = (5, - 1, 4)$ in the standard basis ${e_{1}, e_{2}, e_{3}}$ are $(5, - 1, 4)$ .

Recall Solution

True. Each $e_{i}$ switches on exactly one slot: $w = 5 e_{1} - 1 e_{2} + 4 e_{3}$ , so $[w]_{B} = (5, - 1, 4)$ . This equality of coordinates and entries is special to the standard basis.

Level 2 — Application

L2.1

Find the coordinates of $w = (7, 1)$ in the basis $B = {(1, 1), (1, - 1)}$ .

Recall Solution

Solve $c_{1} (1, 1) + c_{2} (1, - 1) = (7, 1)$ : $c_{1} + c_{2} = 7, c_{1} - c_{2} = 1.$ WHY set up a system: spanning promises a solution exists; independence promises it's unique — so solving is legitimate. Add the two equations: $2 c_{1} = 8 \Rightarrow c_{1} = 4$ . Then $c_{2} = 7 - 4 = 3$ . $[w]_{B} = (4, 3) .$

L2.2

In $P_{2}$ (polynomials of degree $\leq 2$ ) with basis $B = {1, x, x^{2}}$ , give the coordinates of $p (x) = 2 - 4 x + x^{2}$ .

Recall Solution

Read off the coefficients in order $1, x, x^{2}$ : $[p]_{B} = (2, - 4, 1) .$ WHY: independence of ${1, x, x^{2}}$ means " $a + b x + c x^{2} = 0$ for all $x$ " forces $a = b = c = 0$ , so the coefficient list is unique.

L2.3

Reconstruct the vector whose coordinates in $B = {(2, 1), (1, 3)}$ are $[w]_{B} = (3, 1)$ .

Recall Solution

Coordinates tell you the recipe; just execute it: $w = 3 (2, 1) + 1 (1, 3) = (6, 3) + (1, 3) = (7, 6) .$

L2.4

In $B = {(1, 1), (1, - 1)}$ , find the coordinates of the zero vector $(0, 0)$ .

Recall Solution

Solve $c_{1} (1, 1) + c_{2} (1, - 1) = (0, 0)$ : gives $c_{1} + c_{2} = 0$ and $c_{1} - c_{2} = 0$ , so $c_{1} = c_{2} = 0$ . $[0]_{B} = (0, 0) .$ WHY it matters: the zero vector has all-zero coordinates in every basis — this is exactly the independence condition in disguise.

Level 3 — Analysis

L3.1

Show that ${(1, 2), (2, 4)}$ is not a basis of $R^{2}$ , and name which property fails.

Recall Solution

Note $(2, 4) = 2 \cdot (1, 2)$ , so $2 (1, 2) + (- 1) (2, 4) = (0, 0)$ — a nontrivial combination equals $0$ . That breaks independence. Consequently the span is only the line through $(1, 2)$ , so spanning fails too. Not a basis.

L3.2

The set ${(1, 0), (0, 1), (1, 1)}$ spans $R^{2}$ . Exhibit two different representations of $(2, 3)$ , proving coordinates aren't unique here.

Recall Solution

Using only the first two: $(2, 3) = 2 (1, 0) + 3 (0, 1) + 0 (1, 1)$ .
Using the third once: since $(1, 1) = (1, 0) + (0, 1)$ , $(2, 3) = 1 (1, 0) + 2 (0, 1) + 1 (1, 1) .$ Check: $1 (1, 0) + 2 (0, 1) + 1 (1, 1) = (1 + 0 + 1, 0 + 2 + 1) = (2, 3)$ . ✔ Two honest addresses $(2, 3, 0)$ and $(1, 2, 1)$ for the same point → not unique → dependence is the culprit, so it's not a basis.

L3.3

Is $B = {(1, 0, 1), (0, 1, 1), (1, 1, 0)}$ a basis of $R^{3}$ ? Justify with a determinant.

Recall Solution

Stack the vectors as columns and compute the determinant:

=1(1\cdot0-1\cdot1)-0+1(0\cdot1-1\cdot1)=1(-1)+1(-1)=-2.$$ **WHY the determinant:** a nonzero determinant means the columns are independent *and* span $\mathbb R^3$ (the matrix is invertible). Since $-2\ne 0$, **yes, it's a basis**.

L3.4

Find the coordinates of $w = (1, 2, 3)$ in the basis of L3.3.

