Question

#50
https://drive.google.com/file/d/0B7ss_Y8M_VEgYmg4eUU2MzRMZ1E/view?usp=drivesdk

Answer 1

Straight lines goto straight lines
So part a would be some rectangle

Answer 2

My guess for partb is all multiples of identity matrix

Answer 3

For partc, any rank1 matrix A has line as columnspace; so the answer would be all nonzero singular matrices

Answer 4

part b: how about all rotation matrices ? It's still square even if rotated (and then scaled)

Answer 5

This is pretty open ended.

a) Additionally you can get a parallelogram out of it from a shearing transformation.

b) Rotation and reflection matrices are also possible choices that leave the region as a square with stretching transformations (which is what a scalar times an identity matrix is but I prefer to say it in terms of the geometrical picture in my mind instead of the mathematical analytical implementation of it that you'd use for calculation). Also possible pitfall, translation is not a linear transformation but there are tricks to make it one in case you were curious by adding on an extra dimension to the space.

c) Yeah that seems right nothing to add here.

d)The area of that square, before the transformation, is made up of two unit vectors, \(\binom{1}{0}\) and \(\binom{0}{1}\). You might remember that the cross product of two vectors gives you the area of the parallelogram spanned by them with a vector perpendicular to them. Usually people just use this perpendicular aspect of the cross product to find normal vectors, but we're going to use the area of it. You can check that the area is 1 by the cross product, here we use a determinant and tack on a 0 for the z-component that these don't have.

\[\begin{pmatrix}1 \\0 \end{pmatrix} \times \begin{pmatrix}0 \\1 \end{pmatrix} =\begin{vmatrix}i & j & k \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} \]
You can expand the determinant along any row or column, so we pick the easiest, the right column with 2 zeroes in it and get:

\[k\begin{vmatrix}1 & 0 \\ 0 & 1 \end{vmatrix} = \det(u, v) = 1\]

Now ok, I threw away the 'z component k' because it doesn't exist in 2D. REALLY I'm not doing a derivation of anything. I'm just showing that if you take the determinant of 2 vectors you are finding the area of the parallelogram so that you know that it's exactly the same definition as you have learned in vector calculus. The problem is, the determinant is truly more fundamental because if you take the determinant of a 3x3 matrix with 3 vectors in it, you find the volume of the cube/rectangular prism/parallelepiped.

I'm kinda ranting getting off track here haha but the main thing I wanted to get across is that now after the transformation, the two edges of our square have to end up being translated like this as well, so in our determinant,

\[\det(Au, Av)=1\]

That's the requirement of part d of your question. We can take this a step further though, since u and v are just simple vectors, we can put them together side by side to get the identity matrix:

\[[u,v] = I\]

Really our determinant is:

\[\det(Au, Av) = \det(AI) = \det(A)=1\]

So that's our condition... WEW. So sometimes they'll say the determinant is the area of a transformation.

In vector calculus you will see this pop up with the Jacobian, when you change coordinates and do the whole \(dxdy = r dr d \theta\) what this is, is really from a determinant that measures the area/volume of the transformation at every point to make sure your change of coordinates isn't messed up and scales it appropriately. More explicitly,

\[dxdy=\det \begin{pmatrix} \tfrac{\partial x}{\partial r} & \tfrac{\partial x}{\partial \theta} \\ \tfrac{\partial y}{\partial r} & \tfrac{\partial y}{\partial \theta} \end{pmatrix} dr d \theta\]

lol OK maybe went a bit off on this didn't mean to but I'm a fast typer and bored so no skin off my back. GOOD luck reading htat block of text tho ;P