Entry
Math: Matrix: Eigenvector: Eigenvalue: How to calculate: 2 x 2 matrix: Example: [[ 1 0 ] [ 0 1 ]]
Jan 15th, 2006 16:29
Knud van Eeden,
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--- Knud van Eeden --- 15 January 2021 - 04:36 am --------------------
Math: Matrix: Eigenvector: Eigenvalue: How to calculate: 2 x 2 matrix:
Example: [[ 1 0 ] [ 0 1 ]]
===
Steps: Overview:
1. -Given the eigenvector equation
- - -
A . x = lambda . x
-
2. -Choose A to be a 2 x 2 matrix
3. -Or thus written in components
[ a11 a12 ] - -
[ ] . x = lambda . x
[ a21 a22 ]
-
4. -Here the matrix A is given to be
[ 1 0 ]
[ ]
[ 0 1 ]
5. -Or thus
[ 1 0 ] - -
[ ] . x = lambda . x
[ 0 1 ]
6. -Or thus you have to solve the equation, determinant equals zero
| 1 - lambda 0 |
| | = 0
| 0 1 - lambda |
7. To find lambda, the eigenvalue, or thus the scaling factor with
which the eigenvector is multiply, you need to solve the 2nd
degree equation with the given components
a11 = 1
a12 = 0
a21 = 0
a22 = 1
8. The characteristic polynomial
lambda^2 - ( a22 + a11 ) . lambda + a11 . a22 - a12 . a21 = 0
becomes thus here
lambda^2 - ( 1 + 1 ) . lambda + 1 . 1 - 0 . 0 = 0
or thus worked out
lambda^2 - 2 . lambda + 1 = 0
9. Using the root formula to find the corresponding lambda
lambda = ( (a22 + a11) +/- sqrt( (a22 + a11)^2 - 4 . (a11 .
a22 - a12 . a21) ) ) / 2
= ( (1 + 1) +/- sqrt( (1 + 1)^2 - 4 . (1 . 1 - 0 .
0) ) ) / 2
= ( (2) +/- sqrt( (2)^2 - 4 . (1) ) ) / 2
= ( 2 +/- sqrt( 4 - 4 ) ) / 2
= ( 2 +/- sqrt( 0 ) ) / 2
= ( 2 ) / 2
= 1
10. This 2nd degree polynomial has thus 2 solutions, 1 and 1, thus a
multiple same root.
The eigenvalues, or scale factors of the eigenvectors, equal thus
1 and 1.
The scale factor is thus for both eigenvectors equal to 1.
In other words the length of the eigenvectors stays the same.
11. -Once you know the 2 eigenvalues, you can solve for the
2 eigenvectors
1. -Eigenvector1 follows by filling in eigenvalue1 in the
determinant equal 0 equation
[ 1 - lambda1 0 ] [ x1 ] [ 0 ]
[ ] . [ ] = [ ]
[ 0 1 - lambda1] [ x2 ] [ 0 ]
2. -Or thus, by working this out according to the rules of matrix
multiplication, this is equivalent to the following 2 linear
equations in 2 unknowns x1 and x2:
(1 - lambda1) . x1 + 0 . x2 = 0
0 . x1 + (1 - lambda1) . x2 = 0
3. Working this out, using lambda1 = 1, gives:
(1 - 1) . x1 + 0 . x2 = 0
0 . x1 + (1 - 1) . x2 = 0
4. Working this out gives
0 . x1 + 0 . x2 = 0
0 . x1 + 0 . x2 = 0
5. In other words any x1 and any x2 will fullfil this equation
In other words, any vector will be an eigenvector for this
given matrix.
That this is true, follows from the fact that
this matrix A is an identity matrix.
In other words when you apply this matrix
to any vector, the same vector will come out,
and nothing changes. Thus also no direction
change or even length change.
Thus every vector is an eigenvector here.
And the eigenvalue, or thus the scale factor, is 1,
which indicates that the length remains exactly the
same.
===
To check this goto e.g.
http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert.htm
and fill in
1 0
0 1
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Internet: see also:
---
Math: Transformation: Eigenvector: Link: Can you give an overview of
links? pu
http://www.faqts.com/knowledge_base/view.phtml/aid/39001/fid/1856
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