I have a matrix
$$\begin{pmatrix}0&-1\\1&2\end{pmatrix}$$
where I have to find the invariant points for a transformation using this matrix.
I have no problem working through to two equations y = -x which means that the invariant points are all points on line y = -x (y + x = 0).
But question asks for co-ordinates to be expressed as parameter so I expressed my answer as:
$$(-\lambda, \lambda)$$
But book answer shows:
$$(\lambda, -\lambda)$$
Was I wrong? Is there a convention to show x co-ordinate as the positive if corresponding why opposite sign???
OK, so I was thinking that maybe they are equivalent until I looked at the next question: matrix: $$\begin{pmatrix}0.6&0.8\\0.8&-0.6\end{pmatrix}$$
$$\frac{3}{5}x + \frac{4}{5}y = x$$
$$3x + 4y = 5x$$
$$4y = 2x$$
$$y = \frac{1}{2}x$$
and
$$\frac{4}{5}x – \frac{3}{5}y = y$$
$$4x – 3y = 5y$$
$$4x = 8y$$
$$y = \frac{1}{2}x$$
This then gives you two equations same: $$y = \frac{1}{2}x$$
So when x = 1, y = 2
So I expressed as:
$$(\lambda, 2\lambda)$$
But book shows:
$$(2\lambda, \lambda)$$
Is the book wrong?
If the book is correct then what is the explanation?
Best Answer
In the first case, you and the book gave the same answer. When you express something parametrically, you can "change the speed" of the parameter: $$\{(\lambda, -\lambda)\ :\ \lambda\in\mathbb{R}\}$$ is the set of all the points of that form obtained by letting $\lambda$ run through all the real numbers; if you write $$\{(-\lambda,\lambda)\ :\ \lambda\in\mathbb{R}\}$$ you are just "enumerating" the elements in a different way, but you end up with the same collection (set) of elements… to obtain, e.g., $(1,-1)$, you have to set $\lambda=1$ in the first case, $\lambda=-1$ in the second. You could have also written $$\{(2\lambda, -2\lambda)\ :\ \lambda\in\mathbb{R}\}$$ or $$\{3\lambda-1, 1-3\lambda)\ :\ \lambda\in\mathbb{R}\}$$ or, as long as you don't want to do any more linear algebra with such a result $$\{(\lambda^3, -\lambda^3)\ :\ \lambda\in\mathbb{R}\}\;.$$ All these sets are the same, simply described in different ways.
Now, for your second question, you solved the last equation incorrectly: $$y=\frac12 x$$ means that the $y$ is half of the $x$ … so $x=1$, $y=1/2$ is a solution and not $x=1$, $y=2$. In generale, set $x=\lambda$ and work out the value for $y$ … or the opposite, if computations seem easier this way. If $x=\lambda$, $y=\lambda/2$, so $\{(\lambda, \lambda/2)\ :\ \lambda\in\mathbb{R}\}$ is your set of points, which can be equivalently described by $$\{(2\lambda,\lambda)\ :\ \lambda\in\mathbb{R}\}$$ (you find this second form if you set $y=\lambda$ and find $x$ from the equation, in terms of $\lambda$).