Let A = $$\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$$ and define T:R2 → R2 by T(x) = Ax, find the image under T of$$u = \begin{bmatrix}1\\ -3\end{bmatrix}$$ and $$v = \begin{bmatrix}1\\ 5\end{bmatrix}$$

Solution:

Given

A = $$\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$$

$$u = \begin{bmatrix}1\\ -3\end{bmatrix}$$

$$v = \begin{bmatrix}1\\ 5\end{bmatrix}$$

Now, We have to find image of T under u and v

T(u) = Au = $$\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$$ $$\begin{bmatrix}1\\ -3\end{bmatrix}$$

= $$\begin{bmatrix}-3\\ -1\end{bmatrix}$$

T(v) = Av = $$\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$$ $$\begin{bmatrix}1\\ 5\end{bmatrix}$$

= $$\begin{bmatrix}5\\ 1\end{bmatrix}$$