For physics/linear algebra I'm supposed to transform the big sideways F into the small upright F using a matrix, and then do the opposite. I'm wondering if anyone can shed some light on how to do this, I know I have to use sin and cos in some sort of combination but I don't know the reasoning behind it.
Thanks in advance
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Well, you could decompose this transformation into the composition of a rotation and uniform scaling, but there’s a rather simple way to construct the matrices directly if you remember that the columns of a transformation matrix are the images of the basis vectors.
So, what does the point $(1,0)$ in the untransformed F correspond to in the transformed one? That point is the bottom right corner of the F’s vertical stem, which ends up at $(0,6/5)$, so there’s the first column of your matrix. For the second column, you need to see where $(0,1)$ ends up after the transformation. This point is part-way up the left side of the $F$, so it’s going to end up somewhere along the negative $x$-axis, i.e., some point with coordinates $(0,x)$ with $x\lt0$. To work out what $x$ must be, use the ratio of the original height of the F to its transformed “height.”
For the inverse transformation, either invert the matrix that you just computed or go through the same process as above, but this time going from the transformed F back the the original. In this case, the point $(1,0)$ doesn’t lie on the transformed $F$, but $(-1,0)$ does, so take advantage of linearity: for any linear transformation $L:\mathbb R^2\to\mathbb R^2$, $L(-1,0)=-L(1,0)$.
You have a rotation by $90^{\circ} = \frac{\pi}{2}$ radians. You have a dilatation by $2$ because the max of the original is $5$ and the max of the new is $10$.
$$u(t) = 2 \cos(90^{\circ}) x(t) + 2 \sin(90^{\circ}) y(t) = 2 y(t)$$ and $$v(t) = 2 \sin(90^{\circ}) x(t) - 2 \cos(90^{\circ}) y(t) = 2 x(t).$$