Recall Solution

Solve $c_{1} (1, 0, 1) + c_{2} (0, 1, 1) + c_{3} (1, 1, 0) = (1, 2, 3)$ : $c_{1} + c_{3} = 1, c_{2} + c_{3} = 2, c_{1} + c_{2} = 3.$ WHY add all three equations: each equation is a true statement (its left side literally equals its right side), and adding true equations gives another true equation — the same move as putting both feet on a scale to read your combined weight. On the left, every unknown $c_{1}, c_{2}, c_{3}$ appears in exactly two of the three equations, so summing gives $2 c_{1} + 2 c_{2} + 2 c_{3}$ ; on the right we get $1 + 2 + 3 = 6$ . Hence: $2 (c_{1} + c_{2} + c_{3}) = 6 \Rightarrow c_{1} + c_{2} + c_{3} = 3.$ Now subtracting each original equation from this total isolates one unknown at a time (also a legal move — subtracting equal things from equal things keeps equality):

$(c_{1} + c_{2} + c_{3}) - (c_{1} + c_{3}) = c_{2} = 3 - 1 = 2.$
$(c_{1} + c_{2} + c_{3}) - (c_{2} + c_{3}) = c_{1} = 3 - 2 = 1.$
$(c_{1} + c_{2} + c_{3}) - (c_{1} + c_{2}) = c_{3} = 3 - 3 = 0.$ $[w]_{B} = (1, 2, 0) .$ Check: $1 (1, 0, 1) + 2 (0, 1, 1) + 0 (1, 1, 0) = (1, 2, 3)$ . ✔

Level 4 — Synthesis

L4.1

Prove: if $B = {v_{1}, v_{2}, v_{3}}$ is a basis of $V$ , then $B^{'} = {v_{1}, v_{1} + v_{2}, v_{1} + v_{2} + v_{3}}$ is also a basis.

Recall Solution

$V$ is 3-dimensional (it has a basis of size 3), so by the counting theorem it suffices to prove $B^{'}$ is independent — three independent vectors in a 3-dim space automatically span. Suppose $a v_{1} + b (v_{1} + v_{2}) + c (v_{1} + v_{2} + v_{3}) = 0 .$ Collect coefficients of each original basis vector: $(a + b + c) v_{1} + (b + c) v_{2} + c v_{3} = 0 .$ WHY collect: $B$ is independent, so each coefficient must vanish: $a + b + c = 0, b + c = 0, c = 0.$ Back-substitute: $c = 0 \Rightarrow b = 0 \Rightarrow a = 0$ . Only the trivial combination works, so $B^{'}$ is independent, hence a basis. ∎

L4.2

Coordinate change. With $B^{'}$ from L4.1 and $w = 2 v_{1} - v_{2} + 4 v_{3}$ , find $[w]_{B^{'}}$ .

Recall Solution

We want $d_{1}, d_{2}, d_{3}$ with $d_{1} v_{1} + d_{2} (v_{1} + v_{2}) + d_{3} (v_{1} + v_{2} + v_{3}) = 2 v_{1} - v_{2} + 4 v_{3} .$ Group by $v_{1}, v_{2}, v_{3}$ and match (legal because ${v_{1}, v_{2}, v_{3}}$ is independent): $d_{1} + d_{2} + d_{3} = 2, d_{2} + d_{3} = - 1, d_{3} = 4.$ From the bottom up: $d_{3} = 4$ ; then $d_{2} = - 1 - 4 = - 5$ ; then $d_{1} = 2 - (- 5) - 4 = 3$ . $[w]_{B^{'}} = (3, - 5, 4) .$ Check: $3 v_{1} - 5 (v_{1} + v_{2}) + 4 (v_{1} + v_{2} + v_{3}) = (3 - 5 + 4) v_{1} + (- 5 + 4) v_{2} + 4 v_{3} = 2 v_{1} - v_{2} + 4 v_{3} .$ ✔

L4.3

Structural claim. Suppose a set $B$ of $n$ vectors in $V$ has the property that every $w \in V$ has at most one representation as $\sum c_{i} v_{i}$ (uniqueness), but you are not told it spans. Which basis property is guaranteed, and which might fail? Give an example of failure.

Recall Solution

"At most one representation for all $w$ " applied to $w = 0$ says the only way to write $0$ is with all-zero coefficients — that is independence ✔ (guaranteed). Spanning may fail: $B = {(1, 0, 0), (0, 1, 0)}$ in $R^{3}$ gives at-most-one representation everywhere (it's independent), yet $(0, 0, 1)$ has no representation — spanning fails. So uniqueness alone $\Rightarrow$ independence, not basis.

