Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: vector spaces and subspaces (201-SN4-RE)

This sheet does not list the eight axioms again: you have them in your notes, and the exam almost never asks for them. It deals with what actually gets marked on tests in 201-SN4-RE (formerly 201-NYC): the order in which the subspace test is checked, the sentence that proves closure, and the dimension that is counted on a basis.

It is written for CEGEP students in the Science program, and for Social Science students in the mathematics profile taking linear algebra (201-105). Once you have read it, the corrected exercise set of the same chapter puts each habit to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A subspace is PROVED, not observed: three conditions, the first of which, $\vec{0}\in W$ , settles half the questions in one line. And a dimension is counted on a BASIS, never on the number of vectors you happen to have in hand.

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

The essentials

The subspace test, in the order you check it

• $W\subseteq V$ is a subspace if and only if the following THREE conditions all hold.
• 1) $\vec{0}\in W$ . This is the first test, and when it fails it settles the question in one line.
• 2) Closure under addition: if $u\in W$ and $v\in W$ , then $u+v\in W$ .
• 3) Closure under scalar multiplication: if $u\in W$ and $\lambda\in\mathbb{R}$ , then $\lambda u\in W$ .
• Three objects are subspaces WITHOUT any check: $\operatorname{span}\{v_{1},\ \dots,\ v_{k}\}$ , $\ker T$ and $\operatorname{im}\,T$ . Saying so is enough and earns the marks.

The only subspaces of the plane are these three: the zero vector alone, a line THROUGH THE ORIGIN, and the whole plane. Anything not shown here, in particular a shifted line, is not one.

A subspace is never empty: it always contains $\vec{0}$ . That is why the zero vector check comes first, before any manipulation.

Vectors that are not arrows

• $P_{n}$ , the polynomials of degree at most $n$ , with the standard basis $\{1,\ x,\ x^{2},\ \dots,\ x^{n}\}$ : $\dim P_{n}=n+1$ , not $n$ .
• $M_{m\times n}$ , the matrices, with the matrices that have a single entry equal to 1 as a basis: $\dim M_{m\times n}=mn$ . In particular $\dim M_{2\times 2}=4$ .
• The continuous functions on an interval, an infinite-dimensional space, where there is no longer any finite basis.
• In all these spaces, a question of independence or dimension comes down to an ordinary system once each vector is written in the standard basis of the space.

The polynomial

a+bx

reads as the vector

(a, b)

of its coordinates in the basis

\{1,\ x\}

1+x

and

1-x

become

(1, 1)

and

(1, -1)

, not parallel, hence a basis of

P_{1}

(

\dim P_{1}=2

$\{1+x,\ 1-x\}$ in $P_{1}$ reads $\{(1, 1),\ (1, -1)\}$ in the basis $\{1,\ x\}$ : the determinant is $-2\neq 0$ , the set is linearly independent, and it is a basis of $P_{1}$ .

Dimension, rank, kernel and image

• The DIMENSION is the number of vectors in a BASIS, that is, in a set that is both linearly independent and spanning. All bases of the same space have the same number of vectors.
• $\ker T=\{v : T(v)=\vec{0}\}$ is a subspace of the DOMAIN. $\operatorname{im}\,T=\{T(v)\}$ is a subspace of the CODOMAIN.
• RANK-NULLITY THEOREM: $\dim(\ker T)+\dim(\operatorname{im}\,T)=\dim V$ , where $V$ is the DOMAIN. The codomain plays no part.
• The rank of a matrix is the number of pivots in its echelon form, and it is also $\dim(\operatorname{im}\,T)$ and the dimension of the column space.
• GRASSMANN'S FORMULA: $\dim(F+G)=\dim F+\dim G-\dim(F\cap G)$ .

The projection

T(x, y, z)=(x, y, 0)

: kernel, the

z

-axis (orange, dimension

1

); image, the horizontal plane (blue, dimension

2

). And

1+2=3

, the dimension of the DOMAIN.

$T$ is one-to-one if and only if $\ker T=\{\vec{0}\}$ , that is, $\dim\ker T=0$ . The kernel is never empty: at the very least it contains the zero vector.

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. Taking a line that misses the origin for a subspace

the whole question, usually 3 marks, and the answer fit in one line.

