SME
Linear Algebra
Vector spaces and subspaces
Use this for “Find the rank of A = [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]] and a basis for Nul A”, or to test whether the columns of A are independent (rank = number of columns).
Diagram
A is 3 × 4: row operations keep its rank.
Pivots in columns 1, 3: rank 2.
This is an example. Type your own values to start a new problem.
A
=
rank A
Rank of A
worked out (not typed)
dim Col A
Dimension of Col A
worked out (not typed)
dim Nul A
Dimension of Nul A
worked out (not typed)
v₁
Basis vector v of Nul A, component 1
worked out (not typed)
v₂
Basis vector v of Nul A, component 2
worked out (not typed)
v₃
Basis vector v of Nul A, component 3
worked out (not typed)
v₄
Basis vector v of Nul A, component 4
worked out (not typed)
w₁
Basis vector w of Nul A, component 1
worked out (not typed)
w₂
Basis vector w of Nul A, component 2
worked out (not typed)
w₃
Basis vector w of Nul A, component 3
worked out (not typed)
w₄
Basis vector w of Nul A, component 4
worked out (not typed)
Formulas
rank
A
=
rank
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
2 = rank [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
dim Col A = rank A
2 = 2
dim Nul A = 4 − (rank A)
2 = 4 − 2
v₁
=
entry
1
of
v,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
−2 = entry 1 of v, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
v₂
=
entry
2
of
v,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
1 = entry 2 of v, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
v₃
=
entry
3
of
v,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
0 = entry 3 of v, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
v₄
=
entry
4
of
v,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
0 = entry 4 of v, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
w₁
=
entry
1
of
w,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
−1 = entry 1 of w, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
w₂
=
entry
2
of
w,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
0 = entry 2 of w, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
w₃
=
entry
3
of
w,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
−2 = entry 3 of w, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
w₄
=
entry
4
of
w,
a
basis
vector
of
Nul
[[
a₁₁
,
a₁₂
,
a₁₃
,
a₁₄
],
[
a₂₁
,
a₂₂
,
a₂₃
,
a₂₄
],
[
a₃₁
,
a₃₂
,
a₃₃
,
a₃₄
]]
1 = entry 4 of w, a basis vector of Nul [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
Assumptions
•
Nul A is every x with Ax = 0: row reduce A (a zero column after it would never change).
•
The pivot columns of A are a basis of Col A, so dim Col A = rank A.
•
Each free variable gives one basis vector of Nul A, so rank A + dim Nul A = 4, the number of columns.
Step-by-step
WE KNOW
a₁₁ = 1, a₁₂ = 2, a₁₃ = 0, a₁₄ = 1, a₂₁ = 2, a₂₂ = 4, a₂₃ = 1, a₂₄ = 4, a₃₁ = 3, a₃₂ = 6, a₃₃ = 1, a₃₄ = 5
FIND
rank of A (rank A), dimension of Col A (dim Col A), dimension of Nul A (dim Nul A), basis vector v of Nul A, component 1 (v₁), basis vector v of Nul A, component 2 (v₂), basis vector v of Nul A, component 3 (v₃), basis vector v of Nul A, component 4 (v₄), basis vector w of Nul A, component 1 (w₁), basis vector w of Nul A, component 2 (w₂), basis vector w of Nul A, component 3 (w₃), basis vector w of Nul A, component 4 (w₄)
Step 1 · Find rank of A (rank A)
Use rank A = rank [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
Row reduce A to reduced echelon form, one operation a line. Each pivot adds 1 to the rank.
Start: [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]]
R₂ → R₂ − 2R₁: [[1, 2, 0, 1], [0, 0, 1, 2], [3, 6, 1, 5]]
R₃ → R₃ − 3R₁: [[1, 2, 0, 1], [0, 0, 1, 2], [0, 0, 1, 2]]
R₃ → R₃ − R₂: [[1, 2, 0, 1], [0, 0, 1, 2], [0, 0, 0, 0]]
Pivots in columns 1 and 3: rank 2
rank A = 2
Step 2 · Find dimension of Col A (dim Col A)
Use dim Col A = rank A
The pivot columns of A (columns 1 and 3) are a basis of Col A: one basis vector a pivot.