Level 5 — Mastery

L5.1

Full uniqueness proof, your version. Let $B = {v_{1}, \dots, v_{n}}$ be a basis. Prove that the coordinate map $w \mapsto [w]_{B}$ is well-defined (single-valued) by deriving it from independence — no hand-waving.

Recall Solution

Suppose $w$ had two coordinate lists: $w = \sum_{i} c_{i} v_{i} and w = \sum_{i} d_{i} v_{i} .$ Subtract (same vector minus itself is $0$ ): $0 = w - w = \sum_{i} (c_{i} - d_{i}) v_{i} .$ WHY subtract: it converts "two equal sums" into "a combination equal to $0$ ", which is precisely the situation independence controls. By independence, $\sum_{i} (c_{i} - d_{i}) v_{i} = 0$ forces every coefficient $c_{i} - d_{i} = 0$ , i.e. $c_{i} = d_{i}$ for all $i$ . The two lists coincide, so $[w]_{B}$ is single-valued. ∎

L5.2

Degenerate space. What is the basis and dimension of the zero space $V = {0}$ ? Explain why the empty set works.

Recall Solution

The basis is the empty set $\emptyset$ , and $dim V = 0$ .

Spanning: the span of $\emptyset$ is (by convention) ${0}$ — an empty sum equals $0$ . That's the whole space. ✔
Independence: there are no vectors to combine, so there is no nontrivial combination equal to $0$ — vacuously independent. ✔ So $\emptyset$ is a basis of size $0$ , giving $dim {0} = 0$ . Every basis of a space shares this size (see Dimension).

L5.3

Linear maps meet bases. Let $B = {v_{1}, v_{2}}$ be a basis of $V$ and $T : V \to W$ linear. Prove that $T$ is completely determined by the two values $T (v_{1}), T (v_{2})$ .

Recall Solution

Take any $w \in V$ . Because $B$ is a basis, $w$ has a unique coordinate expression $w = c_{1} v_{1} + c_{2} v_{2}$ (uniqueness, from L5.1). Apply linearity of $T$ : $T (w) = T (c_{1} v_{1} + c_{2} v_{2}) = c_{1} T (v_{1}) + c_{2} T (v_{2}) .$ WHY this proves the claim: the right side uses only the fixed data $T (v_{1}), T (v_{2})$ and the (unique) scalars $c_{1}, c_{2}$ . So once those two images are chosen, $T (w)$ is forced for every $w$ . That is exactly why a linear map is stored as a matrix whose columns are the images of basis vectors — see Matrix of a Linear Map. ∎

L5.4

A subtle counterexample. In $R^{2}$ , is $B = {(1, 1), (2, 2)}$ a basis? Then explain precisely what goes wrong with uniqueness for a vector in its span.

Recall Solution

$(2, 2) = 2 (1, 1)$ , so $B$ is dependent — not a basis. Its span is only the line $y = x$ . For a vector on that line, say $(3, 3)$ , uniqueness collapses: $(3, 3) = 3 (1, 1) + 0 (2, 2) = 1 (1, 1) + 1 (2, 2) = (- 1) (1, 1) + 2 (2, 2), \dots$ infinitely many representations, because the redundancy lets you trade one vector for a multiple of the other. This is the geometric face of the L3 trap: dependence $\Rightarrow$ non-unique addresses.

The figure below draws this: both dashed vectors $v_{1} = (1, 1)$ (blue) and $v_{2} = (2, 2)$ (pink) lie along the same dashed line $y = x$ , and the yellow point $w = (3, 3)$ sits on that line. Because $v_{2}$ is just a stretched copy of $v_{1}$ , you can reach $w$ by pushing farther along $v_{1}$ and pulling back along $v_{2}$ (or vice versa) in endlessly many ways — the three recipes printed under the axes all land on the same yellow dot. Contrast this with a real basis, where the two arrows point in genuinely different directions and pin down a single recipe.

Figure — Basis — definition, uniqueness of representation

Connections

Basis — definition, uniqueness of representation — the parent theorem these drill.
Linear Independence — the engine behind every uniqueness step.
Span and Spanning Sets — the existence half tested in L1–L2.
Dimension — the counting shortcut used in L4.
Coordinate Vectors and Change of Basis — L2, L4.2 compute exactly these.
Matrix of a Linear Map — L5.3 is why matrices exist.