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For each of the following linearly dependent sets, find a redundant vector v in S and verify that span(S{v})=span(S)\operatorname{span}(S-\{\mathbf{v}\})=\operatorname{span}(S). (a) S={[4,2,6,1],[1,0,1,2],[0,0,0,0],[6,2,5,5]}S=\{[4,-2,6,1],[1,0,-1,2],[0,0,0,0],[6,-2,5,5]\}, *(b) S={[1,1,0,0],[1,1,1,0],[0,0,6,0]}S=\{[1,1,0,0],[1,1,1,0],[0,0,-6,0]\}, (c) S={[x1,x2,x3,x4]R4xi=±1S=\left\{\left[x_{1}, x_{2}, x_{3}, x_{4}\right] \in \mathbb{R}^{4} | x_{i}=\pm 1\right., for each i}i \}.

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Answered 8 months ago
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Answer for (a): Given that SS be the linearly dependent set of vectors which contains 4 vectors

S={[4,2,6,1],[1,0,1,2],[0,0,0,0],[6,2,5,5]}S=\{[4,-2,6,1],[1,0,-1,2],[0,0,0,0],[6,-2,5,5]\}

We prove that v=[0,0,0,0]\mathbf{v}=[0,0,0,0] be the redundant vector.
If we verify that span(S{v})=span(S)\text{span}(S-\{\mathbf{v}\})=\text{span}(S), then we can say that v=[0,0,0,0]\mathbf{v}=[0,0,0,0] is the redundant vector.
Now we apply the Simplified Span Method to both (S{v})(S-\{\mathbf{v}\}) and SS.
Applying the Span method on (S{v})(S-\{\mathbf{v}\}) and SS if we have same row reduced echelon form (there may be an extra row of zeros) with v\mathbf{v} or with out v\mathbf{v} as one of the rows, then we can say that two spans are equal.

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