MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: basis, dimension and the fundamental subspaces (MATH 133)

This sheet is not a summary of sections 5.2 and 5.4 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on bases, dimension and the fundamental subspaces in MATH 133 at McGill University, and which precise gesture avoids each loss.

Almost every trap below is the same confusion seen from a different side. Row reduction is a counting machine: it tells you how many pivots there are and where they sit, and those two facts survive every row operation. The vectors themselves do not survive, at least not on the column side, so each fundamental subspace has to take its basis from the right matrix. Get that one reflex right and half of the chapter's final-exam questions become bookkeeping.

The thread of the chapter

Row reduction gives you a COUNT and POSITIONS, never the vectors: the dimension survives row operations and the column space does not, so $\text{col}(A)$ is copied from the ORIGINAL pivot columns, $\text{row}(A)$ from the non-zero rows of $R$ , $\text{null}(A)$ from the solved system, and once the dimension is known a family of the right size needs one check, not two.

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

What is on this sheet

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

Topics covered

Basis of a subspace
Dimension of a subspace
Null space
Column space
Row space
Rank of a matrix
Rank-nullity theorem
Pivot columns
Extension of an independent set
Coordinates relative to a basis
Hyperplane
Rank of the transpose

The essentials

Basis and dimension: the number every basis shares

• A basis of a subspace $W$ is a family of vectors of $W$ that is independent AND spans $W$ . Every basis of $W$ has the same number of vectors: that number is $\dim W$ .
• $\dim \mathbb{R}^n = n$ , a plane through the origin has dimension $2$ , a line $1$ , and $\dim\{0\} = 0$ (its basis is empty).
• In a subspace of dimension $m$ : more than $m$ vectors are always dependent, fewer than $m$ never span, and exactly $m$ vectors form a basis as soon as they are independent OR as soon as they span.
• A subspace of $\mathbb{R}^n$ of dimension $n$ is $\mathbb{R}^n$ itself. A subspace cut out by homogeneous equations has dimension $n -$ (rank of their coefficient matrix).
• Coordinates of $v$ in a basis $\{b_1, \dots, b_m\}$ of $W$ : the unique $c_i$ with $v = c_1 b_1 + \dots + c_m b_m$ . There are $m$ of them, the dimension of $W$ , not the length of $v$ .

The dimension is the tool that saves time: once you know it, the rule of $m$ vectors turns a two-part verification into a one-part one. Say the dimension out loud in the answer, it is what makes the shortcut legal.

One matrix, three subspaces, three different sources

• $\text{null}(A) \subseteq \mathbb{R}^n$ : solve $Rx = 0$ , one basis vector per free variable. Row operations do not change the solutions of $Ax = 0$ .
• $\text{row}(A) \subseteq \mathbb{R}^n$ : the non-zero rows of $R$ . Row operations only recombine rows, so the row space does not move.
• $\text{col}(A) \subseteq \mathbb{R}^m$ : the columns of the ORIGINAL $A$ that sit in pivot positions. Row operations DO change the column space; they keep only the relations between columns.
• $\text{rank}(A) = \dim \text{col}(A) = \dim \text{row}(A) = \text{rank}(A^T)$ , and $\text{rank}(A) + \dim \text{null}(A) = n$ , the number of COLUMNS.

Row operations carry the COUNTS from

A

R

; the column space stays behind, so its vectors are copied from

A

, while the rows and the null space are read on

R

Read the positions on $R$ , copy the column vectors from $A$ . The figure is the whole chapter in one picture.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which matrix gives which basis

Worked on the matrix of the example below, $A = \begin{pmatrix} 1 & -1 & 2 & 1 \\ 2 & -2 & 5 & 4 \\ -1 & 1 & -1 & 1 \end{pmatrix}$ , whose reduced form $R = \begin{pmatrix} 1 & -1 & 0 & -3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ has its pivots in columns $1$ and $3$ . The red line is the rule students invent under pressure.

