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Thread: Please verify.

  1. #1
    Aug 2008

    Wink Please verify.

    If $\displaystyle \vec{a}\times 2\vec{j}=0$, what can you say about $\displaystyle \vec{a}$?
    $\displaystyle \vec{a}$ and $\displaystyle \vec{j}$ are parallel; paralelogram they span has area 0

    There's gotta be more?? Please tell me.

    If $\displaystyle \vec{b}\times\vec{j}=2\vec{i}$ and $\displaystyle \langle \vec{b}, \vec{j}\rangle=-1$, what's $\displaystyle \vec{b}$?
    Are $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$ meant as orthonormal vectors $\displaystyle (1,0,0)$ and $\displaystyle (0,1,0)$?

    If so, is this right: $\displaystyle \vec{b}\times\vec{j}=(-b_3)\vec{i}+b_1\vec{k}=2\vec{i} \Rightarrow b_3=-2, b_1=0, b_2\in\mathbb{R}\Rightarrow \vec{b}=(0,p,-2);p\in\mathbb{R}$??? Geometrically sounds right, in right-handed system $\displaystyle \vec{-k}\times\vec{j}=\vec{i}$, but I want to be sure.

    If $\displaystyle \vec{c}=(c_1,c_2,c_3)$ and $\displaystyle \vec{d}=(-1,2,0)$ what is $\displaystyle \vec{c}\times \vec{d}$?
    $\displaystyle =(-2c_3,-c_3,2c_1+c_2)$

    $\displaystyle \vec{a}=(1,1,1,1), \vec{b}=(1,-1,1,-1), \vec{c}=(1,i,-1,-i)$Are $\displaystyle \vec{c},\vec{c},\vec{c}$ in-dependent? Do they make a basis in $\displaystyle \mathbb{C}^4$
    Please, check my scalar products:
    $\displaystyle \langle c,c\rangle=1-1+1-1=0$

    $\displaystyle \langle a,b\rangle=1-1+1-1=0$
    $\displaystyle \langle a,c\rangle=1+i-1-i=0$
    $\displaystyle \langle b,c\rangle=1-i-1+i=0$
    So, they are in-dependent.

    But, they don't make a basis because "there's not enough of them" () for 4D vector space; they don't span the whole space? OK, what's more rigorous answer?

    Let $\displaystyle X=Y=\mathbb{R}^2$ and $\displaystyle V$ be rotation of 90 around origin. Define kernel, image and rank of map $\displaystyle V$.
    In kernel is only $\displaystyle \vec{0}$, right?
    Image is the whole $\displaystyle \mathbb{R}^2$???
    Rank of $\displaystyle V$ must be 2, otherwise it couldn't rotate (how could I better formalise my answer; if the 2 is correct of course)?

    What should $\displaystyle X=Y=\mathbb{R}^2$ mean (what are $\displaystyle X$ and $\displaystyle Y$, obviously not points)?!

    Let $\displaystyle Y$ be linear sub-space inside $\displaystyle \mathbb{K}^n$. Define orthogonal projection $\displaystyle \vec{y}$ of $\displaystyle \vec{z}\in\mathbb{K}^n$ to $\displaystyle Y.$
    (Connects with previous quote.) Let also $\displaystyle \big{\vec{f_1},...,\vec{f_k}\big}$ be ortho-normal basis for $\displaystyle Y$. Express $\displaystyle \vec{y}$ projection (from previous) with this basis.
    Last edited by courteous; Sep 14th 2009 at 02:01 PM. Reason: added sub-space questions
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