01.2. Where the coordinates are calculated

Content and key concepts

On the previous page, the vector and its coordinates.
To give the coordinates, you have to choose the direction of measurement -- the base. The turn of these directions changes the number recording, but not the arrow itself.
One arrow [3.4]; direction [1.0] and [1.1].
Координаты и расчёт геометрического примера приведены в тексте. Для изменения пространственного рисунка включитеJavaScript.
One vector allows different coordinates
The coordinates answer not only the question of where the arrow ends, but also the question "how many times to take each direction chosen."
b
1
,
…
,
b
n
b_1,\ldots,b_n
b
1
​
,
…
,
b
n
​
is called the base
R
n
\mathbb R^n
R
n
if every vector
x
x
x
The only way we're gonna figure this out is,
x
=
c
1
b
1
+
⋯
+
c
n
b
n
.
x=c_1b_1+\cdots+c_nb_n.
x
=
c
1
​
b
1
​
+
⋯
+
c
n
​
b
n
​
.
The existence of such degradation means that the base creates space.
a
1
b
1
+
⋯
+
a
n
b
n
=
0
a_1b_1+\cdots+a_nb_n=0
a
1
​
b
1
​
+
⋯
+
a
n
​
b
n
​
=
0
possible only when
a
1
=
⋯
=
a
n
=
0
a_1=\cdots=a_n=0
a
1
​
=
⋯
=
a
n
​
=
0
If the non-zero combination gives zero, one of the vectors of the base is redundant and the coordinates are no longer unambiguous.
Let's collect base vectors into matrix columns.
B
=
[
b
1
⋯
b
n
]
B=[b_1\ \cdots\ b_n]
B
=
[
b
1
​
⋯
b
n
​
]
and the coefficients per column
c
c
c
Then one geometric arrow connects to the coordinates of the formula.
x
=
B
c
.
x=Bc.
x
=
B
c
.
If the columns form the base,
B
B
B
reversible and
c
=
B
−
1
x
.
c=B^{-1}x.
c
=
B
−
1
x
.
The direction of change is important:
c
c
c
Report on the number of base columns;
B
c
Bc
B
c
It restores world coordinates.
x
x
x
II. Matrix
B
−
1
B^{-1}
B
−
1
He's going back.
Three exact recordings of one shooter
Let the world's arrow be fixed:
x
=
[
3
,
4
]
T
x=[3,4]^T
x
=
[
3
,
4
]
T
.
Standard base
e
1
=
[
1
,
0
]
T
e_1=[1,0]^T
e
1
​
=
[
1
,
0
]
T
,
e
2
=
[
0
,
1
]
T
e_2=[0,1]^T
e
2
​
=
[
0
,
1
]
T
:
x
=
3
e
1
+
4
e
2
,
c
s
t
d
=
[
3
,
4
]
T
.
x=3e_1+4e_2,
\qquad c_{std}=[3,4]^T.
x
=
3
e
1
​
+
4
e
2
​
,
c
s
t
d
​
=
[
3
,
4
]
T
.
In the cosine base.
b
1
=
[
1
,
0
]
T
b_1=[1,0]^T
b
1
​
=
[
1
,
0
]
T
,
b
2
=
[
1
,
1
]
T
b_2=[1,1]^T
b
2
​
=
[
1
,
1
]
T
:
B
=
[
1
1
0
1
]
,
c
=
[
−
1
4
]
,
B
c
=
[
3
4
]
.
B=\begin{bmatrix}1&1\\0&1\end{bmatrix},
\qquad
c=\begin{bmatrix}-1\\4\end{bmatrix},
\qquad
Bc=\begin{bmatrix}3\\4\end{bmatrix}.
B
=
[
1
0
​
1
1
​
]
,
c
=
[
−
1
4
​
]
,
B
c
=
[
3
4
​
]
.
Ratio
−
1
-1
−
1
means one step against
b
1
b_1
b
1
​
and then four steps along
b
2
b_2
b
2
​
III. Numbers
[
−
1
,
4
]
[-1,4]
[
−
1
,
4
]
drawn in the standard base would give a different arrow.
x
x
x
Only with the matrix
B
B
B
.
