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02.5 Turn, stretch, turn

The complex linear transformation can be broken down into two choices of mutually perpendicular directions and a tensile between them. This is a single-dimensional decomposition, the SVD.

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On the previous page, a direction that does not rotate.
The complex linear transformation can be broken down into two choices of mutually perpendicular directions and a tensile between them. This is a single-dimensional decomposition, the SVD.
Simple table: horizontal x3, vertical x1.
Координаты и расчёт геометрического примера приведены в тексте. Для изменения пространственного рисунка включитеJavaScript.
Two spaces require two sets of directions
Eigendecomposition compares the input direction to it in the same space and does not exist in a convenient physical form for any matrix. Singular vale decomposition also exists for a rectangular physical matrix:
A
=
U
Σ
V
T
,
A=U\Sigma V^T,
A
=
U
Σ
V
T
,
where for full shape
A
∈
R
m
×
n
,
U
∈
R
m
×
m
,
Σ
∈
R
m
×
n
,
V
∈
R
n
×
n
.
A\in\mathbb R^{m\times n},
\quad U\in\mathbb R^{m\times m},
\quad \Sigma\in\mathbb R^{m\times n},
\quad V\in\mathbb R^{n\times n}.
A
∈
R
m
×
n
,
U
∈
R
m
×
m
,
Σ
∈
R
m
×
n
,
V
∈
R
n
×
n
.
Columns
V
V
V
Orthonized input orientations, columns
U
U
U
- Orthonized weekend directions, a
σ
i
≥
0
\sigma_i\ge0
σ
i
​
≥
0
Diagonal
Σ
\Sigma
Σ
They are of a scale.
Right singular vectors satisfied
A
T
A
v
i
=
σ
i
2
v
i
.
A^TAv_i=\sigma_i^2v_i.
A
T
A
v
i
​
=
σ
i
2
​
v
i
​
.
If
σ
i
>
0
\sigma_i>0
σ
i
​
>
0
, that's
u
i
=
A
v
i
σ
i
.
u_i=\frac{Av_i}{\sigma_i}.
u
i
​
=
σ
i
​
A
v
i
​
​
.
The number of positive sinular values is equal to the exact ring.
Three action sequences
For the entrance
x
x
x
A
x
=
U
(
Σ
(
V
T
x
)
)
.
Ax=U\bigl(\Sigma(V^Tx)\bigr).
A
x
=
U
(
Σ
(
V
T
x
)
)
.
First
V
T
V^T
V
T
Expresses its entrance to the right ortho-normalized base.
Σ
\Sigma
Σ
It's an independent scale of the coordinates.
U
U
U
Collects the result in the output space. Orthogonal matrices can include a reflection, so each one is not accurately called a turn.
Here we go.
A
=
diag
⁡
(
3
,
1
)
,
U
=
V
=
I
,
Σ
=
diag
⁡
(
3
,
1
)
.
A=\operatorname{diag}(3,1),
\qquad U=V=I,
\qquad \Sigma=\operatorname{diag}(3,1).
A
=
diag
(
3
,
1
)
,
U
=
V
=
I
,
Σ
=
diag
(
3
,
1
)
.
For
x
=
[
1
,
2
]
T
x=[1,2]^T
x
=
[
1
,
2
]
T
trade is equal to
[
1
,
2
]
T
→
V
T
[
1
,
2
]
T
→
Σ
[
3
,
2
]
T
→
U
[
3
,
2
]
T
.
[1,2]^T\xrightarrow{V^T}[1,2]^T
\xrightarrow{\Sigma}[3,2]^T
\xrightarrow{U}[3,2]^T.
[
1
,
2
]
T
V
T
​
[
1
,
2
]
T
Σ
​
[
3
,
2
]
T
U
​
[
3
,
2
]
T
.
Angling gives the optimum low-rank object in the selected standard
If you just leave
σ
1
=
3
\sigma_1=3
σ
1
​
=
3
We'll get it.
A
1
=
[
3
0
0
0
]
.
A_1=\begin{bmatrix}3&0\\0&0\end{bmatrix}.
A
1
​
=
[
3
0
​
0
0
​
]
.
The second input direction is removed. For the matrix, Frobenius error is equal to the discarded singular value:
∥
A
−
A
1
∥
F
=
1
,
∥
A
−
A
1
∥
F
∥
A
∥
F
=
1
10
≈
0.316228.
\|A-A_1\|_F=1,
\qquad
\frac{\|A-A_1\|_F}{\|A\|_F}=\frac1{\sqrt{10}}\approx0.316228.
∥
A
−
A
1
​
∥
F
​
=
1
,
∥
A
∥
F
​
∥
A
−
A
1
​
∥
F
​
​
=
10
​
1
​
≈
0.316228.
For a specific
x
=
[
1
,
2
]
T
x=[1,2]^T
x
=
[
1
,
2
]
T
full exit
[
3
,
2
]
T
[3,2]^T
[
3
,
2
]
T
and the Frank-one exit.
[
3
,
0
]
T
[3,0]^T
[
3
,
0
]
T
; the error of this vector is equal to 2. You can't mix a matrix error and one output error.
In 90 degrees of rotation, both sinular values are equal to 1. The choice of a single Frank-one direction is not unique: there is no single "largest" axis. Different correct SVDs may differ by vapour signs.
u
i
,
v
i
u_i,v_i
u
i
​
,
v
i
​
or a base within an equal sinular subspace, keeping
U
Σ
V
T
=
A
U\Sigma V^T=A
U
Σ
V
T
=
A
.
The browser lab only decides analytically.
2
×
2
2\times2
2
×
2
Case through
A
T
A
A^TA
A
T
A
I. Industrial libraries use more sustainable algorithms, as the apparent construction
A
T
A
A^TA
A
T
A
it's a condition number squared. Check the reconstruction, orthogonality and error, not the specific sign of each singlelar vector.
When compressed, the vertical component is lost. The SVD describes direction and scope; quality for a particular task requires separate verification.
Correction through a narrow intermediate space. How does this question continue the result?
Python examples of Chapter 02
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