1. ## Practice Test(urgent)

Okay, so my final is tomorrow. Thanks to HallsofIvy and Isomorphism I got rid of some confusion but I still lack knowledge on a few more problems. Any help would be greatly appreciated.

1. Let A = $1/2 \begin{bmatrix}1&1\\1&1\end{bmatrix}$ and define $T:R^2 \rightarrow R^2$ by T(X)=AX. Describe geometrically what effect T has on $R^2$ by describing T(C), where C is the unit circle in $R^2$. (using eigenvectors)

2. a) Given $T:R^2\rightarrow R^2$ is linear and T(1,2) = (2,4) and T(2,0) = (0,0), graph T(F) (the image of F under T), where F is the set of points in $R^2$ located on the boundary of the triangle determined by the points (1,2), (2,0), and (0,0).
(I don't know how one would go about graphing so maybe some guidance as to how I'd do so for this problem would be nice)

b) Graph T(G), where G is the set of points strictly inside the triangle F. (In other words, G is the set of points not in F, but bounded by F.) [u](same applies for this problem)

.doc attached with those 2 questions along with one more.

Thanks again.

2. Originally Posted by chickeneaterguy
Okay, so my final is tomorrow. Thanks to HallsofIvy and Isomorphism I got rid of some confusion but I still lack knowledge on a few more problems. Any help would be greatly appreciated.

1. Let A = $1/2 \begin{bmatrix}1&1\\1&1\end{bmatrix}$ and define $T:R^2 \rightarrow R^2$ by T(X)=AX. Describe geometrically what effect T has on $R^2$ by describing T(C), where C is the unit circle in $R^2$. (using eigenvectors)
Have you found the eigenvalues and eigenvectors for A? Remember that if $\lambda$ is an eigenvalue and v a corresponding eigenvector, then $Av= \lambda v$ so that if $\lambda> 1$, any line in the v direction is "stretched" by a factor of $\lambda$. If $\lambda< 1$, it is "squeezed" by that factor. If 0 is an eigenvalue (which is true if and only if det(A)= 0), the entire line in the direction of v is collapsed to 0 and, in fact, the entire plane is collapsed to the line perpendicular to v.

2. a) Given $T:R^2\rightarrow R^2$ is linear and T(1,2) = (2,4) and T(2,0) = (0,0), graph T(F) (the image of F under T), where F is the set of points in $R^2$ located on the boundary of the triangle determined by the points (1,2), (2,0), and (0,0).
(I don't know how one would go about graphing so maybe some guidance as to how I'd do so for this problem would be nice)
A linear function has the nice property that it maps lines to lines. That is, determine what T(1, 2), T(2, 0), and T(0, 0) are and draw the lines between them.

b) Graph T(G), where G is the set of points strictly inside the triangle F. (In other words, G is the set of points not in F, but bounded by F.) [u](same applies for this problem)

.doc attached with those 2 questions along with one more.

Thanks again.
I don't open ".doc" attachments.

3. Originally Posted by HallsofIvy
Have you found the eigenvalues and eigenvectors for A? Remember that if $\lambda$ is an eigenvalue and v a corresponding eigenvector, then $Av= \lambda v$ so that if $\lambda> 1$, any line in the v direction is "stretched" by a factor of $\lambda$. If $\lambda< 1$, it is "squeezed" by that factor. If 0 is an eigenvalue (which is true if and only if det(A)= 0), the entire line in the direction of v is collapsed to 0 and, in fact, the entire plane is collapsed to the line perpendicular to v.
I got (0, $\begin{bmatrix}-1\\1\end{bmatrix}$) and (1, $\begin{bmatrix}1\\1\end{bmatrix}$)

A linear function has the nice property that it maps lines to lines. That is, determine what T(1, 2), T(2, 0), and T(0, 0) are and draw the lines between them.
If a function is linear will T(0,0) always map to (0,0)?

If not, it did(for me) in this case. I might've made a mistake but since T(2,0) also maps to (0,0) then it'd make it a line segment and not a triangle, right?

I don't open ".doc" attachments.
Okay, thanks again. I guess the whole doc thing is a bit out the way. I should probably re-download that latex tutorial but I didn't think I'd have to do with derivatives anymore so I got rid of it.
Here's the question(and subquestions) from the .doc -

4. Originally Posted by chickeneaterguy
I got (0, $\begin{bmatrix}-1\\1\end{bmatrix}$) and (1, $\begin{bmatrix}1\\1\end{bmatrix}$)

If a function is linear will T(0,0) always map to (0,0)?

If not, it did(for me) in this case. I might've made a mistake but since T(2,0) also maps to (0,0) then it'd make it a line segment and not a triangle, right?
Yes, for any linear operator, T(u+ v)= T(u)+ T(v) so, in particular, T(u)= T(u+ 0)= T(u)+ T(0). Since we have T(u)= T(u)+ T(0), we must have T(0)= 0. Since T(2, 0)= 0, also, yes, the entire plane is mapped into a single line.

Okay, thanks again. I guess the whole doc thing is a bit out the way. I should probably re-download that latex tutorial but I didn't think I'd have to do with derivatives anymore so I got rid of it.
Here's the question(and subquestions) from the .doc -

Okay, the definition of "linear transformation" is "T(au+ bv)= aT(u)+ bT(v). It is enough to show that T(au+ v)= aT(u)+ T(v) which is why this problem says "Let each of f and g be in $C^\infty$ and r be in R". You need to show that T(rf+ g)= rT(f)+ T(g). Here, that says that (rf+ g)"- (rf+ g)= r(f"- f)+ (g"- g). Can you can show that? Of course, it is important that (f+ g)'= f'+ g'.

"Ker(T)", the kernel or nullspace of T, is the set of functions such that T(f)= 0. Just go through each of the functions given and see if f"- f= 0 for all x.
Given that the nullspace is two dimensional, it can be spanned by two independent functions in that set. Just choose two independent functions from the given functions that are in the nullspace. (There are actually six correct answers here!)

Any such solution can be written as a linear combination of two independent functions-g(x)= Af(x)+ Bh(x) where "f" and "h" are your two independent functions from (c). Use the fact that g(0)= 1 and g'(0)= 2 to determine A and B. Just as there are six correct answer to (c), there are six correct answers here.