# Invariant lines again

1. Jun 16, 2009

### Gregg

1. The problem statement, all variables and given/known data

Find all invariant lines, of the form y=mx for the matrix transformation.

a) $\left( \begin{array}{cc} 5 & 15 \\ -2 & 8 \end{array} \right)$

b) $\left( \begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array} \right)$

3. The attempt at a solution

$\left( \begin{array}{cc} 5 & 15 \\ -2 & 8 \end{array} \right)\left( \begin{array}{c} x \\ \text{mx} \end{array} \right)=\left( \begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx} \end{array} \right)$

And for it to map onto the line, this is probably the wrong bit.

$\left( \begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx} \end{array} \right)=\left( \begin{array}{c} x \\ \text{mx} \end{array} \right)$

$x=5x+15mx \Rightarrow 4=15m \Rightarrow m=\frac{4}{15}$

I'll stop there because I know it's wrong. So what am I doing wrong, and what false assumptions have been made?

I think if I put y=m(5x+15mx) i'll get the answer but I dont see the logic behind that at all even though it's been explained to be, in my head you could be going y=m(m(m(m(m(m(m(m... forever if you did that.

Edit: Nope didn't work.

2. Jun 16, 2009

### Staff: Mentor

I think that what you are asked to do here is to find all vectors (x, y) for which multiplying on the left by your matrix produces a new vector (kx, ky) that is a multiple of the input vector.

I don't believe that there are any for your first matrix, but I believe there are two for your second matrix.

If my understanding of this problem is correct, it is setting you up for a discussion of eigenvectors. An eigenvector is a nonzero vector x such that Ax = kx. IOW, multiplication of x by the matrix A produces a vector that is a scalar multiple of the input vector; i.e., a vector that has the same direction (possibly pointing the other direction) and therefore, has the same slope.

3. Jun 16, 2009

### Gregg

error in the question, It supposed to be {{5,15},{-2,-8}} not +8!

edit: yes you're right it's the preamble to the chaper on eigenvectors and eigenvalues, diagonalisation etc. It is still troubling me though, how I can find lines that map to themselves (invariant lines) such as y=mx or y=mx+c, but eigenvectors are fine. It's just this that's confusing me. An example would be good. Or the solution to these?

4. Jun 16, 2009

### HallsofIvy

Saying a line is invariant does NOT mean that points are invariant. The line y= mx is invariant as long as a point, (x, mx) is mapped into something of the form (a, ma), not necessarily with a= x.
You do not need
$\left(\begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx}\end{array}\right)=\left(\begin{array}{c} x \\ \text{mx}\end{array}\right)$
rather you need
$\left(\begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx}\end{array}\right)=\left(\begin{array}{c} a \\ \text{ma}\end{array}\right)$
for some number a. More specifically, you need to look at
$$\frac{-2x-8mx}{5x+ 15mx}= \frac{-2-8m}{5+ 15m}= \frac{ma}{a}= m$$
Solve that for m.

5. Jun 16, 2009

### Gregg

Ok so,

$\left( \begin{array}{cc} 5 & 15 \\ -2 & -8 \end{array} \right)\left( \begin{array}{c} x \\ \text{mx} \end{array} \right)=\left( \begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx} \end{array} \right)=\left( \begin{array}{c} x' \\ \text{mx}' \end{array} \right)$

$mx'=m(5x+15mx)$

$-2x-8mx=m(5x+15mx)$

$\Rightarrow 15m^2+13m+2=0$

$m=-\frac{1}{5},-\frac{2}{3}$

6. Jun 16, 2009

### Staff: Mentor

For your 2nd problem, calculate Ax, with x = (-5/4, 1)T. You should get a vector that is a constant multiple of x. In fact, when you multiply Ay, where y is any multiple of the vector I showed for x, you'll get another vector that is the same constant multiple of y. This shows that for this matrix A, the vector (-5/4, 1) is special in the sense that the product of A and this vector is another vector whose only difference is that it is longer than (-5/4, 1). For this matrix, there is another vector that behaves the same way.