MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: subspaces, span and linear independence (MATH 133)

This sheet is not a summary of Nicholson 5.1: you already have the notes. It answers one question only, what makes students lose marks on subspaces, span and linear independence in MATH 133 at McGill University, and which precise gesture avoids each loss.

The chapter is where the course stops asking for numbers and starts asking for proofs, and the rule that organizes every answer fits in five words: yes needs letters, no needs numbers. Everything stays in $\mathbb{R}^n$ ; spaces of polynomials and functions belong to MATH 223 and do not appear on this exam.

The thread of the chapter

Yes needs letters, no needs numbers: a subspace, a span or an independent family is PROVED with arbitrary vectors written with letters, and a failure is SHOWN with one explicit witness, a vector that leaves the set, a right-hand side that makes the system inconsistent, or a dependency relation you can check.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Subspace of R^n
Subspace test
Closure under addition
Closure under scalar multiplication
Null space of a matrix
Image of a matrix
Linear combination
Linear span
Linear independence
Linear dependence relation
Homogeneous linear system
Trivial solution

The essentials

The three-point test, and what each answer must look like

• $U \subseteq \mathbb{R}^n$ is a subspace when: (1) $0 \in U$ ; (2) $u, v \in U \Rightarrow u + v \in U$ ; (3) $u \in U$ , $c \in \mathbb{R} \Rightarrow cu \in U$ .
• A YES is proved for ARBITRARY $u, v$ written with letters: 'let $u = (x_1, y_1, z_1)$ and $v = (x_2, y_2, z_2)$ be in $U$ '.
• A NO is shown with ONE explicit counterexample written with numbers: a named vector of $U$ , and a sum or multiple that leaves $U$ .
• The zero test is a filter: failing it ends the question, passing it decides nothing.
• Always subspaces: a set cut out by homogeneous linear equations, $\text{null}(A)$ , $\text{im}(A)$ , any span, $\{0\}$ and $\mathbb{R}^n$ .

Same slope, one difference:

y = x/2

goes through the origin and is a subspace,

y = x/2 + 2

misses it by

2

and is not, whatever else is true about it.

A line or a plane in $\mathbb{R}^3$ is a subspace exactly when it passes through the origin. Look at the constant term first: a non-zero constant is a no in one line.

Span and independence are both systems

• $b \in \text{span}\{v_1, \dots, v_k\}$ exactly when $(v_1 \; \cdots \; v_k \mid b)$ is consistent: no row $0 = c$ with $c \neq 0$ .
• $\{v_1, \dots, v_k\}$ is independent exactly when $(v_1 \; \cdots \; v_k)c = 0$ has ONLY the trivial solution: a pivot in every column.
• 'Dependent' is answered by a relation $c_1v_1 + \cdots + c_kv_k = 0$ with coefficients not all zero, written with numbers and checked.
• Automatic dependence: a family containing $0$ ; more than $n$ vectors of $\mathbb{R}^n$ . Two vectors are dependent iff one is a multiple of the other.
• $\text{im}(A) = \text{span}$ of the columns of $A$ , because $Ax = x_1c_1 + \cdots + x_nc_n$ .

Put the vectors as COLUMNS. Rows answer a different question, and mixing the two is the commonest setup error of the chapter.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Concluding no without a counterexample

most of the question: the claim is right, the proof is missing

What not to write

“ $U = \{(x, y) : x \ge 0, y \ge 0\}$ is not closed under scalar multiplication, so it is not a subspace.”

What to write

“ $(1, 2) \in U$ but $(-1)(1, 2) = (-1, -2) \notin U$ , so $U$ is not closed under scalar multiplication and is not a subspace.”

Why: 'Not closed' is the conclusion, not the evidence. A marker gives the points for the named vector and the named scalar, because that is what makes the claim checkable. Choose the witness to be as simple as possible: $-1$ as scalar, standard vectors as summands.

2. Taking the zero test for the whole test

the whole question

What not to write

“ $(0, 0)$ is in $\{(x, y) : xy = 0\}$ , so it is a subspace.”

