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  1. #1
    Super Member Showcase_22's Avatar
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    Tests

    I've done these questions but I would like them to be checked.

    Let w=e^{\frac{2i\pi}{3}}. Then iw is an n-th root of unity where n is:

    a). 12
    b). 7
    c). 4
    d). 3
    My answer is c).

    Suppose that u, v and w are unit vectors in \Re ^4. Then ||u+3v||^2+||3u-v||^2 is equal to

    a). 2\sqrt 7
    b). 20
    c). 12
    d). 16
    My answer is b).

    Let T: \Re^2 \rightarrow \Re^3 be given by T(x,y)=(x+y,x+y,x+y). The image of T is:

    a). A point
    b). A line
    c). A Plane
    d). depends on x and y
    My answer is c).

    Can someone check these please. I think they're right.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Showcase_22 View Post
    I've done these questions but I would like them to be checked.



    My answer is c).
    Note that i=e^{i \frac{\pi}{2}}

    So iw=e^{i \frac{7 \pi}{6}}

    We know that an n-th root of the unity is in the form : e^{i \frac{2k \pi}{n}}

    So you have to make appear that factor (2k) in order to find n.

    \frac{7 \pi}{6}=\frac{14 \pi}{12}

    Hence here, k=7 and n=12


    My answer is b).
    Hmmm isn't it rather |u+3v|^2+\dots ?

    Where is w ? Are you sure you copied it well ? because I find something like 40 :s


    My answer is c).
    Not sure about this one
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  3. #3
    Super Member Showcase_22's Avatar
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    Thanks moo. I did do the first one wrong but it's nice to see why!

    I did mistype the next question just by mentioning that a z is in the question. The rest of it is right though. I did it like this:

    ||u||^2+2||u||||3v||+||3v||^2+||3u||^2-2||3u||||v||+||v||^2

    =1+6+9+9-6+1=20

    Question 11 is giving me some difficulty. One of my friend's Thinks it's a line (b). I think it would be a line if all the points x and y satisfied an equation in the first place but otherwise it would be a plane.

    =S
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    Question 11 is giving me some difficulty. One of my friend's Thinks it's a line (b).
    Think of it this way.
    The image of each point T(x,y) = \left( {x + y,x + y,x + y} \right) = \left[ {x + y} \right](1,1,1)
    Every image is a multiple of a single vector.
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  5. #5
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    Thanks moo. I did do the first one wrong but it's nice to see why!

    I did mistype the next question just by mentioning that a z is in the question. The rest of it is right though. I did it like this:

    ||u||^2+2||u||||3v||+||3v||^2+||3u||^2-2||3u||||v||+||v||^2

    =1+6+9+9-6+1=20

    Question 11 is giving me some difficulty. One of my friend's Thinks it's a line (b). I think it would be a line if all the points x and y satisfied an equation in the first place but otherwise it would be a plane.

    =S
    Okay, here is my point
    In a first time, I didn't understand what || stood for.

    Now, I've watched wikipedia and it's just that I missed a possibility -_-

    \vec{u}=(x,y) \implies \|\vec{u}\|=\sqrt{x^2+y^2}

    But you can't say that \|u+v\|^2=\|u\|^2+\|v\|^2+2 \|u\| \|v\|
    What you had to use is the triangle inequality (google for it if you don't know), which is a property for a norm :
    \|u+v\| \leq \|u\|+\|v\|
    But you put an equality sign instead of a \leq, which is only true if there exists k such that u=kv


    Here is how I would do :
    Let u=(u',u'') and v=(v',v'')
    Since u and v are unit vectors, their norm is 1 :
    \sqrt{u'^2+u''^2}=1 and \sqrt{v'^2+v''^2}=1 (we'll use the square of these inequalities)

    So u+3v=(u'+3v',u''+3v'')
    Hence \|u+3v\|^2=(u'+3v')^2+(u''+3v'')^2=u'^2+u''^2+9v'^  2+9v''^2+6u'v'+6u''v'' =(u'^2+u''^2)+9(v'^2+v''^2)+6(u'v'+u''v'')=10+6(u'  v'+u''v'')


    Similarly, we have :
    \|3u-v\|^2=(3u'-v')^2+(3u''-v'')^2=\dots=10-6(u'v'+u''v'')

    Hence \|u+3v\|^2+\|3u-v\|^2=20
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I did mistype the next question just by mentioning that a z is in the question. The rest of it is right though. I did it like this:

    ||u||^2+2||u||||3v||+||3v||^2+||3u||^2-2||3u||||v||+||v||^2

    =1+6+9+9-6+1=20
    As Moo wrote it, it is not always true that \|u+v\|^2=\|u\|^2+2\|u\|\|v\|+\|v\|^2. However, it is true that \|u+v\|^2=\|u\|^2+2u\cdot v+\|v\|^2 because of the definition of the norm ( \|u\|=\sqrt{u\cdot u} where \cdot is the dot product) :

    \begin{aligned}\|u+v\|^2&=(u+v)\cdot(u+v)\\<br />
&=u\cdot u+u\cdot v+v\cdot u+v\cdot v\\<br />
&=\|u\|^2+2u\cdot v+\|v\|^2\end{aligned}.
    Using this identity we get

    <br />
\|u+3v\|^2=\underbrace{\|u\|^2+\|3v\|^2}_{10}+6u\c  dot v<br />
and  \|3u-v\|^2=\underbrace{\|3u\|^2+\|v\|^2}_{10}-6u\cdot v

    thus \|u+3v\|^2+\|3u-v\|^2=20.
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  7. #7
    Super Member Showcase_22's Avatar
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    Thanks guys! I have heard of the triangle inequality, but only in analysis. Since these are multiple choice tests it's easy to say that the method is right just because the answer that you get at the end is one of the choices. I'll definitely remember next time that <br />
\|u+v\|^2=\|u\|^2+2u\cdot v+\|v\|^2<br />
. I think it would be better to get the right answer with the correct working!

    I have an answer for the last question. Apparently it's a line b). I'm trying to work it out using the rank-nullity theorem but it isn't working. =S
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