MAT240H1/Unit 02
Unit 02 · Algebra I

Subspaces, Span & Independence

The three ideas that organize a vector space from the inside: which subsets are themselves spaces, what a set of vectors generates, and when vectors carry genuinely independent information. Together they set up the notion of a basis in Unit 3.

By the end you can
Why it matters & where it comes from
Why it exists

Span and independence are how we measure "how much" a set of vectors captures — without redundancy. They lead directly to the idea of a basis.

In the real world

Linear independence is feature redundancy in data science; the span of a dataset is the subspace it can represent. Dimensionality reduction (PCA) is exactly this language.

Story & open questions

These notions crystallized as mathematicians sought the smallest set of "directions" needed to describe a space — the search that produced the modern theory of dimension.

Subspaces

Definition Subspace

$W\subseteq V$ is a subspace if it is itself a vector space under the operations of $V$.

Theorem Subspace test

$W\subseteq V$ is a subspace if and only if: (i) $0\in W$; (ii) $u+v\in W$ whenever $u,v\in W$; (iii) $av\in W$ whenever $a\in F$, $v\in W$.

Proof idea
The eight axioms are equations that already hold in $V$, so they hold automatically for elements of $W$. The only things that can fail are the closures and the presence of $0$ and inverses — and $-v=(-1)v$ comes from closure under scalars. ∎
Examples and a non-example
Subspaces: the solution set of a homogeneous system $Ax=0$; the symmetric matrices inside $M_{n\times n}(F)$; the polynomials with $p(0)=0$.
Non-example: the union of the two axes in $\mathbb{R}^2$. It passes the zero and scalar tests but fails closure under addition: $(1,0)+(0,1)=(1,1)$ lies on neither axis.
Theorem

Any intersection of subspaces of $V$ is a subspace. (Unions usually are not — the example above.)

Span

Definition Span

A linear combination of $v_1,\dots,v_k$ is $a_1v_1+\cdots+a_kv_k$ with $a_i\in F$. The span of a set $S$ is the set of all linear combinations of finitely many vectors of $S$ (and $\operatorname{span}\,\emptyset=\{0\}$).

Theorem

$\operatorname{span}(S)$ is a subspace of $V$, and it is the smallest one containing $S$: any subspace containing $S$ contains $\operatorname{span}(S)$.

Proof idea
Sums and scalar multiples of linear combinations are again linear combinations — that gives the subspace. And a subspace containing $S$ is closed under exactly the operations used to build combinations, so it swallows all of them. ∎

Linear independence

Definition

$v_1,\dots,v_k$ are linearly independent if $a_1v_1+\cdots+a_kv_k=0$ forces $a_1=\cdots=a_k=0$. Otherwise they are dependent: some nontrivial combination gives $0$.

Theorem Dependence means redundancy

Vectors $v_1,\dots,v_k$ (with $k\ge2$) are dependent iff some $v_j$ is a linear combination of the others — and then removing $v_j$ does not change the span.

Proof
If $a_1v_1+\cdots+a_kv_k=0$ with some $a_j\ne0$, divide by $a_j$ (field!) and solve for $v_j$. Conversely, if $v_j=\sum_{i\ne j}c_iv_i$, then $\sum_{i\ne j}c_iv_i+(-1)v_j=0$ is a nontrivial combination. ∎
Example The field matters

Take $u=(1,1,0)$, $v=(0,1,1)$, $w=(1,0,1)$.

Over $\mathbb{R}$: $a u+b v+c w=0$ gives $a+c=0$, $a+b=0$, $b+c=0$; substituting, $2c=0$, so $a=b=c=0$ — independent.
Over $\mathbb{Z}_2$: $u+v+w=(2,2,2)=(0,0,0)$ — dependent! The step "$2c=0\Rightarrow c=0$" fails when $2=0$. Independence is a statement about the scalars as much as the vectors.
How to run an independence check Write $a_1v_1+\cdots+a_kv_k=0$, read off one linear equation per coordinate, and solve the homogeneous system. Exactly the trivial solution ⇒ independent; any free variable ⇒ dependent, and the free variable hands you an explicit dependence.
Common pitfalls

• A subspace must contain the zero vector — check it first.
• Independence means the only combination giving $0$ is trivial; it is not "no two are parallel" (three coplanar vectors in $\mathbb{R}^3$ can be pairwise non-parallel yet dependent).
• Any set containing the zero vector is dependent: $1\cdot0=0$ nontrivially.
• Adding vectors can only grow the span; removing a dependent vector keeps it unchanged.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Subspace test

Is $\{p\in P(\mathbb{R}):p(0)=0\}$ a subspace of $P(\mathbb{R})$?

