Basis — definition, uniqueness of representation: Worked examples

Before anything, two reminders in plain words.

Finding those scalars almost always means solving a small system of equations — that is the one skill this page drills. Recall that a basis must do two jobs at once, and each job is guaranteed by one property: spanning guarantees a solution exists (every vector has at least one address), while independence guarantees the solution is unique (no vector has two addresses). We lean on both throughout.

The scenario matrix

Every basis problem you will ever meet lives in one of these cells. Figure 1 encodes the same landscape visually — the three coloured regimes (too few, just right, too many) are the spine of everything below, so glance at it before reading the table.

Figure 1 — The whole landscape at a glance. A set of vectors sits in one of three regimes depending on how its size compares to $dim V$ . Red (too few): independent but cannot reach every vector → some addresses are missing (Cell E). Green (just right): exactly $dim V$ independent vectors → a genuine basis, every vector has exactly one address (Cells A, B, F, G, I). Orange (too many): enough to reach everything but redundant → the same vector gets many addresses (Cell D). The green box also swallows the zero-vector case (Cell C), whose address is always all-zeros. Follow the grey arrows: adding a vector pushes you rightward from "just right" into "too many."

#	Case class	What makes it tricky	Regime (Fig 1)	Worked example
A	Standard basis, plain vector	coordinates equal components — the "invisible" case	green	Ex 1
B	Non-standard basis, mixed signs	must solve a $2 \times 2$ system; signs matter	green	Ex 2
C	Degenerate INPUT: the zero vector	coordinates should all be $0$ — but only if $B$ is a basis	green	Ex 3
D	Degenerate SET: dependent "basis"	coordinates are not unique — must detect failure	orange	Ex 4
E	Degenerate SET: too few (doesn't span)	some vectors have no coordinates — detect failure	red	Ex 5
F	Non-arrow space (polynomials)	"vectors" are functions; same machinery still works	green	Ex 6
G	Non-arrow space (matrices)	$2 \times 2$ matrices as a $4$ -dimensional space	green	Ex 7
H	Real-world word problem	translate a mixing/recipe story into coordinates	green	Ex 8
I	Exam twist: verify a set IS a basis in $R^{3}$	check independence + spanning together via a determinant	green	Ex 9

We now walk every cell.

Worked examples

Cell A — the invisible standard basis

Cell B — non-standard basis, mixed signs

Worked example

Ex 2. Let $B = {(2, 1), (- 1, 3)}$ in $R^{2}$ . Find $[w]_{B}$ for $w = (3, - 4)$ . Forecast: will the coordinates be bigger or smaller than the entries $(3, - 4)$ ? Guess a sign for each.

Set up $c_{1} (2, 1) + c_{2} (- 1, 3) = (3, - 4)$ , giving two scalar equations: $2 c_{1} - c_{2} = 3, c_{1} + 3 c_{2} = - 4.$ Why this step? Spanning promises a solution exists; the equations are just "match each coordinate."
From the first equation $c_{2} = 2 c_{1} - 3$ . Substitute into the second: $c_{1} + 3 (2 c_{1} - 3) = - 4 \Rightarrow 7 c_{1} - 9 = - 4 \Rightarrow c_{1} = \frac{5}{7} .$ Why this step? Substitution collapses two unknowns to one — the cleanest route for a $2 \times 2$ .
Then $c_{2} = 2 \cdot \frac{5}{7} - 3 = \frac{10}{7} - \frac{21}{7} = - \frac{11}{7}$ . Why this step? Back-substitute to recover the second coordinate.

Answer: $[w]_{B} = (\frac{5}{7}, - \frac{11}{7})$ .

Verify: $\frac{5}{7} \cdot (2, 1) + (- \frac{11}{7}) \cdot (- 1, 3) = (\frac{10 + 11}{7}, \frac{5 - 33}{7}) = (\frac{21}{7}, \frac{- 28}{7}) = (3, - 4) = w$ . ✓ Coordinates $\neq =$ components — this is the norm, not the exception. Figure 2 shows this same sum geometrically.