What to write

“ $0 \in U$ , but $(1, 0)$ and $(0, 1)$ are in $U$ and $(1, 0) + (0, 1) = (1, 1)$ is not, since $1 \cdot 1 \neq 0$ : not a subspace.”

Left:

u

is in the quadrant,

-u

is not. Right:

(1, 0)

and

(0, 1)

lie on the axes, their sum

(1, 1)

does not. Both sets contain

0

Why: Passing the zero test is necessary, never sufficient. The two sets on the figure both contain $0$ ; one fails scalar multiplication, the other fails addition, so neither closure condition can be skipped.

3. Proving closure with particular vectors

all the marks for that condition

What not to write

“ $(2, 1, 0)$ and $(1, 2, 1)$ are in $x - 2y + 3z = 0$ and so is their sum $(3, 3, 1)$ , hence the set is closed under addition.”

What to write

“Let $u = (x_1, y_1, z_1)$ , $v = (x_2, y_2, z_2) \in U$ . Then $(x_1 + x_2) - 2(y_1 + y_2) + 3(z_1 + z_2) = 0 + 0 = 0$ , so $u + v \in U$ .”

Why: A yes is a claim about EVERY pair of vectors, and two examples cannot cover infinitely many pairs. The quadrant is closed under addition on every example you try with positive entries, and it is still not a subspace.

4. Calling the solution set of Ax = b a subspace

2 marks

What not to write

“The solutions of $x + y - z = 2$ , $x + 2y + z = 3$ form a line, so they form a subspace.”

What to write

“ $A0 = 0 \neq (2, 3)$ , so $0$ is not a solution: the solution set is the line $\text{null}(A)$ shifted by $(1, 1, 0)$ , not a subspace.”

Why: Being a line is not enough, it must be a line THROUGH THE ORIGIN. Only the homogeneous system $Ax = 0$ gives a subspace; $Ax = b$ with $b \neq 0$ gives a parallel copy of it.

5. Deciding span membership by looking for multiples

the whole question, and a wrong answer

What not to write

“ $(0, 1, 2)$ is not a multiple of $(1, 2, -1)$ nor of $(2, 5, 0)$ , so it is not in their span.”

What to write

“Row reducing $(v_1 \; v_2 \mid b)$ gives a consistent system with $a_1 = -2$ , $a_2 = 1$ : $(0, 1, 2) = -2v_1 + v_2$ , so it IS in the span.”

Why: The span contains all COMBINATIONS, not just multiples. A multiple is a combination with one coefficient zero, a tiny corner of the span. The only reliable test is the system.

6. Using the trivial solution as evidence

all the marks for the justification

What not to write

“ $c_1 = c_2 = c_3 = 0$ solves $c_1v_1 + c_2v_2 + c_3v_3 = 0$ , so the vectors are independent.”

What to write

“Row reduction gives a pivot in every column, so $c_1 = c_2 = c_3 = 0$ is the ONLY solution: the vectors are independent.”

Why: The trivial solution exists for every family, dependent or not, so exhibiting it proves nothing. Independence is about uniqueness of that solution, and the proof is the pivot count or a letters argument that forces every coefficient to zero.

7. Applying the two-vector test to three vectors

the whole question

What not to write

“None of $(1, 0, 0)$ , $(0, 1, 0)$ , $(1, 1, 0)$ is a multiple of another, so they are independent.”

What to write

“ $(1, 0, 0) + (0, 1, 0) - (1, 1, 0) = 0$ is a non-trivial relation, so the family is dependent.”

Why: 'Dependent iff one is a multiple of the other' is true for two vectors only. Three vectors are dependent as soon as one is a COMBINATION of the other two, which means they lie in a common plane through the origin.

8. Dividing by a pivot that contains k

1 to 3 marks, one per lost case

What not to write

“ $R_2 \to \frac{1}{k - 4}R_2$ , and then the family is dependent only for $k = 2$ .”

