Question

Find equation of line that matrix maps to same equation

Answer 1

\[\left[\begin{matrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{matrix}\right]\]
Not sure the best approach to tackle this.
any help much appreciated!

Answer 2

That's the formal of going about solving this problem. But I also encourage you to think geometrically about what this matrix is doing. How does it change the geometry of R^3. If you understand that, the line or lines that remain invariant will be immediately obvious.

Answer 3

isn't ax+by+cz=d the equation of a plane?

Answer 4

Yes, sorry. I should have had my coffee first.

The general form a line is tv + w, where v and w are vectors, and t is a parameter.

Answer 5

or x-a/b = y-c/d etc

Answer 6

i can see this matrix is scaling in x and z directions leaving y unchanged
not sure how that is immediately obv what the invariant lines are

Answer 7

The vector form is by far the easiest to deal with here. Call you matrix M, then a line
tv + w is invariant if
M(tv + w) = tMv + Mw = tkv + kw, for some constant k.

Answer 8

That being the case, then the question is what vectors v in general have the property that
Mv = kv for some constant k

Answer 9

thanks James, great help

Answer 10

Have you covered eigenvalues/vectors/spaces in your course?

Answer 11

no, this UK further math
start of the course

Answer 12

Got it.

Answer 13

i'm self studying for fun

Answer 14

good for you.

Answer 15

HI James I'm not sure i understand that expression after thinking about it

M(tv + w) = tMv + Mw = tkv + kw, for some constant k.

as tkv+kw isn't necessarily the same line as tv + w?

Answer 16

M(tv + w) = tMv + Mw = tkv + k'w

I should have written the second k as k'.

If w = 0, then it is the same. If k' = +1 then it is the same line regardless of w.

Answer 17

The question i am looking at is this:
Find equation of the lines that map to themselves under the transformation:
\[\left[\begin{matrix}2 & 0 &0\\ 0 & 1 & 0\\ 0 & 0 & 2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}X \\ Y \\ Z\end{matrix}\right)\]

Answer 18

using James approach:
orig line = w+tv
therefore
x = wx + tvx
y= wy + tvy
z= wz + tvz

which results in
X = 2wx+2tvx
Y = wy+tvy
Z = 2wz+2tvz

what next?

Answer 19

I tried this:
from matrix:
X = 2x, Y=y, Z =2z

using general equation
\[(x-a)/b = (y-c)/d = (z -e)/f\]

2 points that are definitely on the line that could be transformed are:
(a,c,d) and (a+b, c+d, e+f)

and these will map to:
(2a, c, 2d) and (2a+2b, c+d, 2e+2f)

then I use:
\[(x-a)/(x2-x1) = (y-c)/(y2-y1) = (z -e)/(z2-z1)\]

results in:
\[(x-2a)/2b = (y-c)/d = (z -2e)/2f\]

Answer 20

not sure if this is completely wrong approach

Answer 21

So let me be more precise.

A line in \( \mathbb{R}^3 \) (and this whole approach generalizes to n-dimensional Euclidean space, \( \mathbb{R}^n \)) can be represented as the set L which is a function of two vector parameters, \( v, w \in \mathbb{R}^3 \):
\[ L = L(v,w) = \{ tv + w \ | \ t \in\mathbb{R} \} \]

Now let T be a linear transformation of \( \mathbb{R}^3 \) to itself, which we will represent as a 3x3 matrix. Then we say that the line L is invariant under T if
\[ T(L) = L \]

Now consider two cases:
1. If w = 0, the zero vector, then T(L) if
\[ T(v) = kv \ \ \hbox{ for some } k \neq 0 \]
as
\[ T(L) = \{ T(tv) \ | \ t \in \mathbb{R}\} = \{ tT(v)\ | \ t \in \mathbb{R} \} = \{ tkv \ | \ t \in \mathbb{R}\} = \{ tv \ | \ t \in \mathbb{R}\} = L \]

2. If \( w\neq 0 \), the zero vector, then T(L) = L if both
a) T(w) = kw for some non zero k and
b) T(v) = v.

You can work through the logic of that.

Answer 22

Now let's analyze your matrix M. Write as usual \(i, j, k \) for the unit vectors in the x, y an z directions respectively.

You can see that Mi = 2i and Mk = 2k. Hence in fact for any vector \( v = xi + zk\) that sits in the xz-plane,
\[ Mv = M(xi + zk) \] \[ = xMi + zMk \] \[ = x.2i + z.2k \] \[ = 2(xi + zk) \] \[ = 2v \]

On the other hand, \( Mj = j \). Hence any vector v along the x-axis, v = yj has the property that \( Mv = 1v \).

Therefore the general form of the line that will be invariant under M is
\[ tv + w \]
where \( v\) is a vector in the xz-plane and \( w \) is vector along the y-axis.

Answer 23

You can this an infinite set of family of lines. There's a family where w = 0 and there are all the straight lines in the xz-plane that pass through the origin.