What not to write

“ $W=\{(x, y) : y=2x+1\}$ is a line, so it is a subspace of $\mathbb{R}^{2}$ .”

What to write

“ $(0, 0)$ does not satisfy $y=2x+1$ , since $0\neq 1$ : $\vec{0}\notin W$ , so $W$ is not a subspace.”

The two solid points are on the line; their sum, the open point, is not, and neither is the origin: two of the three conditions fail at once on the same figure.

Why: The word “line” is not the test. A line is a subspace exactly when it passes through the origin, and the constant term of its equation tells you so at once.

2. Checking closure on a numerical example instead of proving it

2 marks out of 3: the computation is right, the proof does not exist.

What not to write

“ $u=(1, 2)$ and $v=(3, 6)$ are in $W$ , and so is $u+v=(4, 8)$ : $W$ is closed under addition.”

What to write

“Let $u=(a, 2a)$ and $v=(b, 2b)$ be two ARBITRARY elements of $W$ . Then $u+v=(a+b, 2(a+b))$ , which does have the form $(t, 2t)$ , so $u+v\in W$ .”

Why: An example never proves a universal property, it can only disprove one. The elements must be written in their general form, with letters.

3. Believing that the union of two subspaces is a subspace

the whole proof question, often 4 marks.

What not to write

“ $F$ and $G$ are two subspaces of $\mathbb{R}^{2}$ , so $F\cup G$ is a subspace.”

What to write

“False in general: with $F$ the line $y=\frac{x}{2}$ and $G$ the line $y=-\frac{3x}{2}$ , we have $u=(2, 1)\in F$ and $v=(-1, \frac{3}{2})\in G$ , but $u+v=(1, \frac{5}{2})$ is in neither $F$ nor $G$ . On the other hand, $F\cap G$ and $F+G$ are always subspaces.”

The two solid points each lie on one of the lines; their sum, the open point, lies on neither: the union is not closed under addition.

Why: The union keeps the two sets apart, whereas addition takes you out of each of them. It is precisely to repair this that the SUM $F+G$ is defined.

4. Taking the number of spanning vectors for the dimension

2 marks, and the wrong dimension then contaminates the rank-nullity theorem and Grassmann's formula.

What not to write

“ $W=\operatorname{span}\{v_{1},\ v_{2},\ v_{3}\}$ , so $\dim W=3$ .”

What to write

“ $\dim W$ is the number of vectors in a BASIS. If $v_{3}=v_{1}+v_{2}$ , the set is linearly dependent and $\dim W=2$ . I row reduce the matrix of the three vectors and count the pivots.”

Three spanning vectors, but

v_{3}=v_{1}+v_{2}

stays in the plane of

v_{1}

and

v_{2}

: the span is that plane, of dimension

2

, not

3

. Only the pivots count.

Why: A spanning set can contain any amount of redundancy. Only a linearly INDEPENDENT spanning set has the right number of vectors.

5. Applying the rank-nullity theorem to the codomain

3 marks, plus every later question that uses the basis of the kernel.

What not to write

“ $T:\mathbb{R}^{4}\to\mathbb{R}^{3}$ with $\dim\operatorname{im}\,T=2$ , so $\dim\ker T=3-2=1$ .”

What to write

“The rank-nullity theorem is about the DOMAIN: $\dim\ker T=\dim\mathbb{R}^{4}-\dim\operatorname{im}\,T=4-2=2$ .”

The theorem splits the DOMAIN bar into two pieces, kernel and image. The codomain bar is drawn apart: it supplies no term of the equation.

Why: The kernel lives in the domain, and it is the domain that the theorem splits. The dimension of the codomain only serves as an upper bound on the rank.

6. Writing that the kernel is empty when the transformation is one-to-one

1 mark for rigour, and 2 more if what follows uses

\dim\ker T

in the rank-nullity theorem.

What not to write

“ $T$ is one-to-one, so $\ker T=\varnothing$ .”

What to write

“ $\ker T=\{\vec{0}\}$ : the kernel ALWAYS contains the zero vector, it is never empty. One-to-one means it contains only $\vec{0}$ , so its dimension is $0$ .”

