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Could I solve this matrix equation as stated for non trivial solutions? If so, underwhat conditions and how? $$ \left[ \begin{array}{c} \vec{v_1} \\ \vec{v_2} \\ \vec{v_3} \end{array} \right]_{3 \times 4} \left[ \begin{array}{c} \vec{x} \, \vec{x} \, \vec{x} \end{array} \right]_{4 \times 3} = [\vec{0}]_{3 \times 3}$$

$v_i$'s are row vectors. $x$ are column vectors.

I want $x$ such that when I take $x \cdot v_i$ for $i \in [1, 3]$, I get $0$.

I understand there are smarter ways to solve this problem but this is the first way I thought of and I'd like to see how to follow it through.

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$x \cdot v_i$ should be a scalar $0$, not a vector. –  Ross B. Apr 5 '13 at 17:24

3 Answers 3

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The matrix equation that you've written works, but it's threefold redundant. Each column of zeroes on the RHS of the equation is equated to the same thing. In other words, you could just solve $$ \left[ \begin{array}{c} \vec{v_1} \\ \vec{v_2} \\ \vec{v_3} \end{array} \right] \left[ \vec{x} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]. $$

Geometrically, you're looking for the vector $\vec{x} \in \mathbb{R}^4$ that is orthogonal to the plane spanned by $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$. By the way, there is a generalization of the cross product in $\mathbb{R}^3$ that works in $\mathbb{R}^n$ for finding such a vector.

In $\mathbb{R}^4$, let $\{\vec{e}_1, \ldots, \vec{e}_4\}$ denote the standard basis, and expand the determinant following determinant (treating the $\vec{e}_i$s as formal symbols). $$ \vec{x} = \left| \begin{array}{c} \vec{e}_1 \; \vec{e}_2 \; \vec{e}_3 \; \vec{e}_4 \\ \vec{v_1} \\ \vec{v_2} \\ \vec{v_3}\end{array} \right| $$ Assuming that the three given vectors are linearly independent, all solutions will be scalar multiples of this particular non-zero solution.

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If I really wanted to solve the redundant threefold equation, could I find a nonzero solution? –  jpp Apr 5 '13 at 21:10

You are solving the same equation multiple times. Instead, solve $Vx = \mathbf{0}$, where $V$ is your matrix of row vectors and $\mathbf{0} \in \mathbb{R}^3$.

However, $x=\mathbf{0}_{4\times1}$ is a solution, so I'm not sure you have your problem formulated correctly. Non-zero solutions would involve finding the nullspace of $V$. This can be accomplished through Gaussian elimination or singular value decomposition of $V$.

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Yes, you must find a vector $x$ that is perpendicular to $v(i), i=1,2,3.$ There is always a solution since the $3$ vectors are in a $4$-dimensional space.

The solution is a vector space.

david; barcelona

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