[Math] Linear Transformation – unit vectors and squares

linear algebralinear-transformations

Let the standard matrix representation of a linear transformation $T \colon \mathbb{R}^2 \to \mathbb{R}^2$ be given by
$$
A_T =
\begin{bmatrix}
-1 & 1 \\
\phantom{-}1 & 1
\end{bmatrix}.
$$

  1. Compute the image of the unit vectors $e_1 = (1,0)$, $e_2 = (0,1)$.
  2. Compute the image of the unit square $\{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq 1\}$.
  3. Describe $T$ geometrically.

Source

I'm a little rusty on Linear Transformations and wasn't sure if computing an image of a unit vector was as simple as multiplying it in a matrix format through the transformation or something else?

The unit square I'm also confused on. Is it simply a square made up of the image of unit vecctors? Do you think a sketch would be required.

Thanks

Best Answer

I think you have the right idea for the first part. In particular, the image of the unit vector $e_1$ is given by $$ T(e_1) = \pmatrix{-1&1\\1&1} \pmatrix{1\\0} = \pmatrix{-1\\1} $$ so, the image of $e_1$ is simply $(-1,1)$.

It's clear that you're confused about the second part. The "unit square" here means the set of all vectors (not just unit vectors) whose components are the coordinates of some point in the unit square. That is, we want every vector $(x,y)$ for which $0 \leq x \leq 1$ and $0 \leq y \leq 1$.

In practice, however, it is not necessary to consider every vector in this space. The real approach here is to note that the image of your quadrilateral will be another quadrilateral, and note that each corner of the unit square will be mapped to a corner of the image.

I cannot say whether a sketch will be required. However, I will say that a sketch will be helpful enough that you should make one even if it isn't.

As for the last part: you should think of the transformation as a reflection followed by a scaling.

Related Question