Subspace	Lives in	Basis taken from
$\text{null}(A)$	$\mathbb{R}^4$	solutions of $Rx = 0$
Example: $x_2$ and $x_4$ free: $\{(1, 1, 0, 0), (3, 0, -2, 1)\}$ , dimension $4 - 2 = 2$ .
$\text{row}(A)$	$\mathbb{R}^4$	non-zero rows of $R$
Example: $\{(1, -1, 0, -3), (0, 0, 1, 2)\}$ , dimension $2$ .
$\text{col}(A)$	$\mathbb{R}^3$	pivot columns of $A$
Example: Columns $1$ and $3$ of $A$ : $\{(1, 2, -1), (2, 5, -1)\}$ , dimension $2$ .
$\text{col}(A)$	$\mathbb{R}^3$	pivot columns of $R$ rule that does not exist
Example: $\text{col}(A)$ is the plane $3y_1 - y_2 + y_3 = 0$ , and $(1, 0, 0)$ gives $3 \neq 0$ , $(0, 1, 0)$ gives $-1 \neq 0$ : neither is even in $\text{col}(A)$ . What to do: Use $R$ only to find the pivot positions, then copy those columns from $A$ .

The three correct lines all have dimension $2$ , the rank. The red line has the right SIZE too, which is exactly why it survives until the marker checks a 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. Copying the pivot columns of the REDUCED matrix as a basis of the column space

the whole basis part, typically 3 to 4 marks

What not to write

“ $R = \begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}$ , so a basis of $\text{col}(A)$ is $\{(1, 0)\}$ .”

What to write

“The pivot is in column $1$ , so a basis of $\text{col}(A)$ is column $1$ OF $A$ : $\{(1, 3)\}$ .”

The column space of

A

is the blue line through

(1, 3)

; the pivot column of

R

lies on the orange axis, a different line of the same dimension.

Why: Here $A = \begin{pmatrix} 1 & -2 \\ 3 & -6 \end{pmatrix}$ and every column of $A$ satisfies $y_2 = 3y_1$ ; $(1, 0)$ does not. Row operations preserve the relations between columns and the dimension, not the column space itself.

2. Subtracting the rank from the number of rows

2 marks, and every later answer built on the nullity

What not to write

“ $A$ is $5 \times 7$ of rank $3$ , so $\dim \text{null}(A) = 5 - 3 = 2$ .”

What to write

“Rank plus nullity counts the $7$ columns: $\dim \text{null}(A) = 7 - 3 = 4$ .”

Why: The null space lives in $\mathbb{R}^7$ , one coordinate per unknown, and each free unknown is a column without a pivot. The number $5 - 3 = 2$ does exist: it is $\dim \text{null}(A^T)$ , the count of zero rows of $R$ .

3. Counting the zero rows of R as free variables

1 to 2 marks, plus a null space basis with the wrong number of vectors

What not to write

“ $A$ is $3 \times 5$ and $R$ has one zero row, so the null space has dimension $1$ .”

What to write

“ $R$ has $2$ pivots among $5$ columns, so $3$ free variables and $\dim \text{null}(A) = 3$ .”

Why: Free variables are COLUMNS without a pivot, zero rows are ROWS without a pivot. The two counts are $n - r$ and $m - r$ and they coincide only for a square matrix.

4. Trusting the rows of A that seem to match the non-zero rows of R

the row space part, 2 to 3 marks

What not to write

“The zero row of $R$ is row $3$ , so rows $1$ and $2$ of $A$ form a basis of $\text{row}(A)$ .”

What to write

“A basis of $\text{row}(A)$ is the set of non-zero rows of $R$ : $\{(1, 2, 0, 1), (0, 0, 1, -2)\}$ .”

Why: With $A = \begin{pmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -2 & 6 \\ 1 & 2 & 0 & 1 \end{pmatrix}$ , rows $1$ and $2$ are proportional; the reduction made row $2$ zero and then SWAPPED rows. Row $i$ of $R$ is a combination of rows of $A$ , never row $i$ of $A$ in disguise.

