Worked examples — Dimension — basis cardinality

This page is a shooting gallery. The parent note proved the one theorem: every basis has the same number of vectors, and that number is the dimension. Here we fire every kind of situation at that theorem — small spaces, big spaces, the empty basis, infinite-dimensional spaces, redundant piles, word problems, and an exam trap — so you never meet a case you haven't already seen worked.

Before symbols fly, three plain-word reminders (each defined in the parent, re-anchored here so line one is readable):

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The scenario matrix

Every example below is tagged with a cell from this table. Together they hit every cell.

Cell	Scenario class	What could go wrong / what to test
A	Ordinary finite space, two honest bases	do the counts really match?
B	Degenerate: the zero space $\{0\}$	what is the basis of "nothing but the origin"?
C	Too-big spanning set (dependent pile)	Steinitz still works — does it forbid over-large independent sets?
D	Too-small independent set	independence alone does not pin the dimension
E	Non-arrow space (polynomials / matrices)	dimension of things that aren't ${\mathbb{R}}^n$
F	Limiting / infinite case	what happens when no finite spanning set exists?
G	Real-world word problem	dimension as "number of free dials"
H	Exam twist — subspace inside a bigger space	count without listing a basis

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Cell A — two honest bases, same count

Step 1. Test independence: set $a(2,1)+b(1,3)=(0,0)$. This gives $2a+b=0$ and $a+3b=0$. Why this step? Independence means the only way to build the zero arrow is $a=b=0$. If some non-zero $a,b$ worked, one arrow would be a shadow of the other.

Step 2. Solve. From the first, $b=-2a$; substitute: $a+3(-2a)=a-6a=-5a=0\Rightarrow a=0\Rightarrow b=0$. Why this step? Getting only the trivial solution proves independence — the two arrows point in genuinely different directions (see figure).

Step 3. Two independent vectors in ${\mathbb{R}}^2$ automatically span ${\mathbb{R}}^2$ (independent count $=$ dimension $\Rightarrow$ basis). So $B_2$ is a basis and $|B_2|=2$. Why this step? This is the theorem paying off: once the count matches the dimension, "independent" upgrades to "basis" for free.

Verify: The determinant of the columns is $2\cdot 3-1\cdot 1=5\ne 0$, so the vectors are independent and span. Count $=2$ ✓ — same as the standard basis, exactly as the Dimension Theorem promises.

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Cell B — the zero space, the degenerate input

Step 1. A basis must be independent. But the set $\{0\}$ is dependent: $1\cdot 0=0$ is a non-trivial combination equalling zero. Why this step? The zero vector can never sit in an independent set — you can always scale it by a non-zero number and still get $0$.

Step 2. So the basis cannot contain $0$. The only remaining candidate is the empty set $\varnothing$. Why this step? We are forced by Step 1 to throw out the only element available.

Step 3. Is $\varnothing$ a basis? It is independent (vacuously — no vectors, so no bad combination exists) and it spans $\{0\}$ (the empty sum is defined to be $0$, the only vector present). Why this step? Both defining properties of a basis hold, so $\varnothing$ qualifies, and $|\varnothing |=0$.

Verify: $\dim \{0\}=0$. Sanity check with the theorem: any two bases must have equal size, and here the only basis is $\varnothing$, size $0$ — no contradiction ✓.

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Cell C — a too-big, dependent spanning pile

Step 1. Apply Steinitz using the spanning pile we already know: ${\mathbb{R}}^3$ is spanned by the 3 standard vectors. Why this step? The lemma's ceiling is set by the smallest spanning set we can name, not by the bloated 5-vector one. IN ≤ SPAN, and the tightest SPAN available is $3$.

Step 2. Therefore any independent set has size $\le 3$. So 4 independent vectors are impossible. Why this step? Even though someone advertised a size-5 spanning set, a smaller spanning set (size 3) still forbids 4-independent piles. The lemma does not require the spanning pile to be independent — dependence is allowed.

Step 3. The maximum independent count is exactly $3$, achieved e.g. by $\{(1,0,0),(0,1,0),(0,0,1)\}$. Why this step? Three independent vectors exist (standard basis), and Step 2 says you can't beat 3, so 3 is the ceiling and it's reached.

Verify: $\dim {\mathbb{R}}^3=3$; a size-5 spanning set is legal (it's just dependent) but the independent ceiling stays $3$. Largest independent set $=3$; four independent vectors $\Rightarrow$ impossible ✓.

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Cell D — a too-small independent set

Step 1. Check independence: $a(1,0,0)=(0,0,0)\Rightarrow a=0$. Independent ✓ (a single non-zero vector always is). Why this step? Confirms the set is a legitimate independent pile — the trap is that this feels "enough".

Step 2. Check spanning: can we build $(0,1,0)$ as $a(1,0,0)$? That needs $(a,0,0)=(0,1,0)$ — impossible. Why this step? Spanning demands every vector be reachable. One arrow along the $x$-axis can never reach off-axis vectors.

Step 3. So it spans only a line (a 1-dimensional subspace), not all of ${\mathbb{R}}^3$. It is not a basis of ${\mathbb{R}}^3$; it is a basis of that line. Why this step? Independence gives the lower half of $\text{IN}\le n\le \text{SPAN}$; you need spanning too before the count is pinned at $\dim =3$.

Verify: Size $1<3=\dim {\mathbb{R}}^3$, so it cannot be a basis of ${\mathbb{R}}^3$ — consistent with the theorem: a basis must hit the dimension exactly, not fall short ✓.