Col A = span{⟨1, 2, 3⟩, ⟨0, 1, 1⟩}
dim Col A = 2
Step 3 · Find dimension of Nul A (dim Nul A)
Use dim Nul A = 4 − (rank A)
rank A + dim Nul A = 4, the number of columns: each column without a pivot is a free variable.
dim Nul A = 4 − 2
dim Nul A = 2
Step 4 · Find basis vector v of Nul A, component 1 (v₁)
Use v₁ = entry 1 of v, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
Its row of the reduced matrix reads x₁ + 2x₂ + x₄ = 0, so x₁ = −2 when x₂ = 1 and x₄ = 0.
v: [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]] × ⟨−2, 1, 0, 0⟩ = ⟨0, 0, 0⟩
v₁
=
−2
Step 5 · Find basis vector v of Nul A, component 2 (v₂)
Use v₂ = entry 2 of v, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
x₂ is free: set it to 1 and x₄ = 0.
v₂
=
1
Step 6 · Find basis vector v of Nul A, component 3 (v₃)
Use v₃ = entry 3 of v, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
Its row of the reduced matrix reads x₃ + 2x₄ = 0, so x₃ = 0 when x₂ = 1 and x₄ = 0.
v₃
=
0
Step 7 · Find basis vector v of Nul A, component 4 (v₄)
Use v₄ = entry 4 of v, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
x₄ is another free variable: set it to 0.
v₄
=
0
Step 8 · Find basis vector w of Nul A, component 1 (w₁)
Use w₁ = entry 1 of w, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
Its row of the reduced matrix reads x₁ + 2x₂ + x₄ = 0, so x₁ = −1 when x₄ = 1 and x₂ = 0.
w: [[1, 2, 0, 1], [2, 4, 1, 4], [3, 6, 1, 5]] × ⟨−1, 0, −2, 1⟩ = ⟨0, 0, 0⟩
w₁
=
−1
Step 9 · Find basis vector w of Nul A, component 2 (w₂)
Use w₂ = entry 2 of w, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
x₂ is another free variable: set it to 0.
w₂
=
0
Step 10 · Find basis vector w of Nul A, component 3 (w₃)
Use w₃ = entry 3 of w, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
Its row of the reduced matrix reads x₃ + 2x₄ = 0, so x₃ = −2 when x₄ = 1 and x₂ = 0.
w₃
=
−2
Step 11 · Find basis vector w of Nul A, component 4 (w₄)
Use w₄ = entry 4 of w, a basis vector of Nul [[a₁₁, a₁₂, a₁₃, a₁₄], [a₂₁, a₂₂, a₂₃, a₂₄], [a₃₁, a₃₂, a₃₃, a₃₄]]
x₄ is free: set it to 1 and x₂ = 0.
w₄
=
1
Check
Put the numbers back in. Both sides should match.
Pivots in A: 1 + 1 = 2 ✓
2 = 2 ✓
2 = 4 − 2 ✓
Row 1: 1 × (−2) + 2 × 1 + 0 × 0 + 1 × 0 = 0 ✓
Row 2: 2 × (−2) + 4 × 1 + 1 × 0 + 4 × 0 = 0 ✓
Row 3: 3 × (−2) + 6 × 1 + 1 × 0 + 5 × 0 = 0 ✓
Row 1: 1 × (−2) + 2 × 1 + 0 × 0 + 1 × 0 = 0 ✓
Row 1: 1 × (−1) + 2 × 0 + 0 × (−2) + 1 × 1 = 0 ✓
Row 2: 2 × (−1) + 4 × 0 + 1 × (−2) + 4 × 1 = 0 ✓
Row 3: 3 × (−1) + 6 × 0 + 1 × (−2) + 5 × 1 = 0 ✓
Row 1: 1 × (−1) + 2 × 0 + 0 × (−2) + 1 × 1 = 0 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Are three vectors independent?
Use this for “Are ⟨1, 0, 1⟩, ⟨2, 1, 0⟩ and ⟨0, 1, 1⟩ linearly independent?”
Coordinates in a basis
Use this for “Find the coordinates of x = ⟨5, 1⟩ in the basis b₁ = ⟨1, 1⟩, b₂ = ⟨1, −1⟩.”
Is it a subspace?
Use this for “Is the set of points with y = 3x − 2 a subspace of ℝ²?”