Why: $\{\vec{0}\}$ is a set with one element, of dimension $0$ ; $\varnothing$ is not even a vector space. Confusing the two throws off every dimension count.

7. Concluding it is a subspace because it contains the zero vector

2 marks, and the mistake comes back on every set defined by an inequality or a square.

What not to write

“ $W=\{(x, y, z) : x^{2}+y=0\}$ contains $(0, 0, 0)$ , so it is a subspace.”

What to write

“The zero vector check can only ELIMINATE. Here $(1, -1, 0)\in W$ , but $2\times(1, -1, 0)=(2, -2, 0)$ gives $4-2=2\neq 0$ : not closed under scalar multiplication, so not a subspace.”

In the plane

z=0

W

is the parabola

y=-x^{2}

: it passes through the origin and through

(1, -1)

, but not through

(2, -2)

, twice that point, which falls off it (in red). The zero vector is not enough.

Why: $\vec{0}\in W$ is necessary, never sufficient. Only LINEAR and HOMOGENEOUS equations define subspaces: a square, a product, an absolute value or an inequality kills them.

8. Adding the dimensions of a sum of subspaces

2 marks, and an absurd answer: a dimension

4

in a space of dimension

3

What not to write

“ $F$ and $G$ are two planes in $\mathbb{R}^{3}$ , so $\dim(F+G)=2+2=4$ .”

What to write

“ $\dim(F+G)=\dim F+\dim G-\dim(F\cap G)$ . Since $F+G\subseteq\mathbb{R}^{3}$ , its dimension is at most $3$ , so $\dim(F\cap G)\geq 1$ : two distinct planes through the origin in $\mathbb{R}^{3}$ always meet at least along a line.”

Two distinct planes through the origin in

\mathbb{R}^{3}

meet along a line (in green):

\dim(F\cap G)=1

, so

\dim(F+G)=2+2-1=3

, not

4

Why: The sum counts twice what belongs to both subspaces. Subtracting the intersection is exactly the same correction as for $|A\cup B|$ in counting problems.

Which method to choose

Is it a subspace? Decide on the FORM of the definition

Read how the set is written before computing anything. The form of the definition gives the answer in four cases out of five.

If defined by one or more HOMOGENEOUS LINEAR equations → YES, it is a subspace: it is the null space of the coefficient matrix of the system

Example: $\{(x, y, z) : x+2y-z=0\}$ is a plane through the origin in $\mathbb{R}^{3}$
If an equation with a NON-ZERO constant term → NO: $\vec{0}$ is not in it, and one line is enough to conclude

Example: $y=2x+1$ , or $x+y=3$
If a square, a product of variables, an absolute value or a root → NO in general: test multiplication by $2$ or by $-1$ on a well-chosen element

Example: $x^{2}+y=0$ contains $(1, -1)$ but not $(2, -2)$
If an INEQUALITY → NO: multiplying by $-1$ takes you out of the set

Example: $\{(x, y) : x\geq 0\}$ contains $(1, 0)$ but not $(-1, 0)$
If written as $\operatorname{span}\{\dots\}$ , as $\ker T$ or as $\operatorname{im}\,T$ → YES automatically: saying so is enough, there is nothing to check

This is the most profitable branch of the exam, and the one students use least.

None of the branches applies, for example for a set of matrices or polynomials defined by a condition? Go back to the three conditions in order, writing the elements in their general form.

Which method for a dimension, depending on what is given

Look at how the subspace ARRIVES in the question, because each way of writing it has its own method, and only one.

If $W=\operatorname{span}\{v_{1},\ \dots,\ v_{k}\}$ → put the $v_{i}$ in rows, row reduce, and count the PIVOTS: that is the rank, hence the dimension

Example: three vectors, one of which is the sum of the other two, give $\dim W=2$
If $W$ is the solution set of a homogeneous system $AX=\vec{0}$ → row reduce $A$ : $\dim W=$ number of unknowns $-$ rank, that is, the number of free parameters

Example: one equation in three unknowns gives $\dim W=3-1=2$
If $W=\ker T$ or $W=\operatorname{im}\,T$ , and the other dimension is known → rank-nullity theorem, with the dimension of the DOMAIN

Example: $T:\mathbb{R}^{3}\to\mathbb{R}^{2}$ of rank $1$ gives $\dim\ker T=3-1=2$
If $W=F\cap G$ → combine the equations of both subspaces into one system and solve it

Never intersect spanning sets: it is the EQUATIONS that combine, not the spanning vectors.
If $W=F+G$ → combine the spanning vectors of both and row reduce, or use Grassmann's formula if the other three dimensions are known

A check that works in every case: the dimension found must be an INTEGER between $0$ and the dimension of the surrounding space. A dimension $4$ in $\mathbb{R}^{3}$ signals a mistake, not a discovery.