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Cell E — a space that isn't arrows: $2\times 2$ matrices

Step 1. Propose the pile $E_{11}=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),E_{21}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),E_{22}=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right).$ Why this step? Each $E_{ij}$ is "1 in one slot, 0 elsewhere". Any matrix is $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=a{E}_{11}+b{E}_{12}+c{E}_{21}+d{E}_{22}$, so these clearly span.

Step 2. Independence: if $a{E}_{11}+b{E}_{12}+c{E}_{21}+d{E}_{22}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$, then matching entries forces $a=b=c=d=0$. Why this step? Each basis matrix owns a distinct slot, so no combination can cancel unless every coefficient is zero — genuinely new "directions".

Step 3. Both properties hold, so it's a basis with 4 elements: $\dim V=4$. Why this step? By the Dimension Theorem this $4$ is forced — every other basis of $2\times 2$ matrices also has exactly 4 members.

Verify: $\dim {\mathbb{R}}^{2\times 2}=2\times 2=4$ ✓. General pattern $\dim {\mathbb{R}}^{m\times n}=mn$; here $mn=4$.

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Cell F — the limiting case: no finite spanning set

Step 1. Suppose, for contradiction, that a finite pile $\{p_1,\dots ,p_n\}$ spans $P$. Why this step? We test the limiting behaviour by assuming the finite tools of Steinitz could apply, then look for a crack.

Step 2. Let $d$ be the highest degree among $p_1,\dots ,p_n$. Every combination of them has degree $\le d$. But $x^{d+1}\in P$ has degree $d+1$. Why this step? Combining polynomials never raises the top degree, so a fixed finite pile has a degree ceiling — a genuine limiting obstruction.

Step 3. So $x^{d+1}$ is not spanned. No finite pile spans $P$; $P$ is infinite-dimensional. Why this step? The set $\{1,x,x^2,x^3,\dots \}$ is an infinite basis. The Dimension Theorem still holds in spirit — all bases have the same infinite cardinality — but Steinitz's finite ceiling simply never kicks in.

Verify: For each finite $n$, $\dim P_n=n+1$ (matches parent). As $n$ grows without bound, $\dim P=\infty$ ✓ — the honest limiting value.

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Cell G — word problem: dimension as free dials

Step 1. "None reproducible from the others" = the 5 recipe-vectors are linearly independent in ${\mathbb{R}}^3$. Why this step? Translating the word "irreproducible" into "independent" lets the algebra decide the claim.

Step 2. ${\mathbb{R}}^3$ is spanned by 3 vectors, so IN ≤ SPAN gives independent count $\le 3$. Why this step? Three fruits = three dials = a 3-slot spanning pile. The lemma caps independence at that.

Step 3. $5>3$, so the 5 recipes cannot all be independent — at least one is a mix of the others. Why this step? The claim violates the ceiling; the machine's marketing is mathematically impossible.

Verify: Max independent recipes $=\dim {\mathbb{R}}^3=3$; claiming 5 independent is impossible since $5>3$ ✓. (You can offer 3 truly independent signature blends at most.)

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Cell H — exam twist: dimension of a subspace, no basis listed

Step 1. Solve the constraint for one variable: $z=-x-y$, with $x,y$ free. Why this step? One linear equation removes one degree of freedom; the free variables count the dimension (this is Rank-Nullity Theorem in miniature).

Step 2. Write a general element: $(x,y,-x-y)=x(1,0,-1)+y(0,1,-1)$. Why this step? Every solution is a combination of $(1,0,-1)$ and $(0,1,-1)$ — so these two span $W$.

Step 3. Independence: $x(1,0,-1)+y(0,1,-1)=(0,0,0)$ forces $x=0$ (first slot) and $y=0$ (second slot). Independent ✓. So they form a basis of size 2. Why this step? Both basis properties hold, and by the Dimension Theorem the count $2$ is the honest dimension — no other basis of $W$ can differ.

Verify: One constraint on ${\mathbb{R}}^3$ gives $\dim W=3-1=2$ ✓. Rank-nullity check: the map $(x,y,z)\mapsto x+y+z$ has rank 1, so nullity $=3-1=2=\dim W$ ✓.

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Roundup

Recall Which cell taught what?

Matched every scenario in the matrix? Cell A — tilted basis still counts 2 ::: determinant $\ne 0\Rightarrow$ independent, count $=\dim =2$. Cell B — dimension of $\{0\}$ ::: $0$; basis is the empty set $\varnothing$. Cell C — 5-vector dependent spanning set ::: independent ceiling stays $3$; 4 independent is impossible. Cell D — single independent vector in ${\mathbb{R}}^3$ ::: not a basis; spans only a line, too small. Cell E — $2\times 2$ matrices ::: $\dim =4$; basis $\{E_{11},E_{12},E_{21},E_{22}\}$. Cell F — all polynomials $\mathbb{R}[x]$ ::: infinite-dimensional; no finite pile spans. Cell G — 5 smoothie recipes on 3 fruits ::: at most 3 can be independent, so the claim fails. Cell H — plane $x+y+z=0$ ::: $\dim =2$ via one constraint / rank-nullity.

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Connections

• Steinitz Exchange Lemma — the ceiling used in Cells A, C, G.
• Linear Independence — the "no vector is a shadow" test in every cell.
• Spanning Sets — the slots being filled or falling short.
• Basis of a Vector Space — both-at-once object each example builds.
• Rank-Nullity Theorem — the counting shortcut in Cell H.
• Coordinates and Change of Basis — why same-count bases can look so different (Cell A).