SME
Linear Algebra · Problem type
Powers of a matrix by its eigenvectors
Use this for “A = [[4, 1], [2, 3]] and x = (1, 0). Find A³x by writing x in the eigenvectors of A.”
Diagram
c₁v₁ = ⟨0.6667, 0.6667⟩
|c₁v₁| = 0.9428.
c₂v₂ = ⟨0.3333, −0.6667⟩
|c₂v₂| = 0.7454.
Parallelogram: x = ⟨1, 0⟩
|x| = 1 at 0°.
This is an example. Type your own values to start a new problem.
=
tr A
Trace of A
worked out (not typed)
det A
Determinant of A
worked out (not typed)
Δ
Discriminant tr² − 4 det
worked out (not typed)
λ₁
Larger eigenvalue
worked out (not typed)
λ₂
Smaller eigenvalue
worked out (not typed)
p₁
Slope of v₁
worked out (not typed)
p₂
Slope of v₂
worked out (not typed)
c₁
Coordinate of x along v₁
worked out (not typed)
c₂
Coordinate of x along v₂
worked out (not typed)
Formulas
tr
A
=
a
+
d
7 = 4 + 3
det
A
=
a
×
d
−
b
×
c
10 = 4 × 3 − 1 × 2
Δ
=
(tr A)
2
−
4
×
(det
A)
9
=
(7)
2
−
4
×
10
λ₁
=
tr A + √
Δ
2
5
=
7 +
√
9
2
λ₂
=
tr A − √
Δ
2
2
=
7 −
√
9
2
p₁
=
λ₁
−
a
b
1
=
5 − 4
1
p₂
=
λ₂
−
a
b
−2
=
2 − 4
1
c₂
=
x₂
−
p₁
×
x₁
p₂
−
p₁
1
3
=
0 − 1 × 1
−2 − 1
c₁
=
x₁
−
c₂
2
3
=
1
−
1
3
y₁
=
c₁
×
λ₁
k
+
c₂
×
λ₂
k
86
=
2
3
×
5
3
+
1
3
×
2
3
y₂
=
c₁
×
λ₁
k
×
p₁
+
c₂
×
λ₂
k
×
p₂
78
=
2
3
×
5
3
×
1
+
1
3
×
2
3
×
(−2)
Assumptions
•
A has two different real eigenvalues, so v₁ = (1, p₁) and v₂ = (1, p₂) are a basis and x = c₁v₁ + c₂v₂.
•
Each multiplication by A scales v₁ by λ₁ and v₂ by λ₂, so Aᵏx = c₁λ₁ᵏv₁ + c₂λ₂ᵏv₂.
•
For a large k the term with the larger |λ| takes over, so Aᵏx turns toward that eigenvector.
Step-by-step
WE KNOW
a = 4, b = 1, c = 2, d = 3, x₁ = 1, x₂ = 0, k = 3
FIND
trace of A (tr A), determinant of A (det A), discriminant tr² − 4 det (Δ), larger eigenvalue (λ₁), smaller eigenvalue (λ₂), slope of v₁ (p₁), slope of v₂ (p₂), coordinate of x along v₂ (c₂), coordinate of x along v₁ (c₁), Aᵏx, top entry (y₁), Aᵏx, bottom entry (y₂)
Step 1 · Find trace of A (tr A)
Use tr A = a + d
The trace is the sum of the diagonal entries.
tr A = 4 + 3
tr A = 7
Step 2 · Find determinant of A (det A)
Use det A = a × d − b × c
Down the main diagonal, minus the other diagonal. With the trace it gives det(A − λI).
det A = 4 × 3 − 1 × 2
det([[4
−
λ,
1],
[2,
3
−
λ]])
=
(4
−
λ)(3
−
λ)
−
1
×
2
=
λ
2
−
7λ
+
10
det A = 10
Step 3 · Find discriminant tr² − 4 det (Δ)
Use Δ = (tr A)² − 4 × (det A)
The
discriminant
of
λ
2
−
(tr
A)λ
+
det
A
=
0
says
whether
its
roots
are
real.