For a twisted base.
b
1
=
[
0
,
1
]
T
b_1=[0,1]^T
b
1
​
=
[
0
,
1
]
T
,
b
2
=
[
−
1
,
0
]
T
b_2=[-1,0]^T
b
2
​
=
[
−
1
,
0
]
T
:
B
=
[
0
−
1
1
0
]
,
c
=
[
4
,
−
3
]
T
,
B
c
=
[
3
,
4
]
T
.
B=\begin{bmatrix}0&-1\\1&0\end{bmatrix},
\qquad c=[4,-3]^T,
\qquad Bc=[3,4]^T.
B
=
[
0
1
​
−
1
0
​
]
,
c
=
[
4
,
−
3
]
T
,
B
c
=
[
3
,
4
]
T
.
In all three cases, the numbers and grid change, but the world end of the arrow remains.
(
3
,
4
)
(3,4)
(
3
,
4
)
.
Size and Abandoned Set
The dimension of the final space is equal to the number of vectors in any of its bases. On a plane, two independent directions are sufficient and necessary.
[
1
,
0
]
T
,
[
2
,
0
]
T
[1,0]^T,[2,0]^T
[
1
,
0
]
T
,
[
2
,
0
]
T
The plane does not produce: any combination remains on a horizontal line,
c
1
[
1
,
0
]
T
+
c
2
[
2
,
0
]
T
=
[
c
1
+
2
c
2
,
0
]
T
.
c_1[1,0]^T+c_2[2,0]^T=[c_1+2c_2,0]^T.
c
1
​
[
1
,
0
]
T
+
c
2
​
[
2
,
0
]
T
=
[
c
1
​
+
2
c
2
​
,
0
]
T
.
For
x
=
[
3
,
4
]
T
x=[3,4]^T
x
=
[
3
,
4
]
T
There's no solution because the second co-ordinate of any combination is zero.
[
3
,
0
]
T
[3,0]^T
[
3
,
0
]
T
On the contrary, solutions are endless: for example,
(
c
1
,
c
2
)
=
(
3
,
0
)
(c_1,c_2)=(3,0)
(
c
1
​
,
c
2
​
)
=
(
3
,
0
)
and
(
1
,
1
)
(1,1)
(
1
,
1
)
Therefore, the absence of a backward matrix is manifested either by the lack of coordinates or by their lack of ownership.
Length at cross-references
In a standard orthonormalised base
∥
x
∥
2
2
=
x
T
x
\|x\|_2^2=x^Tx
∥
x
∥
2
2
​
=
x
T
x
In the cosine base, no squares of coefficients can be folded without thought.
c
c
c
. Because
x
=
B
c
x=Bc
x
=
B
c
,
∥
x
∥
2
2
=
(
B
c
)
T
(
B
c
)
=
c
T
B
T
B
c
=
c
T
G
c
,
G
=
B
T
B
.
\|x\|_2^2=(Bc)^T(Bc)=c^TB^TBc=c^TGc,
\qquad G=B^TB.
∥
x
∥
2
2
​
=
(
B
c
)
T
(
B
c
)
=
c
T
B
T
B
c
=
c
T
G
c
,
G
=
B
T
B
.
For example,
G
=
[
1
1
1
2
]
,
c
T
G
c
=
25
,
G=\begin{bmatrix}1&1\\1&2\end{bmatrix},
\qquad
c^TGc=25,
G
=
[
1
1
​
1
2
​
]
,
c
T
G
c
=
25
,
while
(
−
1
)
2
+
4
2
=
17
(-1)^2+4^2=17
(
−
1
)
2
+
4
2
=
17
The Grama Matrix holds the angles and lengths of the reference directions, and the simple sum of squares of coordinates only works for an orthonormal base.
In the lab, change the base at the same time.
x
x
x
Check the equality, not the pattern.
B
c
=
x
Bc=x
B
c
=
x
Then compare the coordinates to the Gram matrix: the metric refers to the geometric object, not automatically to its list of coefficients.
B = [[1, 1], [0, 1]]
x = [3, 4]
c = [x[0] - x[1], x[1]]
recovered = [B[0][0]*c[0] + B[0][1]*c[1],
B[1][0]*c[0] + B[1][1]*c[1]]
assert c == [-1, 4] and recovered == x
The code repeats the operations just taken care of.
Completely checked Python-example Chapter
Available with a separate file; the short piece shown does not replace its implementation.
The shooter himself remained the same, and the directional system that we're describing her through has changed.
How to measure length. How does this question continue?
Python examples of Chapter 01
Academic base