02.3 When the direction is lost

Content and key concepts

On the previous page, two consecutive conversions.
Transformation can fold the plane into a line. The Rang is the number of independent directions left at the exit.
Rule: [x1,x2] [x1+2x2,0].
Координаты и расчёт геометрического примера приведены в тексте. Для изменения пространственного рисунка включитеJavaScript.
The image tells us how many directions have been transformed.
Consider
A
=
[
1
2
0
0
]
.
A=\begin{bmatrix}1&2\\0&0\end{bmatrix}.
A
=
[
1
0
​
2
0
​
]
.
Her columns.
[
1
,
0
]
T
[1,0]^T
[
1
,
0
]
T
and
[
2
,
0
]
T
[2,0]^T
[
2
,
0
]
T
Any exit has a view.
A
x
=
[
x
1
+
2
x
2
,
0
]
T
,
Ax=[x_1+2x_2,0]^T,
A
x
=
[
x
1
​
+
2
x
2
​
,
0
]
T
,
Therefore, the whole plane is folded into the horizontal axis.
Rang
The matrix is equal to the dimension of its column space, that is, the dimensional dimension of the image. Here, the Frank equals 1 although the non-zero cells are two.
For the matrix
[
m
,
n
]
[m,n]
[
m
,
n
]
0
≤
rank
⁡
(
A
)
≤
min
⁡
(
m
,
n
)
.
0\le\operatorname{rank}(A)\le\min(m,n).
0
≤
rank
(
A
)
≤
min
(
m
,
n
)
.
Rank 0 has only a zero matrix.
n
n
n
means that different entrances do not run into each other; complete string Frank
m
m
m
means every exit is achievable.
The kernel describes invisible changes in the inlet
Core
ker
⁡
(
A
)
=
{
v
:
A
v
=
0
}
\ker(A)=\{v:Av=0\}
ker
(
A
)
=
{
v
:
A
v
=
0
}
contains directions that have been completely destroyed by the operator.
A
A
A
v
=
t
[
−
2
,
1
]
T
,
A
v
=
0.
v=t[-2,1]^T,
\qquad Av=0.
v
=
t
[
−
2
,
1
]
T
,
A
v
=
0.
Therefore,
x
x
x
and
x
+
v
x+v
x
+
v
Give one exit:
A
(
x
+
v
)
=
A
x
+
A
v
=
A
x
.
A(x+v)=Ax+Av=Ax.
A
(
x
+
v
)
=
A
x
+
A
v
=
A
x
.
For example,
[
2
,
0
]
T
[2,0]^T
[
2
,
0
]
T
and
[
0
,
1
]
T
[0,1]^T
[
0
,
1
]
T
Both go to
[
2
,
0
]
T
[2,0]^T
[
2
,
0
]
T
The loss of information means not just a reduced picture, but the existence of different entrances with the same result.
For linear display from
R
n
\mathbb R^n
R
n
Theorem of Frank-nullity:
rank
⁡
(
A
)
+
dim
⁡
ker
⁡
(
A
)
=
n
.
\operatorname{rank}(A)+\dim\ker(A)=n.
rank
(
A
)
+
dim
ker
(
A
)
=
n
.
In our example, one output direction and one kernel direction together take into account two inherent degrees of freedom.
Determinant and square matrix reversibility
For
2
×
2
2\times2
2
×
2
det
⁡
[
a
b
c
d
]
=
a
d
−
b
c
.
\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=ad-bc.
det
[
a
c
​
b
d
​
]
=
a
d
−
b
c
.
Non-zero determinant is equivalent to Frank 2 and existence inverse.
A
=
[
2
1
0
1
]
A=\begin{bmatrix}2&1\\0&1\end{bmatrix}
A
=
[
2
0
​
1
1
​
]
determinant is equal to 2, and
A
−
1
=
[
1
/
2
−
1
/
2
0
1
]
.
A^{-1}=\begin{bmatrix}1/2&-1/2\\0&1\end{bmatrix}.
A
−
1
=
[
1/2
0
​
−
1/2
1
​
]
.
He's returning it.
[
4
,
2
]
T
[4,2]^T
[
4
,
2
]
T
in
[
1
,
2
]
T
[1,2]^T
[
1
,
2
]
T
B. Computer practice is usually decided
A
x
=
b
Ax=b
A
x
=
b
It's factoring, not building a full inverse: obvious inverse is more expensive and can exacerbate numerical errors.
The exact and numerical rank answers different questions
Matrix
D
=
diag
⁡
(
1
,
10
−
12
)
D=\operatorname{diag}(1,10^{-12})
D
=
diag
(
1
,
1
0
−
12
)
Mathematically has a Frank 2 because both diagonals are non-zero, but restoring the second coordinate requires a division by
10
−
12
10^{-12}
1
0
−
12
, so the noise of order
10
−
10
10^{-10}
1
0
−
10
to exit becomes an error of order.
10
2
10^2
1
0
2
Into the entrance.
The laboratory defines the numerical Frank as the number of sinkular values greater than
10
−
10
max
⁡
(
1
,
σ
1
)
.
10^{-10}\max(1,\sigma_1).
1
0
−
10
max
(
1
,
σ
1
​
)
.
With a clearly selected tolerance
D
D
D
Another algorithm or other precision can choose a different threshold. It is not possible to issue a rounded zero to prove an exact linear relationship.
Low Frank is sometimes a desired emission that removes noise, but it is not possible to restore an arbitrary destroyed component without further assumptions. In the lab, compare the complete, Frank-one and zero matrices, while observing the grid, kernel, and collion of the entrances.
You can change the two entrances, but there's only one direction left on the exit.
How does this question go on with the result?
Python examples of Chapter 02
Academic base