I have a matrix given: $$A=\begin{pmatrix} 7 & -2 \\ -1 & 8 \end{pmatrix} $$
I have found its characteristic polynomial: $\lambda^2 - 15\lambda +54 = 0$, which gave me $\lambda = 6, 9$.
Now, I used that to find the Diagonal matrix. $$\begin{pmatrix} 6 & 0\\ 0 & 9 \end{pmatrix}$$
Now, to describe the transformation $ T: \mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $T(\vec{x}) = A\vec{x}$ geometrically I would think the following, but I am uncertain.
The diagonal matrix shows the transformation upon the elementary matrix. So, in this case we could say the transformation of a vector in $\mathbb{R}^2$ eg: $\begin{pmatrix}2\\1\end{pmatrix}$ would give us the following vector: $\begin{pmatrix}12\\9\end{pmatrix}$ is this correct?
My reasoning is because I think the x-value of a vector in $\mathbb{R}^2$ is multiplied by magnitude of 6 and the y-value by 9.
Did you do the computation $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}2 & 1 \end{pmatrix}$? That gives $\begin{pmatrix}7(2)-2(1) \\ -1(2)+ 8(1)\end{pmatrix}= \begin{pmatrix}12 \\ 6 \end{pmatrix}$ not $\begin{pmatrix}12 \\ 9 \end{pmatrix}$. You can multiply the matrix $\begin{pmatrix}6 & 0 \\ 0 & 9 \end{pmatrix}$ by a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ only when the given vector can be written as x times the unit vector in the direction of the eigenvector corresponding to eigenvalue 6 and y times then unit vector in the direction of the eigenvector corresponding to eigenvalue 9. To find those solve $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}6x \\ 6y\end{pmatrix}$ and $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}9x \\ 9y\end{pmatrix}$