Figure 2 — Coordinates as a recipe of scaled basis arrows. The blue arrow $u = (2, 1)$ and green arrow $v = (- 1, 3)$ are the two directions of basis $B$ . The orange arrow is the target $w = (3, - 4)$ . To build $w$ we walk $\frac{5}{7}$ of the way along $u$ (dashed blue, landing at the blue dot), then $- \frac{11}{7}$ along $v$ (dashed green — the negative coefficient means we travel opposite to $v$ ). The dashed path lands exactly on the orange arrow's tip: that is what " $[w]_{B} = (\frac{5}{7}, - \frac{11}{7})$ " means as a picture.

Cell C — degenerate input: the zero vector

Cell D — degenerate SET: dependent, coordinates not unique

Worked example

Ex 4. Is $S = {(1, 0), (0, 1), (1, 1)}$ a basis of $R^{2}$ ? Try to coordinate $w = (2, 3)$ . Forecast: three vectors in a 2-D space — smell a problem yet?

Try $c_{1} (1, 0) + c_{2} (0, 1) + c_{3} (1, 1) = (2, 3)$ , i.e. $c_{1} + c_{3} = 2, c_{2} + c_{3} = 3.$ Why this step? Set up the coordinate system honestly and see what happens.
Two equations, three unknowns → $c_{3}$ is free. Pick $c_{3} = 0$ : $(c_{1}, c_{2}, c_{3}) = (2, 3, 0)$ . Pick $c_{3} = 1$ : $(1, 2, 1)$ . Why this step? A free variable means infinitely many solutions — coordinates are not unique.
Check the failure directly with the zero vector: $(1, 0) + (0, 1) - (1, 1) = (0, 0)$ with coefficients $(1, 1, - 1) \neq = (0, 0, 0)$ . Why this step? A non-trivial combination hitting $0$ is the definition of dependence — that is precisely what breaks uniqueness.

Answer: $S$ is not a basis. It spans $R^{2}$ but is dependent, so addresses aren't unique. By Dimension, any basis of $R^{2}$ has exactly $2$ vectors, never $3$ . (This is the orange "too many" regime of Figure 1.)

Verify: both $(2, 3, 0)$ and $(1, 2, 1)$ rebuild $w$ : $(2, 3, 0) \to (2 + 0, 3 + 0) = (2, 3)$ ✓ and $(1, 2, 1) \to (1 + 1, 2 + 1) = (2, 3)$ ✓. Two valid addresses = failure.

Cell E — degenerate SET: too few, doesn't span

Worked example

Ex 5. Is $S = {(1, 0, 0), (0, 1, 0)}$ a basis of $R^{3}$ ? Coordinate $w = (0, 0, 1)$ . Forecast: which direction is completely missing from $S$ ?

Try $c_{1} (1, 0, 0) + c_{2} (0, 1, 0) = (0, 0, 1)$ . Third coordinate: $0 = 1$ . Why this step? Setting up the equation exposes the contradiction immediately.
$0 = 1$ is impossible, so no solution exists. Why this step? No solution means $w$ has no coordinates in $S$ — spanning has failed.

Answer: $S$ is not a basis: it's independent but doesn't span $R^{3}$ (the $z$ -direction is unreachable). This is the red "too few" regime of Figure 1; a basis of $R^{3}$ needs exactly $3$ vectors.

Verify: any combination $c_{1} (1, 0, 0) + c_{2} (0, 1, 0) = (c_{1}, c_{2}, 0)$ always has third entry $0 \neq = 1$ . ✓ contradiction confirmed. This is the exact mirror of Ex 4: too few $\Rightarrow$ existence fails; too many $\Rightarrow$ uniqueness fails.

Cell F — a space with no arrows: polynomials

Worked example

Ex 6. In $P_{2}$ (polynomials of degree $\leq 2$ ), let $B = {1, 1 + x, 1 + x + x^{2}}$ . Find $[p]_{B}$ for $p (x) = 2 + 3 x + 4 x^{2}$ . Forecast: this basis is "triangular" (Figure 3 shows why) — guess whether you should read the coefficients off from the top ( $x^{2}$ ) or the bottom ( $1$ ) first.

