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If I have the set $\{(3, −2, 1, 3),(−1, 3, −3, 4),(3, 8, 7, 0)\}$, how can I extend this into a basis of $\mathbb{R}^4$? I have seen questions in the past asking to reduce a set in order to fit it into a basis, but I am unsure how to do the opposite. Any help with this issue would be greatly appreciated.

sai-kartik
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Jamie Warren
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  • Would this post answer your question? https://math.stackexchange.com/questions/1740134/extending-basis-for-span-of-3-vectors-into-basis-for-r4 – teddy Apr 20 '20 at 02:29
  • That certainly helps. However, once I have it in reduced row echelon form, how do I find the new vector itself? – Jamie Warren Apr 20 '20 at 02:41
  • As the answer says, you look at the pivot columns (which in that example are the first four columns) meaning that the first four columns of the augmented matrix $A$ is your basis for $\mathbb{R}^4$, so in that example, the fourth vector in the basis would be $(1,0,0,0)^T$ . – teddy Apr 20 '20 at 02:56
  • I see. That answer seemed too simple. To clarify, if I wanted to extend into, say, the 5th or 6th dimension, I would use the respective 5th and 6th columns? – Jamie Warren Apr 20 '20 at 02:57
  • I think you would need to view the lower dimensional space ($\mathbb{R}^4$) as being embedded in the higher dimensional space ($\mathbb{R}^5$ or $\mathbb{R}^6$). Kaj Hansen's answer illustrates this quite nicely: https://math.stackexchange.com/questions/2808038/how-to-extend-a-basis-for-mathbbrn-to-a-basis-for-mathbbrn1?rq=1 – teddy Apr 20 '20 at 03:05

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Make a $4\times4$ matrix whose first three rows are your vectors, and whose last row is $(a\ b\ c\ d)$. Compute the determinant of this matrix (by expansion along the bottom row. would be a good way). Then choose any $a,b,c,d$ that make this determinant nonzero.

Gerry Myerson
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