1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\begin{pmatrix}7 & -2 \\ -1 & 8 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}9x \\ 9y\end{pmatrix}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.