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Homework #1: Decomposable Forms

  1. Let $ \vec{u} $ be an ordinary vector in $ \Re^3 $, so that
    \[ \vec{u} = A\hat{i} + B\hat{j} + C\hat{k} \]

    Find $ \vec{v} $ and $ \vec{w} $ such that

    \[ \vec{u} = \vec{v} \times \vec{w} \]

    We observe that the cross product is an alternating linear function, uniquely determined by its action on the basis,

    \begin{align*}<br /> \hat{j} \times \hat{k} &= \hat{i} \\<br /> \hat{k} \times \hat{i} &= \hat{j} \\<br /> \hat{i} \times \hat{j} &= \hat{k}<br /> \end{align*}

    and also that $ \times $ is antisymmetric and distributive. Bring this all together, we can express $ \vec{u} $ as a product of two vectors:

    \begin{align*}<br /> \vec{u} &= A\hat{j} \times \hat{k} + B\hat{k} \times \hat{i} + C\hat{i} \times \hat{j} \\<br /> &= A\hat{j} \times \hat{k} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= (A + B) \hat{j} \times \hat{k} + B \hat{k} \times \hat{j} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= -C \frac{(A + B)}{-C} \hat{j} \times \hat{k} + B \hat{k} \times \hat{j} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= (B\hat{k} - C\hat{j})\times (\hat{i} + \hat{j} - \frac{(A + B)}{C}\hat{k}) \end{align*}

    Notice that if $ C=0 $, the answer is undefined. One, of A,B,C must be nonzero, however, in which case a symmetric argument will suffice. $ \square $


Sun, 2007-02-11 23:24 | eric