Let T : R2 → R2 be the linear transformation which rotates the plane clockwise by 45 degrees, then expands the plane by a factor of 2 in the direction of the x-axis, then finally rotates the plane counterclockwise by 45 degrees.
Find a standard matrix for T.
What does T do to the square whose vertices are (0, 0),(1, 1),(0, 2),(−1, 1)?
I'm struggling with this question at the moment. How would I find the standard matrix of T when I don't know the initial values of R2, before transformation?
For the transformation of the square, I'm not sure what the question is asking. Wouldn't the transformation of the square just be all four vectors rotated by 45 degrees, expanded towards the x-axis by a factor of 2 and then rotated counterclockwise by 45? How would I show that in terms of matrix T.
Help would be appreciated.
Best Answer
Because any vector of $\Bbb R^2$ can be written $\vec v=x\vec e_x+y\vec e_y$ using some basis, a linear transformation $T$ is totally determined if it is known on some basis: $T(\vec v) = xT(\vec e_x)+yT(\vec e_y)$.
So for your linear transformation, let $\vec e_x$ and $\vec e_y$ be unit vectors in the $x$ and $y$ directions, respectively. It is a composition of 3 transformations.
Clockwise rotation: $$\vec e_x\to\frac 1 {\sqrt 2} (\vec e_x - \vec e_y)$$ $$\vec e_y\to\frac 1 {\sqrt 2} (\vec e_x + \vec e_y)$$
Expansion: $$\vec e_x\to 2\vec e_x$$ $$\vec e_y\to\vec e_y$$
Counterclockwise rotation: $$\vec e_x\to\frac 1 {\sqrt 2} (\vec e_x + \vec e_y)$$ $$\vec e_y\to\frac 1 {\sqrt 2} (-\vec e_x + \vec e_y)$$
Putting all of that together yields: $$T(\vec e_x)=\frac 1 2 (2(\vec e_x + \vec e_y) - (-\vec e_x + \vec e_y))=\frac 3 2 \vec e_x + \frac 1 2 \vec e_y$$ $$T(\vec e_y)=\frac 1 2 (2(\vec e_x + \vec e_y) + (-\vec e_x + \vec e_y))=\frac 1 2 \vec e_x + \frac 3 2 \vec e_y$$
Now, if you represent your basis vectors as column vectors $\vec e_x = \bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \bigr)$ and $\vec e_y = \bigl( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \bigr)$, then the matrix for your linear transformation is the matrix whose 1st and 2nd columns are $T(\vec e_x)$ and $T(\vec e_y)$, respectively. (Columns in the same order as the ones in the unit vectors, so that matrix multiplication results in the correct linear combination of the transformed unit vectors.) So: $$T=\frac 1 2 \begin{pmatrix} 3 & 1 \\ 1 & 3 \\ \end{pmatrix} $$
And there is your result.
In the second part of the question, $T$ acts on the vertices of the square as you describe (really it acts on every point of the square) so you just have to describe the shape that it transforms into.