Write $p = a \cdot 1 + b \cdot (1 + x) + c \cdot (1 + x + x^{2})$ and expand: $p = (a + b + c) + (b + c) x + c x^{2} .$ Why this step? Two polynomials are equal only when every coefficient matches (that's independence of $1, x, x^{2}$ ), so we group by power.
Match highest power first: $c = 4$ . Then $b + c = 3 \Rightarrow b = - 1$ . Then $a + b + c = 2 \Rightarrow a = 2 - (- 1) - 4 = - 1$ . Why this step? The "triangular" shape — each basis polynomial adds exactly one new power (Figure 3) — lets us solve top-down with no algebra tangle; a common exam shortcut.

Answer: $[p]_{B} = (a, b, c) = (- 1, - 1, 4)$ .

Verify: $- 1 \cdot 1 - 1 \cdot (1 + x) + 4 \cdot (1 + x + x^{2}) = (- 1 - 1 + 4) + (- 1 + 4) x + 4 x^{2} = 2 + 3 x + 4 x^{2} = p$ . ✓ Same machinery as $R^{3}$ — that's the whole point of a basis: abstract spaces get concrete addresses. See Coordinate Vectors and Change of Basis.

Figure 3 — Why the basis is "triangular." Each basis polynomial is a row; each power $1, x, x^{2}$ is a column. A filled cell means "this power appears." The filled cells form a lower-triangular staircase: $1$ uses only the $1$ -column; $1 + x$ adds the $x$ -column; $1 + x + x^{2}$ adds the $x^{2}$ -column. Because only the last basis polynomial reaches the $x^{2}$ column, its coefficient $c$ is pinned first by the top power; then $x$ is pinned, then $1$ — exactly the top-down solve of step 2.

Cell G — a space of matrices

Worked example

Ex 7. The space $M_{2 \times 2}$ of $2 \times 2$ real matrices has the standard basis $B = {E_{11}, E_{12}, E_{21}, E_{22}}$ , where $E_{ij}$ has a $1$ in slot $(i, j)$ and $0$ elsewhere. Find the coordinates of $A = (70 - 2 5)$ . Forecast: how many coordinates does a $2 \times 2$ matrix have? (Hint: how many slots?)

Write $A = c_{1} E_{11} + c_{2} E_{12} + c_{3} E_{21} + c_{4} E_{22}$ . Why this step? Each $E_{ij}$ switches on exactly one slot — the matrix version of the standard basis.
Read slot by slot: $c_{1} = 7, c_{2} = - 2, c_{3} = 0, c_{4} = 5$ . Why this step? Just as with $e_{i}$ in $R^{3}$ , coordinates equal the entries for this standard basis.

Answer: $[A]_{B} = (7, - 2, 0, 5)$ , a vector in $R^{4}$ .

Verify: rebuild: $7 E_{11} - 2 E_{12} + 0 E_{21} + 5 E_{22} = (70 - 2 5) = A$ . ✓ So $dim M_{2 \times 2} = 4$ : a matrix "is" a 4-vector once you fix a basis (this is what powers the Matrix of a Linear Map).

Cell H — a real-world word problem

Worked example

Ex 8. A juice bar mixes two base syrups. Syrup $u = (3, 1)$ contains (grams sugar, grams vitamin-C) per scoop; syrup $v = (1, 2)$ likewise. A customer wants a drink with exactly $(11, 12)$ grams. How many scoops of each? Forecast: more of the sugary syrup or the vitamin syrup? Guess before solving.

This asks: find coordinates of the target $(11, 12)$ in the basis $B = {u, v} = {(3, 1), (1, 2)}$ . Set up $c_{1} (3, 1) + c_{2} (1, 2) = (11, 12)$ : $3 c_{1} + c_{2} = 11, c_{1} + 2 c_{2} = 12.$ Why this step? "Mix two ingredients to hit a target" is literally a coordinate problem in disguise.
From the first, $c_{2} = 11 - 3 c_{1}$ . Substitute: $c_{1} + 2 (11 - 3 c_{1}) = 12 \Rightarrow c_{1} - 6 c_{1} + 22 = 12 \Rightarrow - 5 c_{1} = - 10 \Rightarrow c_{1} = 2$ . Why this step? Substitution again — reliable for $2 \times 2$ .
Then $c_{2} = 11 - 3 (2) = 5$ . Why this step? Back-substitute for the second scoop count.

