1. ## Matrix problem

M = $\left[ \begin{array}{ c c } 4 & -5 \\ 6 & -9 \end{array} \right]$

A transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is represented by the matrix M. There is a line through the origin for which every point on the line is mapped onto itself under $T$.

Find a cartesian equation of this line.

2. Originally Posted by rednest
M = $\left[ \begin{array}{ c c } 4 & -5 \\ 6 & -9 \end{array} \right]$

A transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is represented by the matrix M. There is a line through the origin for which every point on the line is mapped onto itself under $T$.

Find a cartesian equation of this line.
You need to find the eigenvalues of the matrix. so the two solutions to $\left| \begin{array}{ c c } 4 - \lambda & -5 \\ 6 & -9 - \lambda \end{array} \right| =0$ and one of the eigenvectors is mapped onto it self. the rest of the problem is fairly routine.

Bobak

3. Hello,

Originally Posted by rednest
M = $\left[ \begin{array}{ c c } 4 & -5 \\ 6 & -9 \end{array} \right]$

A transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is represented by the matrix M. There is a line through the origin for which every point on the line is mapped onto itself under $T$.

Find a cartesian equation of this line.
Actually, you just have to find $X=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ which is the eigenvector associated to the eigenvalue 1, and such that :

$MX=X$ (translation from the text)

and that's a characteristic of the eigenvalue 1 (when checking with bobak's formula, the determinant is indeed 0 when $\lambda=1$)