Consider the following $5$ vectors in $\\mathbb{R^4}$ .
\n\\[\\begin{align} \\textbf{v}_1&=\\var{rowvector(v1)}\\\\ \\textbf{v}_2&=\\var{rowvector(v2)}\\\\ \\textbf{v}_3&=\\var{rowvector(v3)}\\\\ \\textbf{v}_4&=\\var{rowvector(v4)}\\\\ \\textbf{v}_5&=\\var{rowvector(v5)}\\end{align}\\]
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Given $5$ vectors in $\\mathbb{R^4}$ determine if a spanning set for $\\mathbb{R^4}$ or not by looking for any simple dependencies between the vectors.
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\nNot linearly independent as any set of more than $4$ vectors in $\\mathbb{R^4}$ is linearly dependent.
\n\n(b)
\nNote that \\[{v}_{\\var{t0}} =\\simplify{ {f1} * v_1+ {f2} * v_2 + {f3} * v_3 + {f4} * v_4}.\\]
\nNote also {test}
\n\nBecause 2 of our 5 vectors can be written as linear combinations of the other 3, there are not 4 linearly independent vectors in the list. Hence not all vectors in $\\mathbb{R^4}$ can be written as a linear combination of these vectors, so they are not spanning.
\n\n
(c)
\nThis set {contains} a linearly independent subset of $4$ vectors as it is {nt} spanning.
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\n\n
2. Do the above vectors form a spanning set of $\\mathbb{R}^4$? [[1]]
\n\n
3. Does the set $\\{\\textbf{v}_1,\\;\\textbf{v}_2,\\;\\textbf{v}_3,\\;\\textbf{v}_4,\\;\\textbf{v}_5\\}$ contain a linearly independent subset which forms a basis of $\\mathbb{R}^4$? [[2]]
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