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 subset is a subspace

When to use it: The question says “show that $W$ is a subspace of $V$ ”, or “is $W$ a subspace?”.

1 State the inclusion, “ $W$ is a subset of $\mathbb{R}^{3}$ ”, then exhibit the zero vector and check that it satisfies the condition, showing the computation.
2 Take TWO arbitrary elements, in literal form: “let $u=(a, b, a+b)$ and $v=(a', b', a'+b')$ be two elements of $W$ ”.
3 Compute $u+v$ and show that it has EXACTLY the required form, naming the new parameters.
4 Take $\lambda\in\mathbb{R}$ and do the same work on $\lambda u$ .
5 Conclude with the full sentence, which restates the three conditions in order.

Concluding sentence

$W$ contains the zero vector, is closed under addition and is closed under scalar multiplication: $W$ is a subspace of $\mathbb{R}^{3}$ .

The trap: Taking two PARTICULAR elements, with numbers. The proof is then worthless, and it is the most frequent mistake on an average paper. Letters are compulsory.

Marking: 0.5 mark for the zero vector, 1 mark for addition, 1 mark for scalar multiplication, 0.5 mark for the concluding sentence.

Finding a basis and the dimension of the kernel

When to use it: The question asks to “find $\ker T$ ”, for “a basis of the kernel”, or for “the solutions of $AX=\vec{0}$ ”.

1 Write $T(v)=\vec{0}$ as a homogeneous system, with its coefficient matrix.
2 Row reduce with Gaussian elimination and identify the PIVOT variables and the FREE variables, naming them.
3 Introduce one parameter per free variable and express all the other unknowns in terms of these parameters.
4 Write the general solution as a linear combination of fixed vectors, one vector per parameter: these vectors span the kernel.
5 Justify that this set is linearly independent, which is immediate from where the $1$ s and $0$ s sit in the parameter rows, then conclude on the dimension.

Concluding sentence

$\ker T=\operatorname{span}\{(-2, 1, 0),\ (1, 0, 1)\}$ ; this set is linearly independent, so it is a basis of $\ker T$ , and $\dim\ker T=2$ .

The trap: Stopping at the description “ $\ker T$ is the set of $(x, y, z)$ such that $x+2y-z=0$ ”. The question asks for a BASIS, that is, explicit vectors, and the parametrization is the only way to produce them.

Marking: 1 mark for the system and the row reduction, 1 mark for the parametrization, 0.5 mark for the basis vectors, 0.5 mark for the dimension.

Check before you hand in

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

Test the zero vector first, always

Replace every variable by $0$ in the definition of the set. If the equation becomes false, you are done: the set is not a subspace, and the answer fits in one line.

On $x+y=3$ , $0+0=0\neq 3$ : the question is settled without writing a single $u+v$ .
Multiply an element by $-1$

Take a non-zero element of the set, multiply it by $-1$ , and check whether it stays in. This test catches inequalities, squares and absolute values in one go.

$\{(x, y) : x\geq 0\}$ contains $(1, 0)$ but not $(-1, 0)$ : it is not a subspace, even though it contains $\vec{0}$ .
Close the rank-nullity loop

Add the dimensions you found for the kernel and the image: the sum must be exactly the dimension of the DOMAIN. It is a free check on two results at once.

$T:\mathbb{R}^{3}\to\mathbb{R}^{2}$ with $\dim\ker T=2$ and $\dim\operatorname{im}\,T=1$ : the sum is $3$ , everything is consistent. A sum of $2$ signals a missed pivot.
Bound the dimension by the surrounding space

The dimension of a subspace of $\mathbb{R}^{n}$ is an integer between $0$ and $n$ , and the rank of an $m\times n$ matrix is at most the smaller of the two numbers $m$ and $n$ .