What to write

“Swap $R_2 \leftrightarrow R_3$ to keep a pivot free of $k$ ; the last pivot is $-(k - 2)(k - 3)$ , so the family is dependent exactly for $k = 2$ and $k = 3$ .”

Why: Dividing by $k - 4$ silently assumes $k \neq 4$ , and a quadratic factor has two roots, not one. Keep parameters out of the pivots when a swap allows it, and solve the final condition completely.

Which method to choose

Is this set a subspace? Decide on the form of its definition

Read how the set is defined before testing anything

If a linear equation with a non-zero constant, such as $y = 2x + 1$ or $x + 2y - z = 5$ → no, in one line: $0$ does not satisfy it

Example: $0 \neq 2(0) + 1$
If an inequality, such as $x \ge 0$ → no: multiply a vector of the set by $-1$

Example: $(1, 2) \mapsto (-1, -2)$
If a square, a product or an absolute value: $z = x^2$ , $xy = 0$ , $|x| = |y|$ → no, find the witness: scale by $2$ for a square, add two simple vectors for a product or an absolute value

Example: $(1, 0) + (0, 1) = (1, 1)$ leaves $xy = 0$
If homogeneous linear equations, even with a parameter on an unknown: $x + y + z = kz$ → yes: prove the three conditions with letters

Example: $x + y + (1 - k)z = 0$ for every $k$
If $\text{null}(A)$ , $\text{im}(A)$ , or a span → yes: quote or reprove the theorem

Example: $A(x + y) = Ax + Ay = 0$

A parameter on the right-hand side is decided by the zero test: $x + 2y - z = k^2 - 4$ is a subspace exactly when $k^2 = 4$ , so for $k = 2$ AND $k = -2$ .

What does the span look like? Count the pivots

Put the vectors of $\mathbb{R}^3$ as columns and row reduce

One pivot: a line. Two pivots: a plane through

O

. A third vector OUTSIDE that plane gives three pivots and all of

\mathbb{R}^3

; a third vector inside it changes nothing.

If one pivot, the vectors all multiples of one non-zero vector → a line through the origin

Example: $(2, -1, 3)$ and $(-4, 2, -6)$
If two pivots → a plane through the origin: carry $(x, y, z)$ in the last column to get its equation

Example: $(1, 0, 2)$ , $(0, 1, -1)$ give $2x - y - z = 0$
If three pivots → all of $\mathbb{R}^3$ : the system is consistent for every $b$

Example: add $(0, 0, 1)$ to the two above
If fewer pivots than vectors → dependent: read the relation off the free variable

Example: $2(1, -1, 2) - (2, 1, 1) + (0, 3, -3) = 0$

Count what is independent, never what is listed: three vectors can span a plane, and two vectors can span a line.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving that a set is a subspace

When to use it: 'Show that $U$ is a subspace of $\mathbb{R}^n$ ', with $U$ defined by homogeneous conditions

1 Zero: substitute $0$ into every condition and write the resulting equality.
2 Addition: 'let $u = (x_1, \dots)$ and $v = (x_2, \dots)$ be in $U$ ', write what that means, then compute the condition for $u + v$ and regroup it as (condition for $u$ ) plus (condition for $v$ ).
3 Scalars: 'let $c \in \mathbb{R}$ ', compute the condition for $cu$ and factor out $c$ .
4 Conclude with a sentence that names the three conditions.

Concluding sentence

“ $0 \in U$ , and $U$ is closed under addition and scalar multiplication, so $U$ is a subspace of $\mathbb{R}^3$ .”

The trap: Writing the addition step with numbers. Two examples are evidence, never a proof.

Marking: Typically 1 mark for zero, 2 for addition, 2 for scalars; a numerical example earns none of the last four.