Answer
Yes: the zero polynomial qualifies, and if $p(0)=q(0)=0$ then $(p+q)(0)=0$ and $(ap)(0)=0$.
Practice 2 · Independence

Are $(1,1,0),(0,1,1),(1,0,-1)$ independent in $\mathbb{R}^3$?

Answer
No — the third equals the first minus the second, so the set is dependent.
Practice 3 · Span

What is $\operatorname{span}\{(1,0),(0,1)\}$ in $\mathbb{R}^2$? And $\operatorname{span}\{(1,2)\}$?

Answer
All of $\mathbb{R}^2$; and the line $\{(t,2t):t\in\mathbb{R}\}$ through the origin.
Practice 4 · Union

Show the union of the $x$-axis and $y$-axis is not a subspace of $\mathbb{R}^2$.

Solution
$(1,0)$ and $(0,1)$ belong, but their sum $(1,1)$ lies on neither axis — closure under addition fails.
Practice 5 · Intersection

Prove that the intersection of two subspaces of $V$ is a subspace.

Solution
$0$ lies in both. If $u,v$ lie in both, then $u+v$ and $av$ lie in each (both are subspaces), hence in the intersection. The three-part test passes.
Practice 6 · Polynomials

Prove $1,x,x^2$ are independent in $P(\mathbb{R})$.

Solution
If $a+bx+cx^2$ is the zero polynomial, all its coefficients vanish by definition of polynomial equality — so $a=b=c=0$. (Or: evaluate at $x=0$ to get $a=0$, then at $x=\pm1$ to force $b=c=0$.)
Practice 7 · Field dependence

Show $(1,1,0),(0,1,1),(1,0,1)$ are independent over $\mathbb{R}$ but dependent over $\mathbb{Z}_2$.

Solution
Over $\mathbb{R}$ the system $a+c=a+b=b+c=0$ forces $2c=0$, so all vanish. Over $\mathbb{Z}_2$ the sum of the three vectors is $(2,2,2)=(0,0,0)$ — a nontrivial dependence, because $2=0$ there.
Practice 8 · New from old

Suppose $u,v,w$ are independent in a real vector space. Prove $u+v,\ v+w,\ u+w$ are independent.

Solution
If $a(u+v)+b(v+w)+c(u+w)=0$, regroup: $(a+c)u+(a+b)v+(b+c)w=0$. Independence gives $a+c=a+b=b+c=0$; subtracting equations yields $b=c$, then $2b=0$, so $a=b=c=0$. (Over $\mathbb{Z}_2$ this fails — the three sums are dependent there.)

Expanded topic checklist

Course coverage. This unit is part of the complete MAT240H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT240H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The spine is: subspace test, span as construction, independence as uniqueness, then basis as both spanning and independent.

Study routine. For every set of vectors, ask two separate questions: can it create every vector, and can it create zero in a nontrivial way?

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Subspace test

Is $\{(x,y):x+y=0\}$ a subspace of $F^2$?

Guide
Yes. It contains $(0,0)$ and is closed under addition and scalar multiplication because the equation is homogeneous.
Extension 2 · Not subspace

Is $\{(x,y):x+y=1\}$ a subspace?

Guide
No. It does not contain the zero vector and is not closed under scalar multiplication.
Extension 3 · Independence

Test whether $(1,0),(1,1)$ are independent in $F^2$.

Guide
If $a(1,0)+b(1,1)=(0,0)$, then $(a+b,b)=(0,0)$, so $b=0$ and $a=0$.
Extension 4 · Span

Describe span$\{(1,1),(1,-1)\}$ over a field of characteristic not $2$.

Guide
They span all of $F^2$ because any $(x,y)$ is $((x+y)/2)(1,1)+((x-y)/2)(1,-1)$.
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