02.4. Direction which does not rotate

Content and key concepts

On the previous page, when the direction is lost.
We'll look for a non-zero arrow that changes only the scale after it's been converted. It's called its own direction.
The table doubles the horizontal, the vertical doesn't change.
Для изменения рисунка включитеJavaScript; расчёт и теория доступны далее.
Some directions the matrix does not turn
Non-zero
v
v
v
eigenvector square matrix
A
A
A
if
A
v
=
λ
v
.
Av=\lambda v.
A
v
=
λ
v
.
The matrix can stretch, squeeze or reverse this vector, but the result remains on the same straight line.
A
0
=
λ
0
A0=\lambda0
A
0
=
λ
0
True for anyone.
λ
\lambda
λ
and doesn't show direction.
The carry-over gives
(
A
−
λ
I
)
v
=
0.
(A-\lambda I)v=0.
(
A
−
λ
I
)
v
=
0.
A non-zero solution exists only when
A
−
λ
I
A-\lambda I
A
−
λ
I
Absorbed:
det
⁡
(
A
−
λ
I
)
=
0.
\det(A-\lambda I)=0.
det
(
A
−
λ
I
)
=
0.
For
2
×
2
2\times2
2
×
2
characteristic polinomial is
λ
2
−
tr
⁡
(
A
)
λ
+
det
⁡
(
A
)
=
0.
\lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0.
λ
2
−
tr
(
A
)
λ
+
det
(
A
)
=
0.
Two Directions of the Triangular Matrix
For
A
=
[
2
1
0
1
]
A=\begin{bmatrix}2&1\\0&1\end{bmatrix}
A
=
[
2
0
​
1
1
​
]
eigenvalues are equal to diagonal values 2 and 1.
λ
=
2
\lambda=2
λ
=
2
That's right.
v
1
=
[
1
,
0
]
T
v_1=[1,0]^T
v
1
​
=
[
1
,
0
]
T
:
A
v
1
=
[
2
,
0
]
T
=
2
v
1
.
Av_1=[2,0]^T=2v_1.
A
v
1
​
=
[
2
,
0
]
T
=
2
v
1
​
.
Prix
λ
=
1
\lambda=1
λ
=
1
That's right.
v
2
=
[
1
,
−
1
]
T
v_2=[1,-1]^T
v
2
​
=
[
1
,
−
1
]
T
:
A
v
2
=
[
1
,
−
1
]
T
=
v
2
.
Av_2=[1,-1]^T=v_2.
A
v
2
​
=
[
1
,
−
1
]
T
=
v
2
​
.
Arbitrary
x
=
[
1
,
2
]
T
x=[1,2]^T
x
=
[
1
,
2
]
T
goes to
[
4
,
2
]
T
[4,2]^T
[
4
,
2
]
T
And it doesn't stay collinear with the original direction. The presence of eigenvectors does not turn every entrance into eigenvironment.
There may be no real directions of your own.
Turn 90 degrees.
R
=
[
0
−
1
1
0
]
,
λ
2
+
1
=
0.
R=\begin{bmatrix}0&-1\\1&0\end{bmatrix},
\qquad \lambda^2+1=0.
R
=
[
0
1
​
−
1
0
​
]
,
λ
2
+
1
=
0.
His eigenvalues are equal
±
i
\pm i
±
i
So there's no non-zero direction on the material plane that would stay on its line, but the conversion is completely reversible and retains the length. The absence of a real eigenvector does not mean Frank loss.
The repeated eigenvalue also does not guarantee a complete base of its own.
S
=
[
1
1
0
1
]
S=\begin{bmatrix}1&1\\0&1\end{bmatrix}
S
=
[
1
0
​
1
1
​
]
has eigenvalue 1 times 2, but condition
(
S
−
I
)
v
=
0
(S-I)v=0
(
S
−
I
)
v
=
0
Demands
v
2
=
0
v_2=0
v
2
​
=
0
Independent eigen orientation is one thing, so the matrix does not diagonalize over
R
\mathbb R
R
Two eigenvectors.
For a symmetrical matrix, the situation is better.
H
=
[
2
1
1
2
]
H=\begin{bmatrix}2&1\\1&2\end{bmatrix}
H
=
[
2
1
​
1
2
​
]
direction
[
1
,
1
]
T
[1,1]^T
[
1
,
1
]
T
and
[
1
,
−
1
]
T
[1,-1]^T
[
1
,
−
1
]
T
Orthogonal and eigenvalues are equal to 3 and 1. A real symmetrical matrix allows an orthonormalized base of its own.
Hessian is symmetric under standard conditions; its eigenvalues are described as local curves along eigen directions. This helps to analyse concentration optimization. Eigenvalues arbitrary web matrix are not in itself an assessment of model quality.
A = [[2, 1], [0, 1]]
v = [1, -1]
Av = [sum(a*b for a, b in zip(row, v)) for row in A]
assert Av == v
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 arrow remained on its straight line. Other arrows may change direction, and zero arrows are not taken for their own vector.
How does this question go on with the result?
Python examples of Chapter 02
Academic base