SME
Linear Algebra · Problem type
Singular values of a 2 × 2 matrix
Use this for “Find the singular values of A = [[3, 0], [4, 5]].”
Diagram
A = [[3, 0], [4, 5]] sends (1, 0) to (3, 4) and (0, 1) to (0, 5): its columns.
Area of the image = |det A| × area of the figure: |3 × 5 − 0 × 4| × 1 = 15.
The unit circle becomes an ellipse.
This is an example. Type your own values to start a new problem.
A
=
a₁₁
a₂₁
a₁₂
a₂₂
σ₁
=
σ₁
,
σ₂
=
σ₂
m₁₁
AᵀA, row 1, column 1
worked out (not typed)
m₁₂
AᵀA, off the diagonal
worked out (not typed)
m₂₂
AᵀA, row 2, column 2
worked out (not typed)
det A
Determinant of A
worked out (not typed)
μ₁
Larger eigenvalue of AᵀA
worked out (not typed)
μ₂
Smaller eigenvalue of AᵀA
worked out (not typed)
Formulas
m₁₁
=
a₁₁
2
+
a₂₁
2
25
=
3
2
+
4
2
m₁₂
=
a₁₁
×
a₁₂
+
a₂₁
×
a₂₂
20 = 3 × 0 + 4 × 5
m₂₂
=
a₁₂
2
+
a₂₂
2
25
=
0
2
+
5
2
det
A
=
a₁₁
×
a₂₂
−
a₁₂
×
a₂₁
15 = 3 × 5 − 0 × 4
μ₁
=
(
m₁₁
+
m₂₂
+
√(
(
m₁₁
−
m₂₂
)
2
+
4
×
m₁₂
2
))
÷
2
45
=
(25
+
25
+
√(
(25 − 25)
2
+
4
×
20
2
))
÷
2
μ₂
=
(det A)
2
÷
μ₁
5
=
(15)
2
÷
45
σ₁
=
√
μ₁
6.7082
=
√
45
σ₂
=
√
μ₂
2.2361
=
√
5
Assumptions
•
AᵀA is symmetric with eigenvalues μ₁ ≥ μ₂ ≥ 0. The singular values are σ₁ = √μ₁ and σ₂ = √μ₂.
•
A maps the unit circle to an ellipse with semi-axes σ₁ and σ₂: σ₁ is the most A stretches a unit vector.
•
A = UΣVᵀ: the columns of V are unit eigenvectors of AᵀA, Σ = diag(σ₁, σ₂), and each column of U is Avᵢ ÷ σᵢ.
Step-by-step
WE KNOW
a₁₁ = 3, a₁₂ = 0, a₂₁ = 4, a₂₂ = 5
FIND
AᵀA, row 1, column 1 (m₁₁), AᵀA, off the diagonal (m₁₂), AᵀA, row 2, column 2 (m₂₂), determinant of A (det A), larger eigenvalue of AᵀA (μ₁), smaller eigenvalue of AᵀA (μ₂), larger singular value (σ₁), smaller singular value (σ₂)
Step 1 · Find AᵀA, row 1, column 1 (m₁₁)
Use m₁₁ = a₁₁² + a₂₁²
Row 1 of Aᵀ is column 1 of A, so this entry is column 1 dotted with itself.
m₁₁
=
3
2
+
4
2
[[3, 4], [0, 5]] × [[3, 0], [4, 5]] = [[3 × 3 + 4 × 4, 3 × 0 + 4 × 5], [0 × 3 + 5 × 4, 0 × 0 + 5 × 5]] = [[25, 20], [20, 25]]
m₁₁
=
25
Step 2 · Find AᵀA, off the diagonal (m₁₂)
Use m₁₂ = a₁₁ × a₁₂ + a₂₁ × a₂₂
Column 1 of A dotted with column 2. It sits in both corners, so AᵀA is symmetric.
m₁₂
=
3
×
0
+
4
×
5
m₁₂
=
20
Step 3 · Find AᵀA, row 2, column 2 (m₂₂)
Use m₂₂ = a₁₂² + a₂₂²
Column 2 of A dotted with itself.
m₂₂
=
0
2
+
5
2
m₂₂
=
25
Step 4 · Find determinant of A (det A)
Use det A = a₁₁ × a₂₂ − a₁₂ × a₂₁
Down
the
main
diagonal,
take
away
the
other
diagonal.
det(AᵀA)
=
det
Aᵀ
×
det
A
=
(det A)
2
.
det A = 3 × 5 − 0 × 4
det A = 15
Step 5 · Find larger eigenvalue of AᵀA (μ₁)
Use μ₁ = (m₁₁ + m₂₂ + √((m₁₁ − m₂₂)² + 4 × m₁₂²)) ÷ 2
The
quadratic
formula
on
μ
2
−
(
m₁₁
+
m₂₂
)μ
+
det(AᵀA)
=
0,
with
the
+
sign
for
the
larger
root.
μ₁
=
(25
+
25
+
√(
(25 − 25)
2
+
4
×
20
2
))
÷
2
√(
(25 − 25)
2
+
4
×
20
2
)
=
√
0 + 1,600
=
40
μ₁
=
45
Step 6 · Find smaller eigenvalue of AᵀA (μ₂)
Use μ₂ = (det A)² ÷ μ₁
The
two
eigenvalues
multiply
to
det(AᵀA)
=
(det A)
2
,
so
divide
that
by
μ₁
.
μ₂
=
(15)
2
÷
45
μ₂
=
225
45
μ₂
=
5
Step 7 · Find larger singular value (σ₁)
Use σ₁ = √μ₁
For a unit eigenvector v of AᵀA, |Av|² = vᵀAᵀAv = μ, so A stretches it by √μ.
σ₁
=
√
45
σ₁
=
6.7082
Step 8 · Find smaller singular value (σ₂)
Use σ₂ = √μ₂
The same for the other eigenvector, the shortest stretch of the ellipse.
σ₂
=
√
5
σ₂
=
√
5
=
2.2361
check:
σ₁
×
σ₂
=
15
=
|det
A|
✓
σ₂
=
2.2361
Check
Put the numbers back in. Both sides should match.
25
=
3
2
+
4
2
✓
20 = 3 × 0 + 4 × 5 ✓
25
=
0
2
+
5
2
✓
15 = 3 × 5 − 0 × 4 ✓
45
=
(25
+
25
+
√(
(25 − 25)
2
+
4
×
20
2
))
÷
2
✓
5
=
(15)
2
÷
45
✓
6.7082
=
√
45
✓
2.2361
=
√
5
✓
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