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I have a set of $m$ vectors $\{x_i\}$, $x_i \in R^n$. How can I obtain a basis for $R^n/span(\{x_i\})$?

Him
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1 Answers1

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Find a basis of $\text{span(\{x_i\})}=:W$, say $\{x_1,x_2,... ,x_k\}$, and a basis of $\mathbb{R}^n$ of the form $\{x_1,x_2,...,x_k\}\cup\{y_1,y_2,...,y_{n-k}\}$. Then the classes $y_j+W, 1\leq j\leq n-k$, are a basis of $\mathbb{R}^n/W$.

Jens Schwaiger
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  • Yes, but how do we obtain the ${y_i}$?. In 3D (n=3), if m=2, we can take the cross product. If m=1, and $x_0=[a,b,c]$ then take $y_0=[−b,a,0]$ and $y_1=x_0 \times y_0$ (with some linear independence assumptions). Does this algorithm generalize? – Him Oct 19 '18 at 08:56
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    A standard methode is based in the exchange theorem of Steinitz. – Jens Schwaiger Oct 19 '18 at 16:59