A rank of $3$ claimed for a $2\times 5$ matrix is impossible: it is at most $2$ , whatever the entries of the matrix.
Substitute the basis vectors back into the system

Each vector of the kernel basis must satisfy ALL the equations of the original system, not just the first. Three quick substitutions validate the whole parametrization.

$(-2, 1, 0)$ in $x+2y-z$ gives $-2+2-0=0$ : the vector is indeed in the kernel.

The typical problem, taken apart

Kernel, image and the rank-nullity theorem, on a matrix of rank 1

Let $T:\mathbb{R}^{3}\to\mathbb{R}^{2}$ be defined by $T(x, y, z)=(x+2y-z, 2x+4y-2z)$ .

1) Write the matrix of $T$ and find its rank.

2) Find a basis and the dimension of $\operatorname{im}\,T$ .

3) Find a basis and the dimension of $\ker T$ , then check the rank-nullity theorem.

Step 1

$A=\begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix}$ , obtained by reading the coefficients of each component, row by row.

Why

The matrix is the only object you know how to compute with. Everything that follows, rank, kernel and image, is read from it and not from the formula for $T$ .

Step 2

$R_{2}\to R_{2}-2R_{1}$ gives $\begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}$ : only ONE pivot remains, so $\operatorname{rank}(A)=1$ .

Why

The second row was twice the first, which is not always visible at a glance. Row reduction reveals it and fixes the dimension of the image.

Step 3

The columns of $A$ are $(1, 2)$ , $(2, 4)$ and $(-1, -2)$ , all multiples of $(1, 2)$ . So $\operatorname{im}\,T=\operatorname{span}\{(1, 2)\}$ and $\dim\operatorname{im}\,T=1$ .

Why

The image is spanned by the COLUMNS, never by the rows. Its dimension equals the rank, which gives a second reading of the same number and an immediate check.

Step 4

The system $AX=\vec{0}$ reduces to the single equation $x+2y-z=0$ . The unknowns $y$ and $z$ are FREE: set $y=s$ and $z=t$ , so $x=-2s+t$ .

Why

One equation, three unknowns: two free parameters, so $\dim\ker T=2$ before even writing the vectors. The number of parameters IS the dimension.

Step 5

$(x, y, z)=(-2s+t, s, t)=s(-2, 1, 0)+t(1, 0, 1)$ , so $\ker T=\operatorname{span}\{(-2, 1, 0),\ (1, 0, 1)\}$ .

Why

One vector per parameter: that is the step that turns a description into a BASIS. Without it, only half the question has been answered.

Step 6

These two vectors are not parallel, because of where the $1$ and the $0$ sit in the last two components: the set is linearly independent, so $\dim\ker T=2$ .

Why

Independence of the vectors produced by a parametrization is always immediate from these positions, and pointing it out earns half a mark with no computation.

Step 7

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

Why

Substituting back tests BOTH components, whereas the parametrization only used the first equation. It is the only check that catches a forgotten row.

Step 8

Rank-nullity theorem: $\dim\ker T+\dim\operatorname{im}\,T=2+1=3=\dim\mathbb{R}^{3}$ , the domain.

Why

The final check closes both questions at once. If it came out at $2$ , the image or the kernel would be wrong, and you would know it before handing in.

The conclusion, written out

$\operatorname{rank}(A)=1$ , $\operatorname{im}\,T=\operatorname{span}\{(1, 2)\}$ of dimension $1$ , $\ker T=\operatorname{span}\{(-2, 1, 0),\ (1, 0, 1)\}$ of dimension $2$ , and $2+1=3$ , as the rank-nullity theorem requires.

The classic mistake on this problem: Claiming $\dim\operatorname{im}\,T=2$ because the codomain is $\mathbb{R}^{2}$ , then concluding $\dim\ker T=3-2=1$ . The two rows of $A$ are proportional: the image is only a line.