5. Declaring a basis because the number of vectors is right

the whole question

What not to write

“ $(1, 2, 0)$ , $(0, 1, 1)$ and $(1, 3, 1)$ are three vectors of $\mathbb{R}^3$ , so they form a basis.”

What to write

“Three vectors in $\mathbb{R}^3$ , of dimension $3$ : they form a basis if they are independent. Here $(1, 3, 1) = (1, 2, 0) + (0, 1, 1)$ , so they do NOT.”

Why: The rule of $m$ vectors removes ONE of the two checks, never both. The right size is a necessary condition; independence or spanning still has to be shown.

6. Counting equations instead of independent equations

2 marks, and a basis with one vector missing

What not to write

“Three equations in $\mathbb{R}^4$ , so the subspace has dimension $4 - 3 = 1$ .”

What to write

“The third equation is the sum of the first two, the coefficient matrix has rank $2$ , so the dimension is $4 - 2 = 2$ .”

Why: A subspace defined by equations is a null space, and its dimension is $n$ minus the RANK. A redundant equation removes nothing, exactly as a redundant row reduces to zero.

7. Giving a vector of a plane three coordinates

the coordinate part, 2 marks

What not to write

“ $B = \{(1, 0, 1), (0, 1, 1)\}$ and $v = (2, 3, 5)$ , so $[v]_B = (2, 3, 5)$ .”

What to write

“ $v = 2(1, 0, 1) + 3(0, 1, 1)$ , so $[v]_B = (2, 3)$ : two coordinates, because $B$ has two vectors.”

Why: Coordinates are the coefficients of a combination of basis vectors, one per basis vector. The entries of $v$ are its coordinates in the STANDARD basis of $\mathbb{R}^3$ , which is not the basis asked for.

8. Completing a family with a vector that is already in its span

the completion, 2 to 3 marks

What not to write

“ $(2, 1, 3)$ is not a multiple of $(1, 0, 1)$ nor of $(0, 1, 1)$ , so the three vectors form a basis of $\mathbb{R}^3$ .”

What to write

“ $(2, 1, 3) = 2(1, 0, 1) + (0, 1, 1)$ lies in their plane $z = x + y$ . Take $(0, 0, 1)$ instead: $1 \neq 0 + 0$ , so it is outside and the three form a basis.”

Why: A new vector must avoid the SPAN of the family, not just each vector of it. Testing pairs is the independence test of two vectors, and it says nothing about three.

Which method to choose

Which gesture the question is asking for

Read what the statement GIVES: a span, equations, a family of the right size, a size and a rank, or a basis and a vector

Inside a plane, of dimension

2

: one vector only spans a line, two independent vectors are a basis, and a third vector is always a combination of two, here their sum.

If a span, “find a basis of $\text{span}\{v_1, \dots, v_k\}$ ” → vectors as COLUMNS, reduce, keep the ORIGINAL vectors in pivot positions

Example: $v_3 = 2v_1 - v_2$ is dropped, $\{v_1, v_2, v_5\}$ kept, dimension $3$
If homogeneous equations, “the set of $x$ such that ...” → it is a null space: solve, one basis vector per free variable

Example: $x_1 - x_2 + 2x_3 - x_4 = 0$ in $\mathbb{R}^4$ : dimension $4 - 1 = 3$
If $m$ vectors in a subspace whose dimension is known to be $m$ → ONE check: independence, or spanning, whichever is cheaper

Example: $(1, 1, 1, 2)$ and $(0, 1, 1, 1)$ in a subspace of dimension $2$ : not proportional, so a basis
If an independent family, “extend to a basis of $\mathbb{R}^n$ ” → append $e_1, \dots, e_n$ after it, reduce, keep the original pivot columns

Example: $\{(1, 1, 0, 2), (0, 1, 1, 1)\}$ completes with $e_1$ and $e_2$
If only the SIZE of a matrix and its rank, or a statement about one subspace → rank plus nullity $= n$ , $\dim \text{col} = \dim \text{row} = r$ , no computation

Example: $6 \times 8$ of rank $5$ : nullity $3$ , $\dim \text{null}(A^T) = 1$
If a basis $B$ and a vector $v$ , “find $[v]_B$ ” → solve the system whose columns are the basis vectors and whose right-hand side is $v$

Example: $(3, 5, 4)$ in $\{(0, 1, 1), (1, 0, 1), (1, 1, 0)\}$ : $(3, 1, 2)$

If none of these fits, the question is almost always the three fundamental subspaces of one matrix: reduce once, then read $\text{null}$ and $\text{row}$ on $R$ and copy $\text{col}$ from $A$ .