Δ
=
(7)
2
−
4
×
10
Δ
=
49
−
4
×
10
Δ
=
49
−
40
Δ
=
9
Step 4 · Find larger eigenvalue (λ₁)
Use λ₁ = (tr A + √Δ) ÷ 2
The
quadratic
formula
on
λ
2
−
(tr
A)λ
+
det
A
=
0,
with
the
+
sign
for
the
larger
root.
λ₁
=
7 +
√
9
2
λ₁
=
7 + 3
2
λ₁
=
10
2
λ₁
=
5
Step 5 · Find smaller eigenvalue (λ₂)
Use λ₂ = (tr A − √Δ) ÷ 2
The same formula with the − sign gives the smaller root.
λ₂
=
7 −
√
9
2
λ₂
=
7 − 3
2
λ₂
=
4
2
λ₂
=
2
Step 6 · Find slope of v₁ (p₁)
Use p₁ = (λ₁ − a) ÷ b
Row
1
of
(A
−
λI)v
=
0
reads
(
a
−
λ)x
+
by
=
0.
Put
x
=
1
and
solve
for
y.
p₁
=
5 − 4
1
Check: [[4, 1], [2, 3]] × ⟨1, 1⟩ = ⟨5, 5⟩ = 5 × ⟨1, 1⟩
p₁
=
1
Step 7 · Find slope of v₂ (p₂)
Use p₂ = (λ₂ − a) ÷ b
Row
1
of
(A
−
λI)v
=
0
reads
(
a
−
λ)x
+
by
=
0.
Put
x
=
1
and
solve
for
y.
p₂
=
2 − 4
1
Check: [[4, 1], [2, 3]] × ⟨1, −2⟩ = ⟨2, −4⟩ = 2 × ⟨1, −2⟩
p₂
=
−2
Step 8 · Find coordinate of x along v₂ (c₂)
Use c₂ = (x₂ − p₁ × x₁) ÷ (p₂ − p₁)
x
=
c₁
(1,
p₁
)
+
c₂
(1,
p₂
)
gives
x₁
=
c₁
+
c₂
and
x₂
=
c₁
p₁
+
c₂
p₂
.
Take
p₁
times
the
first
from
the
second.
c₂
=
0 − 1 × 1
−2 − 1
c₂
=
−
1
−3
c₂
=
1
3
Step 9 · Find coordinate of x along v₁ (c₁)
Use c₁ = x₁ − c₂
The
top
entries:
c₁
+
c₂
=
x₁
.
c₁
=
1
−
1
3
c₁
=
2
3
Step 10 · Find Aᵏx, top entry (y₁)
Use y₁ = c₁ × λ₁ᵏ + c₂ × λ₂ᵏ
Aᵏ scales v₁ by λ₁ᵏ and v₂ by λ₂ᵏ. The top entry of each vᵢ is 1.
y₁
=
2
3
×
5
3
+
1
3
×
2
3
y₁
=
2
3
×
125
+
1
3
×
8
y₁
=
83
1
3
+
2
2
3
y₁
=
86
Step 11 · Find Aᵏx, bottom entry (y₂)
Use y₂ = c₁ × λ₁ᵏ × p₁ + c₂ × λ₂ᵏ × p₂
The bottom entries of the same two scaled eigenvectors.
y₂
=
2
3
×
5
3
×
1
+
1
3
×
2
3
×
(−2)
2
3
×
5
3
×
⟨1,
1⟩
+
1
3
×
2
3
×
⟨1,
−2⟩
=
⟨86,
78⟩
y₂
=
78
Check
Put the numbers back in. Both sides should match.
7 = 4 + 3 ✓
10 = 4 × 3 − 1 × 2 ✓
9
=
(7)
2
−
4
×
10
✓
5
=
7 +
√
9
2
✓
2
=
7 −
√
9
2
✓
1
=
5 − 4
1
✓
−2
=
2 − 4
1
✓
1
3
=
0 − 1 × 1
−2 − 1
✓
2
3
=
1
−
1
3
✓
86
=
2
3
×
5
3
+
1
3
×
2
3
✓
78
=
2
3
×
5
3
×
1
+
1
3
×
2
3
×
(−2)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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