Finding invariant lines of a matrix Watch

Retsek
Badges: 17
Rep:
?
#1
Report Thread starter 4 weeks ago
#1
I know how to find lines of invariant points, that's fairly straight forward:

\begin{pmatrix}a & b \\c & d \end{pmatrix} \begin{pmatrix}x \\y \end{pmatrix} = \begin{pmatrix}x \\y \end{pmatrix}

But invariant lines make less sense to me, they make sense right up to the very end of the method I'm using (I found it online and it doesn't fully explain the last stage).

\begin{pmatrix}a & b \\c & d \end{pmatrix}  \begin{pmatrix}x \\mx+c \end{pmatrix} = \begin{pmatrix}X \\mX+c \end{pmatrix}

Now I would multiply that out and then substitute the equation that equals X into the other to get an equation in x, c and m

In the case of \begin{pmatrix}2&1 \\-1&0 \end{pmatrix} I end up with the equation (m^2 + 2m + 1)x + (m+1)c=0

I let  x = 0 and find that c = 0 or m = -1

If m = -1
(1-2+1)x+(-1+1)c=0
Thus  0 = 0 which is true...

If  c=0
(m^2+2m+1)x =0
 m = -1

Thus the equation of the invariant line is y = -x + c
(but I'm not sure why we're allowed to keep c there as a variable)

Another scenario is with finding the invariant lines of \begin{pmatrix}2&-1 \\-1&2 \end{pmatrix}

I end up with (m^2-1)x + (m+1)c=0
Again I let x=0 in which case c=0 or m=-1 (coincidence that its the same)

If m=-1
((-1)^2-1)x-(-1+1)c=0
 0 = 0 which again is true... but what is the meaning of this result?

If c=0
(m^2-1)x=0
m = 1

Unlike the other example these results seem to conflict each other and I'm not sure how to resolve these into an equation. Can someone please shine a light on what I'm actually trying to do here? The guide I found online doesn't go into detail about what to do in different scenarios, and I can't find any others, they all instead explain how to find a line of invariant points. (It does explain the scenario where one of the assumptions of m or c leads to something untrue (like x=k where k is a constant but x by definition is a variable.)
Last edited by Retsek; 4 weeks ago
0
reply
Notnek
  • Study Forum Helper
Badges: 20
Rep:
?
#2
Report 4 weeks ago
#2
(Original post by Retsek)
z
I think it's best to ask questions one at a time here otherwise there's too much for a helper to look through!

So for your first one you've shown that if y=mx+c is an invariant line then m and c must satisfy

(m+1)^2x + (m+1)c=0

for all x (since you need it to be true for the whole line).

Are you happy with this so far?


E.g. if you had the line y=x+1 then plugging m=1 and c=1 in would give on the left hand side,

4x+2

This is only equal to 0 if x=-\frac{1}{2} but you need it to be true for the whole line so y=x+1 doesn't work.

What about y=-x+1?

You'd get 0+0=0 on the left hand side and this is 0 whatever x is so this works. But notice that this works whatever c is as well.

So any line of the form y=-x+c will be an invariant line because if you plug in m=-1 then you get 0, whatever c is.

Let me know if there's any part of this that you don't understand. If you're happy with this then have another look at the second part and let me know if you're still confused.
Last edited by Notnek; 4 weeks ago
0
reply
Gregorius
Badges: 14
Rep:
?
#3
Report 4 weeks ago
#3
(Original post by Retsek)
Unlike the other example these results seem to conflict each other and I'm not sure how to resolve these into an equation. Can someone please shine a light on what I'm actually trying to do here? The guide I found online doesn't go into detail about what to do in different scenarios, and I can't find any others, they all instead explain how to find a line of invariant points. (It does explain the scenario where one of the assumptions of m or c leads to something untrue (like x=k where k is a constant but x by definition is a variable.)
I hope I don't confuse things by introducing another method, but there is a nice systematic way of finding invariant lines that I haven't often seen at "A" level.

The first thing that you do is to look for invariant lines that go through the origin. These are simply eigenvectors of the matrix. If one of the eigenvectors happens to have an eigenvalue of 1, then this particular invariant line is a line of invariant points.

If there isn't an eigenvalue of 1, you stop there. If there is an eigenvalue of 1, you then start looking for a left eigenvector, that is, a solution to the equation

 \displaystyle \begin{pmatrix}m & n \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}m & n \end{pmatrix}

Then the family of lines  m x + n y = c is a family of invariant lines, including those not going through the origin.

A bit of a terse explanation, but I can spell out in more detail if there's interest.
0
reply
Retsek
Badges: 17
Rep:
?
#4
Report Thread starter 4 weeks ago
#4
(Original post by Notnek)
Z
Its hard to explain what I think I've figured out so I'll just ask, for the second example are the invariant lines
y=x and y=-x+c
0
reply
RDKGames
  • Study Forum Helper
Badges: 20
Rep:
?
#5
Report 4 weeks ago
#5
(Original post by Retsek)
Its hard to explain what I think I've figured out so I'll just ask, for the second example are the invariant lines
y=x and y=-x+c
Yep.
0
reply
Retsek
Badges: 17
Rep:
?
#6
Report Thread starter 4 weeks ago
#6
(Original post by RDKGames)
Yep.
Now I just have to explain my messy home-brew method to my teacher when she marks my work lol
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA A-level Business Paper 1 go?

Loved the paper - Feeling positive (60)
16.13%
The paper was reasonable (191)
51.34%
Not feeling great about that exam... (80)
21.51%
It was TERRIBLE (41)
11.02%

Watched Threads

View All