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Let $v_1 = (1,-1,2,-2)$ and $v_2 = (0,0,-1,1)$.

Let $v_3 = (1,0,0,0)$ and $v_4 =(1,1,1,1)$.

Show that $\mathcal{B} = \{v_1, v_2, v_3, v_4 \}$ is a basis of $\mathbb{R}^4$.


After showing that $\mathcal{B} $ is set of independent vectors and that $\operatorname{Vect}(v_1, v_2, v_3, v_4)$ is a subspace of $\mathbb{R}^4$ and its dimension is the dimension of $\mathbb{R}^4$. I don't see how to proceed to get $\mathcal{B}$ is a basis of $\mathbb{R}^4$.

Thank you for your help.

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    $\begingroup$ @amWhy, your edit erased the definition of $v_1$ and $v_2$. $\endgroup$ – PersonX May 18 '18 at 21:04
  • $\begingroup$ Could you show what you used to show that $\mathcal{B}$ is a set of independent vectors, etc, $\endgroup$ – Namaste May 18 '18 at 21:04
  • $\begingroup$ Four independent vectors that span $\mathbb R^{4}$...What does it mean for vectors to span a vector space? I think you're info meets the definition. $\endgroup$ – Namaste May 18 '18 at 21:10
  • $\begingroup$ @amWhy I have found that $ av_1 + bv_2 + cv_3 + dv_4 = 0 \implies a = b = c = d = 0$. which means the set of vectors are independent. $\endgroup$ – Zouhair El Yaagoubi May 18 '18 at 21:11
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    $\begingroup$ Four independent vectors in $\mathbb R^4$ must span $\mathbb R^4.$ See @gimusi's answer. $\endgroup$ – Namaste May 18 '18 at 21:19
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Since $\mathcal{B}$ is a set of $4$ linearly independent vectors (i.e. span $\mathbb{R}^4$) it is by definition a basis of $\mathbb{R}^4$.

Indeed note that forall $b\in\mathbb{R}^4$ the following system

$$\begin{pmatrix}1 & 0 & 1 & 1 \\ -1 & 0 & 0 & 1 \\ 2 & -1 & 0 & 1 \\ -2 & 1 & 0 & 1 \end{pmatrix}x=b$$

has an unique solution then $\mathcal{B}$ span $\mathbb{R}^4$.

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  • $\begingroup$ @amWhy Never mind! Thanks $\endgroup$ – gimusi May 18 '18 at 21:18
  • $\begingroup$ There, you're the original and correct answer. $\endgroup$ – Namaste May 18 '18 at 21:21
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In a four dimensional vector space every set of four linearly independent vectors constitutes a basis for the space.

Since you have proved $$ B=\{ v_1, v_2, v_3, v_4\}$$ are linearly independent vectors, $B$ is a basis for $\mathbb {R}^4 $

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    $\begingroup$ This is a duplicate of the earlier posted answer (earlier answer posted 12 minutes prior to your post). No need to duplicate a post already saying what you try to say again. $\endgroup$ – Namaste May 18 '18 at 21:14
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To show that $\mathcal{B}$ is a set of linearly independent vectors, it's sufficient to show that :

$$\begin{pmatrix}1 & -1 & 2 & -2 \\ 0 & 0 & -1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix}$$

cannot be "Gauss-Eliminated" aka it's irreducible. Since Gauss-Elimination is a starter-pack tool in Linear Algebra, you should definitely be familiar with it and by matrix operations (row swaps - subtractions), you should easily conclude it. If one is able to bring this matrix to an echelon form consisting of the $4$ unit vectors of $\mathbb R^4$, it means that not only these vectors are linearly independent (since its dimension won't be reduced) but also that they span $\mathbb R^4$.

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    $\begingroup$ It's an alternative way of proving both questions. Why downvote without reason? $\endgroup$ – Rebellos May 18 '18 at 21:19

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