Proving that a family built from an independent one is independent

When to use it: ' $\{u, v, w\}$ is independent; show that $\{u + v, v + w, u + w\}$ is independent'

1 Start from a general relation: 'suppose $a(u + v) + b(v + w) + c(u + w) = 0$ '.
2 Regroup by the original vectors: $(a + c)u + (a + b)v + (b + c)w = 0$ .
3 Use the hypothesis, by name: 'since $\{u, v, w\}$ is independent, $a + c = a + b = b + c = 0$ '.
4 Solve the small system and conclude $a = b = c = 0$ .

Concluding sentence

“The only relation is the trivial one, so $\{u + v, v + w, u + w\}$ is linearly independent.”

The trap: Skipping the regrouping, or forgetting the sentence that invokes the hypothesis: that sentence is the only place the proof uses it.

Marking: Typically 1 mark for the general relation, 1 for the regrouping, 1 for invoking independence, 2 for solving.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The witness really belongs to the set

Before writing that a sum or a multiple leaves the set, check that the vectors you started from are IN it. A counterexample built on a vector outside the set proves nothing.

$(1, 1)$ on $y = |x|$ : $1 = |1|$ , yes; then $(-1, -1)$ : $-1 \neq 1$ , out.
Substitute the combination back

A claimed combination or dependency relation is checked coordinate by coordinate, including the coordinate you did not use to find the coefficients.

$-2(1, 2, -1) + (2, 5, 0) = (0, 1, 2)$ : three coordinates, three checks.
The plane equation contains every generator

An equation found for a span must be satisfied by each vector you spanned from. If one fails, the row reduction carrying $(x, y, z)$ went wrong.

$5x - 2y + z$ on $(1, 2, -1)$ gives $0$ and on $(2, 5, 0)$ gives $0$ .
Count vectors against the space

More vectors than coordinates means dependent before any computation. If your reduction says three vectors of $\mathbb{R}^2$ are independent, there is an arithmetic slip.

$(1, 2)$ , $(3, -1)$ , $(5, 3)$ must satisfy a relation: $2u + v - w = 0$ .

The typical problem, taken apart

A parameter, a relation and the plane it leaves

Let $v_1 = (1, 1, 1)$ , $v_2 = (1, 2, 3)$ and $v_3 = (2, k, 4)$ .

Find every $k$ for which $\{v_1, v_2, v_3\}$ is dependent, give the dependency relation, and describe the span for that $k$ , with a vector that lies outside it.

At the critical

k

v_3

lies in the plane of

v_1

and

v_2

: three vectors, one plane, and a relation among them.

Step 1

Columns: $\begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & k \\ 1 & 3 & 4 \end{pmatrix}$ . $R_2 \to R_2 - R_1$ gives $(0 \; 1 \; k - 2)$ and $R_3 \to R_3 - R_1$ gives $(0 \; 2 \; 2)$ .

Why

Vectors as columns, because the unknowns are the coefficients of the relation. Each operation is named: the method marks are given for the names.

Step 2

$R_3 \to R_3 - 2R_2$ gives $(0 \; 0 \; 6 - 2k)$ . Dependent exactly when $6 - 2k = 0$ , that is $k = 3$ .

Why

The pivot used is 1, free of k, so nothing was divided away. The last entry decides everything: non-zero means three pivots and only the trivial solution.

Step 3

For $k = 3$ : $c_3 = t$ is free, $c_2 = -(k - 2)t = -t$ , $c_1 = -c_2 - 2c_3 = -t$ . With $t = -1$ : $v_1 + v_2 - v_3 = 0$ .

Why

Dependent is answered with a relation, not only with the word. Picking t to make the coefficients small keeps the check easy.

Step 4

Check: $(1 + 1 - 2, 1 + 2 - 3, 1 + 3 - 4) = (0, 0, 0)$ .

Why

Every coordinate, including the third, which played no role in finding the coefficients.

Step 5

Span: $v_3 = v_1 + v_2$ , so it is $\text{span}\{v_1, v_2\}$ , of the form $(s + t, s + 2t, s + 3t)$ , which satisfies $x - 2y + z = 0$ . Outside it: $(1, 0, 0)$ , since $1 - 0 + 0 = 1 \neq 0$ .