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.

Bases of the three fundamental subspaces

When to use it: “Find a basis and the dimension of $\text{null}(A)$ , $\text{col}(A)$ and $\text{row}(A)$ ”, or any one of the three

1 Reduce $A$ to $R$ and name every operation: $R_2 \to R_2 - 2R_1$ . Method marks are given line by line.
2 Circle the pivot positions and state the rank: “pivots in columns $1$ and $3$ , so $\text{rank}(A) = 2$ ”.
3 Null space: write the free variables as parameters, solve for the pivot variables, and split the general solution into one vector per parameter.
4 Column space: name the columns and the matrix, “columns $1$ and $3$ of $A$ ”, then write the vectors of $A$ .
5 Row space: the non-zero rows of $R$ , and say why: row operations do not change the row space.
6 Close with the check $\text{rank} + \text{nullity} = n$ , and substitute one null space vector into the ORIGINAL $A$ .

Concluding sentence

“The pivots of $R$ are in columns $1$ and $3$ , so a basis of $\text{col}(A)$ is formed by columns $1$ and $3$ of $A$ , and $\dim \text{col}(A) = \dim \text{row}(A) = 2$ , $\dim \text{null}(A) = 4 - 2 = 2$ .”

The trap: Writing the column space basis without saying which matrix the columns come from. Even when the vectors are right, the marker cannot tell whether the method was.

Marking: Typically 2 marks for the reduction, then 2 per subspace (1 for the basis, 1 for the dimension), and 1 for the rank plus nullity check. A wrong column space basis costs its 2 marks even when the reduction is perfect.

Proving that a family is a basis with the dimension

When to use it: “Show that $\{w_1, \dots, w_m\}$ is a basis of $W$ ” when $\dim W$ is known or easy to find

1 Check that each $w_i$ lies in $W$ , by substitution into the equations of $W$ .
2 State $\dim W = m$ , with its reason: a basis already found, or $n$ minus the number of independent equations.
3 Prove ONE property: independence (usually cheaper for two or three vectors) or spanning.
4 Conclude by quoting the theorem: $m$ independent vectors in a subspace of dimension $m$ form a basis.

Concluding sentence

“The vectors lie in $W$ , they are independent, and $\dim W = 2$ ; two independent vectors in a subspace of dimension $2$ form a basis of $W$ .”

The trap: Forgetting the first step. Two independent vectors of R^4 are a basis of SOME plane, not necessarily of W.

Marking: Usually 1 mark for membership, 1 for the dimension, 1 for independence and 1 for the conclusion that names the theorem.

Check before you hand in

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

The dimensions add up

$\dim \text{null}(A) + \dim \text{row}(A)$ must equal $n$ , the number of columns, and $\dim \text{col}(A)$ must equal $\dim \text{row}(A)$ .

A $3 \times 4$ matrix with a column space basis of $2$ vectors and a null space basis of $1$ vector: $2 + 1 \neq 4$ , a free variable has been lost.
Null space vectors go back into A

Multiply each basis vector of $\text{null}(A)$ by the ORIGINAL $A$ : every entry must be $0$ . It catches sign errors when solving for the pivot variables.

$(3, 0, -2, 1)$ into the first row $(1, -1, 2, 1)$ : $3 + 0 - 4 + 1 = 0$ . A $2$ there means a sign flipped.
Column space vectors are columns of A

Every vector of your $\text{col}(A)$ basis must appear, entry for entry, as a column of $A$ , and has $m$ entries. Row space vectors have $n$ entries.