Then as you vary w you're moving backwards and forwards along the y-axis and each point on the y axis gives a new plane parallel to the xz-plane. Each one of those planes has an infinite number of straight-lines in it, each one of which intersects the y-axis.

Answer 24

This property that any vector at all is left invariant by a linear transformation is actually very special, so in most cases when asked a question like this, if there are any straight lines that are invariant at all, they will normally pass through the origin.

But even now, the property that there are any straight lines that are invariant is also special. Here's a 2x2 matrix
0 -1
1 0

that is the linear transformation of rotating 90 degrees anti clockwise in the plane. This transformation leaves no vectors invariant, except the zero vector. And consequently, this linear transformation leaves no line invariant either.

Answer 25

Hey James this is way more complex/formal than math I've dealt with before but I think I can follow!
One question

You say
2. If w≠0, the zero vector, then T(L) = L if both
a) T(w) = kw for some non zero k and
b) T(v) = v.

wouldn't that result in T(L) = kw + v
whereas our line is tv +w

I could see how it would be true in the case where v = w

Answer 26

I think I've followed the logic
pretty amazed by that
thanks Steve

Answer 27

In the case where w is not the zero vector, then
T(tv + w) = T(tv) + T(w) , by linearity of T (or matrix multiplication properties if you like)
= tT(v) + T(w) , by linearity of T (ditto)
= tkv + w , by hypothesis: T(v) = kv and T(w) = w

Now as k is not zero (also by hypothesis), it follows that
\[ T(L) = \{ tkv + w \ | \ t \in \mathbb{R} \} = \{ tv + w \ | \ t \in \mathbb{R} \} = L \]
In other words, the k just scales the t up or down, or even backward is k < 0, but tkv sweeps out the same values as tv.

Answer 28

I think it was this that confused me
a) T(w) = kw for some non zero
if w is kw then for k other than 1 then it wouldn't be the same line as our original line
i.e. tv+w

Answer 29

Yes, it will be the same line. Let v be some (non-zero) vector. Then the set
\[ \{ tv \ | \ t \ \hbox{ is a real number } \} = \{ tv \ | \ t \in \mathbb{R} \} \]
is the line passing through the origin in the direction v. You can think of t as a parameter that scales up and down the line.

When t = 0, we're at the origin; when t = 1, we're at the point at the 'end' of the vector v; when t = 2, we're at twice this or at the 'end' of the vector 2v; when t = 17.5 we're at 17.5 times that; when t = -1, we're at the point at the end of the vector -v; when t = -102, we're at the point at the end of the vector -102v.

You should draw yourself a diagram in the plane and see how this works. For example, suppose v = 2i + j. Now draw the straight line passing through the origin and (2,1). Then every point on that line is equal to tv for some value of t.

Now, the parameter t is somewhat arbitrary. If I scale t by 2 for instance, the set of all points
\[ \{ 2tv \ | \ t \in \mathbb{R} \} \]
is exactly the same; it's just that now instead of t = 1 corresponding to v, t =1/2 corresponds to v.

Now I can scale t by ANY factor k, provided k isn't zero, and then the set of points
\[ \{ ktv \ | \ t \in \mathbb{R} \} \]
is still the same.

Answer 30

I see the vector equation of the line as in the diagram

Answer 31

so if w is scaled how can it represent the same line, i must be missing something

Answer 32

w is NOT scaled. The condition is that w is invariant. It is v that is scaled

Answer 33

oh, whoops. I see. My fault.

Answer 34

is there typo going on here

Answer 35

I should have written
2. If w≠0, the zero vector, then T(L) = L if both
a) T(v) = kv for some non zero k and
b) T(w) = w.

Answer 36

at least i'm not going mad

Answer 37

thank you james for your patience and time!!
your a star!

Answer 38

It's an interesting problem. If I ever taught HS-level maths, I'd find a way to work it in.

Answer 39

i've learnt a lot, i'm hoping to do a Math degree, i suspect this is a glimpse at that world

Answer 40

Are you going to take the Cambridge-Warwick exam?

Answer 41

It's a serious entrance exam.

Answer 42

tbh i just enjoy Math in my spare time

Answer 43

like some people do crosswords that kind of thing

Answer 44

If you live in the UK and you want to study mathematics, Cambridge, Oxford, Warwick are all great places to study.

Answer 45

And to get into Cambridge or Warwick, there is an entrance exam, which is quite rigorous for its level.

Answer 46

i've heard Cambridge is the best, at least on paper

Answer 47

where did you study

Answer 48

Yes, that's right. If I had to force rank them, I'd say Cambridge, Oxford, Warwick. But all of them are excellent.

I studied and worked at The Australian National University, and over the years we had a number of visiting scholars from all of those places, so we know each other well.

Answer 49

do you know any good books to get up to scratch for uni math

Answer 50

The best is "Calculus" by Spivak.

Answer 51

It's a text used in rigorous first year courses.

But I'm sure there are also some good pre-university texts that I don't know about.

Answer 52

thanks James I'll check that out.