Bases for Vector Space V=\mathbb{R}^3

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SUMMARY

The discussion centers on determining which sets S are bases for the vector space V=\mathbb{R}^3. The conclusion is that set (b) is not a basis due to its lack of linear independence, as it does not span the entire space V. Sets (a), (c), and (d) are confirmed to be bases since they consist of three linearly independent vectors, which is essential for spanning the three-dimensional space.

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Homework Statement



Which of the following sets S are bases for the vector space V=\mathbb{R}^3?

(a) S=\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\}

(b) S=\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\}

(c) S=\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \right\}

(d) S=\left\{ \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} , \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \right\}

The Attempt at a Solution



By my reckoning, the only set that isn't a basis is (b) as it isn't linearly independent and \text{Span}(S)\neq V. Correct?
 
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Since there are only 3 and the dimension of R3 is 3, you are correct that whether or not they are linearly independent determines whether they are a spanning set (basis).
 

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