Dimension — basis cardinality — Maths notes

WHAT are we even claiming?

WHY do we care? Without this theorem, "dimension" would be ambiguous — you couldn't say $\mathbb{R}^3$ is 3-dimensional, because maybe some weird basis has 5 vectors. The theorem guarantees the answer is unique.

HOW we prove it — Derivation from scratch

The whole result rests on one workhorse lemma. Derive it and you've derived everything.

Deriving the Exchange Lemma

Let $S = \{w_1,\dots,w_n\}$ span $V$ , and let $L = \{v_1,\dots,v_m\}$ be linearly independent. Goal: show $m \le n$ .

Idea: swap the $v$ 's into the spanning set one at a time, kicking out a $w$ each time, keeping a spanning set of size $n$ throughout. If we never run out of $w$ 's to kick out, then $m \le n$ .

Step 1. Since $S$ spans $V$ , write $v_1 = a_1 w_1 + a_2 w_2 + \dots + a_n w_n.$ Why this step? Spanning means every vector, including $v_1$ , is a combination of the $w$ 's.

Step 2. Some $a_i \ne 0$ (else $v_1 = 0$ , impossible for an independent set). Reorder so $a_1 \ne 0$ . Solve: $w_1 = \frac{1}{a_1}\Big(v_1 - a_2 w_2 - \dots - a_n w_n\Big).$ Why this step? This shows $w_1$ is redundant — we can express it using $v_1$ and the other $w$ 's. So $\{v_1, w_2,\dots,w_n\}$ still spans $V$ .

Step 3 (induction). Suppose after $k$ swaps we have a spanning set $\{v_1,\dots,v_k, w_{k+1},\dots,w_n\}.$ Write $v_{k+1}$ in terms of it: $v_{k+1} = \underbrace{b_1 v_1 + \dots + b_k v_k}_{\text{old }v\text{'s}} + \underbrace{c_{k+1}w_{k+1} + \dots + c_n w_n}_{\text{remaining }w\text{'s}}.$

Step 4 (key). Some $c_j \ne 0$ . Why? If all $c_j = 0$ , then $v_{k+1}$ would be a combination of $v_1,\dots,v_k$ — contradicting that $L$ is independent. So a $w$ must still be present to kick out. This is exactly where independence of $L$ is used. Reorder, kick out $w_{k+1}$ , swap in $v_{k+1}$ . Spanning set of size $n$ survives.

Step 5 (conclude). Each swap consumes one $v$ and one $w$ . We must have at least as many $w$ 's as $v$ 's, i.e. $m \le n$ . $\blacksquare$

From lemma to the Dimension Theorem

Let $B_1$ (size $p$ ) and $B_2$ (size $q$ ) both be bases of $V$ .

$B_1$ is independent, $B_2$ spans $\Rightarrow$ apply lemma: $p \le q$ .
$B_2$ is independent, $B_1$ spans $\Rightarrow$ apply lemma: $q \le p$ .

Therefore $p = q$ . $\blacksquare$ The number is the dimension.

Why this step? A basis is simultaneously independent and spanning, so each one plays both roles in the lemma — giving inequalities in both directions.

Worked Examples

Steel-man the mistakes

Forecast-then-Verify

Flashcards

What is the dimension of a vector space?

The cardinality (number of vectors) of any basis — well-defined because all bases have equal size.

State the Dimension Theorem (invariance of basis cardinality).

Any two bases of the same finite-dimensional space have the same number of vectors.

What does the Steinitz Exchange Lemma state?

V

is spanned by

n

vectors, every linearly independent set has at most

n

vectors (independent ≤ spanning).

In the exchange lemma proof, where is independence of

L

used?

To guarantee some coefficient on a remaining

w

is nonzero, so a

w

can always be swapped out.

Does the spanning set in Steinitz need to be linearly independent?

No — it only needs to span; it may be dependent.

How do you derive the Dimension Theorem from the exchange lemma?

Apply the lemma both ways:

B_1

indep ≤

B_2

spanning gives

p≤q

; reverse roles gives

q≤p

; so

p=q

Why can't 4 vectors form a basis of

\mathbb{R}^3

\mathbb{R}^3

is spanned by 3 vectors, so any independent set has ≤3; 4 vectors must be dependent, hence not a basis.

\dim P_n

(polynomials of degree ≤ n) =

n+1

(basis

\{1,x,\dots,x^n\}

Recall Feynman: explain to a 12-year-old

Imagine you build a LEGO castle. You can use red bricks or blue bricks, but to build that exact same castle you always need the same number of bricks. The bricks are your basis vectors, the castle is the space. No matter which color (which basis) you pick, the count of bricks is fixed — that count is the dimension. The trick to prove it: take your independent bricks and one-by-one swap them into a pile that already builds the castle; you never run out of room, which proves you can't have more independent bricks than the pile had. Do it both directions and the two counts must match.

Connections

Linear Independence — supplies the "some coefficient ≠ 0" step.
Spanning Sets — supplies the slots the lemma fills.
Basis of a Vector Space — the both-at-once object the theorem measures.
Steinitz Exchange Lemma — the engine of this proof.
Rank-Nullity Theorem — dimension counting in action for linear maps.
Coordinates and Change of Basis — why a fixed dimension lets us use coordinate vectors.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek vector space ke andar tum basis kai tarah se chun sakte ho — standard axes, ya tilted axes, koi farak nahi padta. Lekin kamaal ki baat yeh hai ki har basis mein vectors ki count bilkul same hoti hai. Usi fixed number ko hum dimension kehte hain. Jaise $\mathbb{R}^3$ ko represent karne ke liye humesha exactly 3 independent directions chahiye — chahe tum koi bhi 3 directions chuno.

Iska proof ek single lemma pe khada hai — Steinitz Exchange Lemma. Yeh kehta hai: agar space ko $n$ vectors span karte hain, toh koi bhi linearly independent set mein zyada se zyada $n$ vectors ho sakte hain. Simple line mein: independent ≤ spanning. Logic yeh hai — apne independent vectors ko ek-ek karke spanning pile mein ghusao, har baar ek purana vector bahar nikaal do. Jab tak tumhare paas independent vectors hain, koi na koi purana vector nikalne ke liye bachega (yahin pe independence use hoti hai). Isse saabit ho jaata hai ki independent vectors, spanning vectors se zyada nahi ho sakte.

Ab dimension theorem is lemma se turant nikal aata hai: do alag-alag bases lo, $B_1$ aur $B_2$ . $B_1$ independent hai aur $B_2$ span karta hai, toh $|B_1| \le |B_2|$ . Roles ulta karo, toh $|B_2| \le |B_1|$ . Dono milake $|B_1| = |B_2|$ . Bas — har basis ka size same, aur wahi dimension hai.

Yeh kyun important hai? Kyunki iske bina "dimension" word ka koi matlab hi na hota. Yeh theorem guarantee deta hai ki dimension ek well-defined unique number hai, jisse coordinates, rank-nullity, change of basis — sab kuch consistent banta hai. Exam mein common trick: "kya 4 vectors $\mathbb{R}^3$ ka basis ban sakte hain?" — turant lemma lagao, answer NO.

Dimension — basis cardinality

WHAT are we even claiming?

HOW we prove it — Derivation from scratch

Deriving the Exchange Lemma

From lemma to the Dimension Theorem

Worked Examples

Steel-man the mistakes

Forecast-then-Verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Linear Algebra (Full)

Connections