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$v_1=(1, 1, 2, 4), v_2=(2, -1, -5, 2),v_3 = (1, -1, -4, 0), v_4=(2, 1, 1, 6).$

I started with writing down the matrix and through row reduction ended with the solution $x_1=4/3 + (1/3)x_3$ and $x_2 = 1/3 - (2/3)x_3$, $x_3$ is free. I think the basis is supposed to be $\{v_1, v_2\}$, but I'm not sure if this is correct. They are linearly independent, but how do the two vectors generate $\mathbb{R}^4$?

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2 Answers 2

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The easy way to approach this problem is to write down a $4\times4$ matrix with the given vectors in the rows of the matrix. Then do elementary row operations to reduce the matrix to row echelon form. Let's do that...

$$\begin{bmatrix}1 & 1 & 2 & 4 \\ 2 & -1 & -5 & 2 \\ 1 & -1 & -4 & 0 \\ 2 & 1 & 1 & 6 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 0 & -3 & -9 & -6 \\ 0 & -2 & -6 & -4 \\ 0 & -1 & -3 & -2 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 0 & -3 & -9 & -6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

Since the last two rows are all zeros, we know that the given set of four vectors is linearly dependent and the sub-space spanned by the given vectors has dimension 2. Only two of the four original vectors were linearly independent.

Since we put the four vectors into the rows of the matrix and elementary row operations do not change the row space of the matrix (the space spanned by the rows of the matrix), the two remaining non-zero row vectors span the row space of the matrix. That is, the vectors $\begin{pmatrix}1,1,2,4\end{pmatrix} \;\mathrm{and}\; \begin{pmatrix}0,-3,-9,-6\end{pmatrix}$ are linearly independent and span the sub-space $\mathrm{span}\left\{v_1, v_2,v_3,v_4 \right\}$ - so, they form a basis.

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  • $\begingroup$ @Brad.S i feel the vectors are linearly independent $\endgroup$
    – user210387
    Commented Apr 24, 2015 at 19:26
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    $\begingroup$ @Rememberme, why ? He just proved they aren't. $\endgroup$
    – galois
    Commented Jul 20, 2015 at 6:40
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    $\begingroup$ I've been struggling to find a proof of why this method works and, in particular, that the nonzero rows forma basis for the given subspace. Can anyone point me to a reference? Thanks! $\endgroup$ Commented Feb 4, 2021 at 20:31
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Two vectors cannot generate $\mathbb{R}^4$ since $\dim\left(\mathbb{R}^4\right)=4$. If you want to find a basis for $S=\mathrm{Span}(v_1,v_2,v_3,v_4)$ you can write the vectors as rows of a $4\times 4$ matrix, do row reduction, and when you are done, the non-zero rows are a basis for $S$ (this is because row reduction does not change the row space). Note that in this fashion you get some basis for $S$, not necessarily a subset of $\{v_1,v_2,v_3,v_4\}$. If you want to get a basis for $S$ that is made up from vectors of $\{v_1,v_2,v_3,v_4\}$, you need to find which vectors in $\{v_1,v_2,v_3,v_4\}$ are linear combinations of others in $\{v_1,v_2,v_3,v_4\}$ (this involves solving the set of linear equations defined by $\alpha_1v_1+\alpha_2v_2+\alpha_3v_3+\alpha_4v_4=0$, that is, reducing the matrix in which the columns are $v_1,v_2,v_3,v_4$.)

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