A basis vector $(1, 0, 0)$ for a matrix none of whose columns is $(1, 0, 0)$ : it was copied from $R$ .
Coordinates rebuild the vector

Recombine: $c_1 b_1 + \dots + c_m b_m$ must give back $v$ , and the number of coordinates is the number of basis vectors.

$2(1, 0, 1) + 3(0, 1, 1) = (2, 3, 5)$ confirms $[v]_B = (2, 3)$ .

The typical problem, taken apart

The three subspaces of one matrix, from a single reduction

Let $A = \begin{pmatrix} 1 & -1 & 2 & 1 \\ 2 & -2 & 5 & 4 \\ -1 & 1 & -1 & 1 \end{pmatrix}$ .

Find a basis and the dimension of $\text{null}(A)$ , $\text{col}(A)$ and $\text{row}(A)$ , and check the rank theorem.

Step 1

$R_2 \to R_2 - 2R_1$ gives $(0, 0, 1, 2)$ ; $R_3 \to R_3 + R_1$ gives $(0, 0, 1, 2)$ ; $R_3 \to R_3 - R_2$ gives a zero row; $R_1 \to R_1 - 2R_2$ gives $(1, -1, 0, -3)$ . So $R = \begin{pmatrix} 1 & -1 & 0 & -3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ .

Why

Every operation is named because the marks are given for the path. The zero row is information, not a nuisance: it says that the third row of $A$ was a combination of the first two, so the rank is $2$ , not $3$ .

Step 2

Pivots in columns $1$ and $3$ : $\text{rank}(A) = 2$ . Free variables $x_2 = s$ , $x_4 = t$ ; then $x_1 = s + 3t$ and $x_3 = -2t$ , so $x = s(1, 1, 0, 0) + t(3, 0, -2, 1)$ . Basis of $\text{null}(A)$ : $\{(1, 1, 0, 0), (3, 0, -2, 1)\}$ , dimension $2$ .

Why

One vector per free variable, and they are independent for free: each has a $1$ in its own free slot and a $0$ in the other. Using $R$ is legitimate here, because $Ax = 0$ and $Rx = 0$ have exactly the same solutions.

Step 3

Basis of $\text{row}(A)$ : the non-zero rows of $R$ , $\{(1, -1, 0, -3), (0, 0, 1, 2)\}$ , dimension $2$ .

Why

Row operations recombine rows reversibly, so the row space of $R$ IS the row space of $A$ , and its non-zero rows are visibly independent thanks to their leading $1$ s.

Step 4

Basis of $\text{col}(A)$ : columns $1$ and $3$ OF $A$ , $\{(1, 2, -1), (2, 5, -1)\}$ , dimension $2$ .

Why

This is the step the whole sheet is about. $R$ gave the positions; the vectors come from $A$ . All four columns of $A$ satisfy $3y_1 - y_2 + y_3 = 0$ , while the pivot columns of $R$ , $(1, 0, 0)$ and $(0, 1, 0)$ , give $3$ and $-1$ : copying them would give a basis of the wrong plane.

Step 5

Check: $\text{rank} + \text{nullity} = 2 + 2 = 4$ , the number of columns. And $A(3, 0, -2, 1)$ gives $3 - 4 + 1 = 0$ , $6 - 10 + 4 = 0$ , $-3 + 2 + 1 = 0$ .

Why

The count check is free and catches a lost free variable; the substitution into the ORIGINAL $A$ , not into $R$ , catches a sign error in the null space. Thirty seconds for two independent safety nets.

The conclusion, written out

“ $\text{rank}(A) = 2$ . $\text{null}(A)$ has basis $\{(1, 1, 0, 0), (3, 0, -2, 1)\}$ , $\text{row}(A)$ has basis $\{(1, -1, 0, -3), (0, 0, 1, 2)\}$ , and $\text{col}(A)$ has basis $\{(1, 2, -1), (2, 5, -1)\}$ , columns $1$ and $3$ of $A$ ; all three have dimension $2$ , and $2 + 2 = 4$ .”