Why

The geometric description closes the question, and the explicit outside vector proves the span is not all of space: a no, so a number.

The conclusion, written out

“The family is dependent only for $k = 3$ , where $v_1 + v_2 - v_3 = 0$ ; its span is the plane $x - 2y + z = 0$ , which does not contain $(1, 0, 0)$ .”

The classic mistake on this problem: Stopping at 'for $k = 3$ there is a free variable, so dependent' with no relation, then calling the span $\mathbb{R}^3$ because there are three vectors.

Learn by heart

• Subspace: $0 \in U$ , closed under $+$ , closed under scalars. All three, for ALL vectors.
• Yes needs letters; no needs one numerical witness.
• A line or plane is a subspace iff it passes through the origin; $Ax = b$ with $b \neq 0$ never gives one.
• $b \in \text{span}\{v_i\}$ iff $(v_1 \; \cdots \; v_k \mid b)$ is consistent.
• Independent iff a pivot in EVERY column of $(v_1 \; \cdots \; v_k)$ ; dependent means a relation with coefficients not all zero.
• Contains $0$ : dependent. More than $n$ vectors in $\mathbb{R}^n$ : dependent. Sub-family of independent: independent.
• Two vectors: dependent iff parallel. Three or more: that test says NOTHING.

Frequently asked questions

How do I show that a set is not a subspace in linear algebra?

Give one explicit counterexample with numbers. Check first whether the zero vector is in the set; if not, you are done in one line. Otherwise name a vector of the set and a scalar, often minus one, whose product leaves the set, or two vectors whose sum leaves it. Saying that the set is not closed without naming the vectors earns almost nothing.

Is a set that contains the zero vector automatically a subspace?

No. Containing the zero vector is necessary but not sufficient. The graph of y equals the absolute value of x contains the origin but fails scalar multiplication, and the union of the two coordinate axes contains the origin but fails addition. You must still prove closure under addition and under scalar multiplication for arbitrary vectors.

How do I check if a vector is in the span of other vectors?

Write the vectors as the columns of a matrix, add the target vector as an augmented column, and row reduce. If a row reads zero equals a non-zero number, the vector is not in the span. Otherwise it is, and back substitution gives the coefficients, which you should check by recomputing the combination.

What is the difference between linearly independent and spanning?

Independence asks whether the only combination giving zero is the one with all coefficients zero; spanning asks whether every vector of the space can be reached. With the vectors as columns, independence means a pivot in every column, and spanning the whole space means a pivot in every row. A family can have one property without the other.

Why are more than three vectors in R3 always dependent?

Looking for a relation among them is a homogeneous system with three equations and more than three unknowns. It has at most three pivots, so at least one free variable, and choosing that variable non-zero gives a relation whose coefficients are not all zero. For example, the three standard vectors minus the vector one, one, one add up to zero.

Practise it

Corrected exercises: Subspaces, span and linear independence, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: subspaces, span and linear independence (MATH 133)

The essentials

The three-point test, and what each answer must look like

Span and independence are both systems

The mistakes that cost marks

1. Concluding no without a counterexample

2. Taking the zero test for the whole test

3. Proving closure with particular vectors

4. Calling the solution set of Ax = b a subspace

5. Deciding span membership by looking for multiples

6. Using the trivial solution as evidence

7. Applying the two-vector test to three vectors

8. Dividing by a pivot that contains k

Which method to choose

Is this set a subspace? Decide on the form of its definition

What does the span look like? Count the pivots

How the answer is expected to be written

Proving that a set is a subspace

Proving that a family built from an independent one is independent

Check before you hand in

The typical problem, taken apart

A parameter, a relation and the plane it leaves

Learn by heart

Frequently asked questions

How do I show that a set is not a subspace in linear algebra?

Is a set that contains the zero vector automatically a subspace?

How do I check if a vector is in the span of other vectors?

What is the difference between linearly independent and spanning?

Why are more than three vectors in R3 always dependent?

Corrected exercises: Subspaces, span and linear independence, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?