SME
Linear Algebra · Problem type
Kernel and range of a transformation
Use this for “T(x) = Ax with A = [[1, 2, −1], [2, 4, 1], [3, 6, 0]]. Find the kernel and the range of T. Is T one-to-one? Onto?”
Diagram
A is 3 × 3: 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
Dimension of the range
worked out (not typed)
dim ker T
Dimension of the kernel
worked out (not typed)
C
One-to-one and onto?
worked out (not typed)
d₁
Kernel direction d, component 1
worked out (not typed)
d₂
Kernel direction d, component 2
worked out (not typed)
d₃
Kernel direction d, component 3
worked out (not typed)
n₁
Normal n to the range, component 1
worked out (not typed)
n₂
Normal n to the range, component 2
worked out (not typed)
n₃
Normal n to the range, component 3
worked out (not typed)
Formulas
rank
A
=
rank
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
2 = rank [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
dim ker T = 3 − (rank A)
1 = 3 − 2
C
=
one-to-one
and
onto
if
dim
ker
T
=
0,
else
neither
1 > 0, so neither
d₁
=
entry
1
of
d,
with
Ad
=
0
for
A
=
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
−2 = entry 1 of d, with Ad = 0 for A = [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
d₂
=
entry
2
of
d,
with
Ad
=
0
for
A
=
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
1 = entry 2 of d, with Ad = 0 for A = [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
d₃
=
entry
3
of
d,
with
Ad
=
0
for
A
=
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
0 = entry 3 of d, with Ad = 0 for A = [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
n₁
=
entry
1
of
the
cross
product
of
the
two
pivot
columns
of
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
−3 = entry 1 of the cross product of the two pivot columns of [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
n₂
=
entry
2
of
the
cross
product
of
the
two
pivot
columns
of
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
−3 = entry 2 of the cross product of the two pivot columns of [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
n₃
=
entry
3
of
the
cross
product
of
the
two
pivot
columns
of
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
3 = entry 3 of the cross product of the two pivot columns of [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
Assumptions
•
ker T is every x with Ax = 0. The range of T is every Ax: the span of the columns of A.
•
rank A + dim ker T = 3, the number of columns, and the pivot columns of A are a basis of the range.
•
For a 3 × 3 matrix, T is one-to-one exactly when it is onto: both mean rank 3, so ker T = {0}.
•
With rank 2 the kernel is a line along d and the range is a plane through 0, orthogonal to n.
Step-by-step
WE KNOW
a₁₁ = 1, a₁₂ = 2, a₁₃ = −1, a₂₁ = 2, a₂₂ = 4, a₂₃ = 1, a₃₁ = 3, a₃₂ = 6, a₃₃ = 0
FIND
dimension of the range (rank A), dimension of the kernel (dim ker T), one-to-one and onto? (C), kernel direction d, component 1 (d₁), kernel direction d, component 2 (d₂), kernel direction d, component 3 (d₃), normal n to the range, component 1 (n₁), normal n to the range, component 2 (n₂), normal n to the range, component 3 (n₃)
Step 1 · Find dimension of the range (rank A)
Use rank A = rank [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Row reduce A, one operation a line: the range is the span of the pivot columns. Rank 2: the range is a plane through 0 and the kernel is a line.
Start: [[1, 2, −1], [2, 4, 1], [3, 6, 0]]
R₂ → R₂ − 2R₁: [[1, 2, −1], [0, 0, 3], [3, 6, 0]]
R₃ → R₃ − 3R₁: [[1, 2, −1], [0, 0, 3], [0, 0, 3]]
R₂
→
(
1
3
)R₂:
[[1,
2,
−1],
[0,
0,
1],
[0,
0,
3]]
R₁ → R₁ + R₂: [[1, 2, 0], [0, 0, 1], [0, 0, 3]]
R₃ → R₃ − 3R₂: [[1, 2, 0], [0, 0, 1], [0, 0, 0]]
Pivots in columns 1 and 3: rank 2
rank A = 2
Step 2 · Find dimension of the kernel (dim ker T)
Use dim ker T = 3 − (rank A)
rank A + dim ker T = 3: each column without a pivot is a free variable, one direction of the kernel.
dim ker T = 3 − 2
dim ker T = 1
Step 3 · Find one-to-one and onto? (C)
Use C = one-to-one and onto if dim ker T = 0, else neither
Some x ≠ 0 has Ax = 0 = A0, so two inputs share an image: not one-to-one. Rank under 3 leaves vectors outside the range: not onto.
1 > 0: neither
C
=
one-to-one
and
onto
if
1
=
0,
else
neither
C
=
neither
Step 4 · Find kernel direction d, component 1 (d₁)
Use d₁ = entry 1 of d, with Ad = 0 for A = [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Its row of the reduced matrix reads x₁ + 2x₂ = 0, so x₁ = −2 when x₂ = 1.
d₁
=
−2
Step 5 · Find kernel direction d, component 2 (d₂)
Use d₂ = entry 2 of d, with Ad = 0 for A = [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
x₂ is the free variable: set it to 1.
d₂
=
1
Step 6 · Find kernel direction d, component 3 (d₃)
Use d₃ = entry 3 of d, with Ad = 0 for A = [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Its row of the reduced matrix reads x₃ = 0, so x₃ = 0 when x₂ = 1.
Ad = ⟨1 × (−2) + 2 × 1 + (−1) × 0, 2 × (−2) + 4 × 1 + 1 × 0, 3 × (−2) + 6 × 1 + 0 × 0⟩ = ⟨0, 0, 0⟩
ker T = every multiple t⟨−2, 1, 0⟩
d₃
=
0
Step 7 · Find normal n to the range, component 1 (n₁)
Use n₁ = entry 1 of the cross product of the two pivot columns of [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
The range is spanned by the pivot columns, columns 1 and 3 of A. Their cross product is orthogonal to both, so it is normal to the range plane.
⟨1, 2, 3⟩ × ⟨−1, 1, 0⟩ = ⟨2 × 0 − 3 × 1, 3 × (−1) − 1 × 0, 1 × 1 − 2 × (−1)⟩ = ⟨−3, −3, 3⟩
n₁
=
−3
Step 8 · Find normal n to the range, component 2 (n₂)
Use n₂ = entry 2 of the cross product of the two pivot columns of [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
The next entry of the same cross product.
n₂
=
−3
Step 9 · Find normal n to the range, component 3 (n₃)
Use n₃ = entry 3 of the cross product of the two pivot columns of [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
The next entry of the same cross product.
Range: −3y₁ − 3y₂ + 3y₃ = 0
Divided by −3: y₁ + y₂ − y₃ = 0
n₃
=
3
Check
Put the numbers back in. Both sides should match.
Pivots in A: 1 + 1 = 2 ✓
1 = 3 − 2 ✓
1 > 0, so neither ✓
Row 1: 1 × (−2) + 2 × 1 + −1 × 0 = 0 ✓
Row 2: 2 × (−2) + 4 × 1 + 1 × 0 = 0 ✓
Row 3: 3 × (−2) + 6 × 1 + 0 × 0 = 0 ✓
(−3) = 2 × 0 − 3 × 1 ✓
(−3) = 3 × (−1) − 1 × 0 ✓
3 = 1 × 1 − 2 × (−1) ✓
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