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Two vectors

$a = (1,-2,2,-3)$ and $b=(2,-3,2,4)$

I need to complete the basis for $\Bbb{R}^4$

I know that there's a way of solving 2-equations linear system like $$\begin{pmatrix}a \\ b\ \end{pmatrix}\cdot x = 0$$ to get vector $c$, and then solving 3-equations linear system to get vector $d$ But are not there other ways? I heard that gram schmid proccess could be helpful, but I am not sure how to apply it here as well ass if this idea is correct at all.

  • I had a feeling of dejavu, but the other question does have better and more detailed answers already... – mdave16 Sep 16 '17 at 21:37

2 Answers2

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One way to find other two vectors is to consider some simple vectors, such as $(1,0,0,0)$, $(0,1,0,0)$ etc. and check whether they are linearly dependent on $a$ and $b$ or not.

Math Lover
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  • on a similar note, consider ${a,b,}\cup B$ where $B$ is a standard basis, and then sifting? This will result in a basis which include $a$ and $b$. – mdave16 Sep 16 '17 at 21:19
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Your idea is correct: that corresponds to put $\mathbf a \cdot \mathbf c= \mathbf b \cdot \mathbf c=0$. Only that , when you do it again for $\mathbf d$, either you impose also $\mathbf c \cdot \mathbf d = 0$, or you check that the found $\mathbf c$ and $\mathbf d$ be independent.

Actually, imposing $\mathbf c$ to be normal to $\mathbf a$ and $\mathbf b$ is too a strict condition: it would be sufficient that they are mutually independent. For that you can construct a $4 \times 4$ matrix, with the first two columns occupied by $\mathbf a$ and $\mathbf b$, and then try and fill the other columns as suggested in the previous answer, checking that the determinant be non-null.

The Gram-Schmidt process is not applicable in your problem. It is useful, once you have $4$ independent vector, to construct from them a normal basis.

G Cab
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