The classic mistake on this problem: Answering $\text{col}(A) = \text{span}\{(1, 0, 0), (0, 1, 0)\}$ , the plane $y_3 = 0$ . It has the right dimension, so the rank theorem check passes, and it is still the wrong subspace: $(1, 2, -1)$ , the first column of $A$ , is not in it.

Learn by heart

• A basis is independent AND spanning; all bases of $W$ have $\dim W$ vectors; $\dim\{0\} = 0$ .
• In dimension $m$ : $m$ independent vectors are a basis, $m$ spanning vectors are a basis, more than $m$ are dependent.
• $\text{null}(A)$ : solve $Rx = 0$ . $\text{row}(A)$ : non-zero rows of $R$ . $\text{col}(A)$ : pivot columns of the ORIGINAL $A$ .
• $\text{rank}(A) = \dim \text{col}(A) = \dim \text{row}(A) = \text{rank}(A^T)$ .
• $\text{rank}(A) + \dim \text{null}(A) = n$ , the number of COLUMNS. Zero rows of $R$ count $\dim \text{null}(A^T) = m - r$ .
• Subspace given by equations: dimension $= n -$ rank of the coefficient matrix, never $n -$ number of equations.
• Completing a basis: append $e_1, \dots, e_n$ AFTER the family, reduce, keep the original pivot columns.

Frequently asked questions

How do I find a basis for the column space of a matrix?

Row reduce the matrix to find which columns hold a pivot, then go back to the ORIGINAL matrix and copy those columns. They form a basis of the column space, and their number is the rank. Never copy the columns of the reduced matrix: row operations change the column space, so those vectors are usually not even in it, even though there are the right number of them.

Why is the row space basis taken from the reduced matrix but the column space basis from the original?

Row operations replace rows by combinations of rows, which never changes the row space, so the non-zero rows of the reduced form are a basis of it. The same operations mix the entries inside each column, which does change the column space. What they preserve on the column side is only the linear relations between columns, which is why the reduced form tells you which original columns to keep.

What does the rank-nullity theorem say, and is n the number of rows or columns?

For any matrix, the rank plus the dimension of the null space equals n, the number of COLUMNS, which is the number of unknowns. Each column either holds a pivot or corresponds to a free variable. So a 5 by 7 matrix of rank 3 has a null space of dimension 4. The number of rows minus the rank counts something else, the zero rows of the reduced form.

Do I need to check both independence and spanning to show vectors form a basis?

Not when you know the dimension. In a subspace of dimension m, a family of exactly m vectors of that subspace is a basis as soon as it is independent, or as soon as it spans, so one check is enough. Say the dimension and quote that rule in your answer. With a different number of vectors, the answer is no immediately: too many are dependent, too few cannot span.

Practise it

Corrected exercises: Basis, dimension and the fundamental subspaces, 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: basis, dimension and the fundamental subspaces (MATH 133)

The essentials

Basis and dimension: the number every basis shares

One matrix, three subspaces, three different sources

The rules in table form

Which matrix gives which basis

The mistakes that cost marks

1. Copying the pivot columns of the REDUCED matrix as a basis of the column space

2. Subtracting the rank from the number of rows

3. Counting the zero rows of R as free variables

4. Trusting the rows of A that seem to match the non-zero rows of R

5. Declaring a basis because the number of vectors is right

6. Counting equations instead of independent equations

7. Giving a vector of a plane three coordinates

8. Completing a family with a vector that is already in its span

Which method to choose

Which gesture the question is asking for

How the answer is expected to be written

Bases of the three fundamental subspaces

Proving that a family is a basis with the dimension

Check before you hand in

The typical problem, taken apart

The three subspaces of one matrix, from a single reduction

Learn by heart

Frequently asked questions

How do I find a basis for the column space of a matrix?

Why is the row space basis taken from the reduced matrix but the column space basis from the original?

What does the rank-nullity theorem say, and is n the number of rows or columns?

Do I need to check both independence and spanning to show vectors form a basis?

Corrected exercises: Basis, dimension and the fundamental subspaces, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?