Answer: $2$ scoops of $u$ and $5$ scoops of $v$ . (Both non-negative — physically sensible.)

Verify (with units): sugar $= 2 (3) + 5 (1) = 11$ g ✓; vitamin-C $= 2 (1) + 5 (2) = 12$ g ✓. Because $u, v$ are independent, this recipe is the unique one — the customer can't be surprised by a second valid mix.

Cell I — exam twist: prove a set IS a basis of $R^{3}$

Worked example

Ex 9. Show $B = {(1, 2, 0), (0, 1, 1), (1, 0, 1)}$ is a basis of $R^{3}$ , then coordinate $w = (2, 3, 1)$ . Forecast: in $R^{3}$ , a set of exactly $3$ vectors is a basis iff one single number is non-zero — which number?

Stack the vectors as columns and compute the determinant $det 120011101 .$ Why this step? With exactly $dim R^{3} = 3$ vectors, "independent" and "spanning" become the same condition, and both hold iff the determinant $\neq = 0$ . One check does the work of two. (See Dimension.)
Expand along the first row: $1 \cdot (1 \cdot 1 - 0 \cdot 1) - 0 + 1 \cdot (2 \cdot 1 - 1 \cdot 0) = 1 + 2 = 3 \neq = 0.$ Why this step? Non-zero determinant $\Rightarrow$ columns are independent $\Rightarrow$ (with the right count) a basis.
Now coordinate $w$ : solve $c_{1} (1, 2, 0) + c_{2} (0, 1, 1) + c_{3} (1, 0, 1) = (2, 3, 1)$ , i.e. $c_{1} + c_{3} = 2, 2 c_{1} + c_{2} = 3, c_{2} + c_{3} = 1.$ From the first, $c_{3} = 2 - c_{1}$ . From the third, $c_{2} = 1 - c_{3} = 1 - (2 - c_{1}) = c_{1} - 1$ . Substitute into the second: $2 c_{1} + (c_{1} - 1) = 3 \Rightarrow 3 c_{1} = 4 \Rightarrow c_{1} = \frac{4}{3}$ . Why this step? Chain the easy equations to isolate one unknown.
Then $c_{2} = \frac{4}{3} - 1 = \frac{1}{3}$ and $c_{3} = 2 - \frac{4}{3} = \frac{2}{3}$ . Why this step? Back-substitute for the rest.

Answer: $B$ is a basis (det $= 3$ ), and $[w]_{B} = (\frac{4}{3}, \frac{1}{3}, \frac{2}{3})$ .

Verify: $\frac{4}{3} (1, 2, 0) + \frac{1}{3} (0, 1, 1) + \frac{2}{3} (1, 0, 1) = (\frac{4 + 2}{3}, \frac{8 + 1}{3}, \frac{1 + 2}{3}) = (2, 3, 1) = w$ . ✓

Recall Which cell is this problem? (self-test)

Given a problem, ask in order: How many vectors vs. $dim V$ ? ::: More than $dim$ → dependent (Cell D); fewer → can't span (Cell E); equal → possible basis, check det (Cell I). Is the input the zero vector? ::: Coordinates are all $0$ in any genuine basis (Cell C). Is the "vector" a polynomial or matrix? ::: Same machinery — match coefficients / slots (Cells F, G). Is it a mixing/recipe story? ::: It's coordinates in disguise — set up the system (Cell H).

Connections

Linear Independence — the reason coordinates are unique (Cells C, D, I).
Span and Spanning Sets — the reason coordinates exist (Cell E).
Dimension — the "right count" that decides Cells D, E, I.
Coordinate Vectors and Change of Basis — every answer here is a coordinate vector $[w]_{B}$ .
Subspaces — polynomial and matrix examples live in bigger spaces.
Matrix of a Linear Map — Cell G's matrix-as-4-vector is the seed of this idea.