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Given a vector space $V$ with dimension $n$, if I am given $m<n$ linearly independent vectors $a_1,a_2...,a_m$, will there always be a vector $v \in V$ which orthogonal/normal to all the given $m$ vectors. If it is always possible to find such a vector how do I find one ? In simplest case $n=3$ and $m=2$ I can see it is true and I can find the desired vector by cross product. But can't prove in general. If possible I would want a hint as my doubt seems easy and I want to figure it out on my own.

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    $\begingroup$ Use a Gram-Schmitt process. Start with any vector not in the span of the $a_1, ..., a_m$. Such a starting vector must exist because the vector space has a basis, at least one member of which is not in the span of the $a_j$ (because if it were for all basis vectors, then $\dim V \leq m < n = \dim V$.) $\endgroup$ – Simon S May 31 '15 at 11:28
  • $\begingroup$ Mind your spelling: Gram-Schmidt $\endgroup$ – Steven Van Geluwe May 31 '15 at 12:46
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Yes, this is always possible. Choose an arbitrary vector $\vec{w}$ that lies in $V \setminus \text{span}(\vec{a}_1,\dots,\vec{a}_m)$. The orthogonal projection of $\vec{w}$ on $\vec{a}_1$ is \begin{equation} \vec{w}_1=\frac{\vec{a}_1 \cdot \vec{w}}{\vec{a}_1 \cdot \vec{a}_1} \vec{a}_1 = \frac{\vec{a}_1 \cdot \vec{w}}{\left| \vec{a}_1 \right|^2} \vec{a}_1 \end{equation} You can check that the vector $\vec{w}-\vec{w}_1$ is orthogonal to $\vec{a}_1$.

In the same manner, project $\vec{w}-\vec{w}_1$ orthogonally on $\vec{a}_2$: \begin{equation} \vec{w}_2=\frac{\vec{a}_2 \cdot (\vec{w}-\vec{w_1})}{\left| \vec{a}_2 \right|^2} \vec{a}_2 \end{equation}
You can again check that the vector $(\vec{w}-\vec{w}_1)-\vec{w}_2$ is orthogonal to $\vec{a}_2$. However, there is no garantee that $(\vec{w}-\vec{w}_1)-\vec{w}_2$ is orthogonal to $\vec{a}_1$, except if the given vectors $\vec{a}_1,\dots,\vec{a}_m$ are already orthogonalised (which can be done by the Gram-Schmidt process).

Repeat tis strategy: calculate $\vec{w}_3,\vec{w}_4,\dots,\vec{w}_m$ in the same way. The vector you were looking for is $\vec{w}-\vec{w}_1-\vec{w}_2-\dots-\vec{w}_m$.

I hope this answer is helpful for you.

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