Learn by heart

• Three conditions, in this order: $\vec{0}\in W$ , then closure under $+$ , then closure under $\lambda\cdot$ .
• $\vec{0}\notin W$ settles the question in one line. $\vec{0}\in W$ proves NOTHING.
• $\operatorname{span}\{\dots\}$ , $\ker T$ and $\operatorname{im}\,T$ are subspaces with no check needed.
• The intersection of two subspaces is a subspace; the union, almost never.
• Dimension $=$ number of vectors in a BASIS $=$ number of pivots after row reduction.
• $\dim P_{n}=n+1$ and $\dim M_{m\times n}=mn$ .
• Rank-nullity theorem: $\dim\ker T+\dim\operatorname{im}\,T=\dim(\text{DOMAIN})$ .
• $T$ one-to-one is equivalent to $\ker T=\{\vec{0}\}$ , of dimension $0$ , never to $\varnothing$ .
• Grassmann: $\dim(F+G)=\dim F+\dim G-\dim(F\cap G)$ .
• The image is spanned by the COLUMNS; the rank is read on the reduced rows. The two numbers agree.

Frequently asked questions

How do you show that a set is a subspace?

Check three things, in this order. First, that the zero vector belongs to the set. Next, that the sum of two arbitrary elements, written with letters and not with numbers, stays in it. Finally, that the product of an element by an arbitrary scalar stays in it too. The three together are enough, and the first one alone eliminates half the cases.

Why is a line that does not pass through the origin not a subspace?

Because a subspace always contains the zero vector, and the zero vector is the origin. A line whose equation has a non-zero constant term does not pass through the origin, so it fails the first test. The sum of two of its points does not stay on it either, but there is no need to check that: the first condition is enough.

What is the difference between the kernel and the image?

The kernel lives in the domain: it is made of the vectors that the transformation sends to the zero vector. The image lives in the codomain: it is made of all the values actually reached. The rank-nullity theorem links their dimensions, and the two add up to the dimension of the domain.

How do you find the dimension of a span?

Put the spanning vectors in the rows of a matrix and row reduce with Gaussian elimination. The number of non-zero rows you get, that is, the number of pivots, is the rank, and that rank is the dimension you want. The number of spanning vectors you started with tells you nothing until the redundancy has been removed.

Does the rank-nullity theorem use the dimension of the codomain?

No. It splits the dimension of the domain into two pieces, the dimension of the kernel and that of the image. The dimension of the codomain is good for one thing only, bounding the rank from above, since the image sits inside it. Mixing the two up is the most expensive mistake of the chapter.

Do polynomials really form a vector space?

Yes. You can add two polynomials and multiply a polynomial by a number, and all the usual rules hold. The set of polynomials of degree at most n has dimension n plus one, with the monomials as a basis, from the constant term up to degree n. Every question about it then comes down to an ordinary system.

Practise it

Corrected exercises: Vector spaces and subspaces, 201-SN4-RE

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.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn4-vector-spaces-subspaces. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: vector spaces and subspaces (201-SN4-RE)

The essentials

The subspace test, in the order you check it

Vectors that are not arrows

Dimension, rank, kernel and image

The mistakes that cost marks

1. Taking a line that misses the origin for a subspace

2. Checking closure on a numerical example instead of proving it

3. Believing that the union of two subspaces is a subspace

4. Taking the number of spanning vectors for the dimension

5. Applying the rank-nullity theorem to the codomain

6. Writing that the kernel is empty when the transformation is one-to-one

7. Concluding it is a subspace because it contains the zero vector

8. Adding the dimensions of a sum of subspaces

Which method to choose

Is it a subspace? Decide on the FORM of the definition

Which method for a dimension, depending on what is given

How the answer is expected to be written

Proving that a subset is a subspace

Finding a basis and the dimension of the kernel

Check before you hand in

The typical problem, taken apart

Kernel, image and the rank-nullity theorem, on a matrix of rank 1

Learn by heart

Frequently asked questions

How do you show that a set is a subspace?

Why is a line that does not pass through the origin not a subspace?

What is the difference between the kernel and the image?

How do you find the dimension of a span?

Does the rank-nullity theorem use the dimension of the codomain?

Do polynomials really form a vector space?

Corrected exercises: Vector spaces and subspaces, 201-SN4-RE

See also

Looking for a 201-SN4-RE tutor in Montreal?