## Practice Midterm SOLUTION: #6 (part two) |
abbyto1607 |
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Part two asks us to determine if ${(1,1,2,0),(1,0,3,1),(2,1,4,3),(1,-2,4,5)}$ is a basis for $R^{4}$. The two conditions for a basis still holds. The set must be linearly independent, and span all of the vector space. First check if the set is linearly independent by Putting the set into a matrix, and reducing to RREF. If the matrix can be reduced to an identity matrix, the set is linearly independent. Some confusion may arise as two whether a vector is a column or row in the matrix. Since each vector is a "direction" or "variable", we can see that it will be a column. The matrix should look like $\left( \begin{array}{ccc} 1&1&2&1\\ 1&0&1&-2\\ 2&3&4&4\\ 0&1&3&5 \end{array} \right)$ This reduces to $\left( \begin{array}{ccc} 1&0&0&-3\\ 0&1&0&-2\\ 0&0&1&1\\ 0&0&0&0 \end{array} \right)$ The last column does not have a leading one in it and therefore corresponds to a redundant vector. $\left( \begin{array}{ccc} 1\\ -2\\ 4\\ 5 \end{array} \right)$ is a linear combination of the other vectors. The set is not linearly independent, and therefore not a basis for $R^{4}$ |

MichaelCWeyant: May 5, 2015, 6:13 p.m. |
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So, does this technically mean the other three vectors that we are given in this problem make up the basis of R4? |

henry_lee: May 5, 2015, 6:35 p.m. |
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^Kinda adding on to that, I know the problem asks for a basis in R4, and therefore that set is not, but is the set of the first 3 vectors a set in R3 then? |

MichaelCWeyant: May 5, 2015, 6:45 p.m. |
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Also I believe the second row should be (0 1 0 2)-no negative in the last column |

davidLN: May 5, 2015, 7:03 p.m. |
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If the basis was linearly independent, how would we show it spans the entire vector space? |

davidLN: May 5, 2015, 7:07 p.m. |
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Ah, never mind. I guess if it was linearly independent, you would have an identity matrix. This implies invertibility and that it spans the entire space. |

DonnaBranchevsky: May 5, 2015, 8:17 p.m. |
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henry_lee I believe it is. A basis in R3 would need 3 linearly independent vectors, which we have here. |

aleksanderjanczewski: May 5, 2015, 9:47 p.m. |
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@davidLN: If the basis was linearly independent and was made of say 4 vectors, then this all vectors would determine different "direction". If you have four different direction it means that you are in R4. If only 3 vectors are independent then you are in R3 and etc. |