# Xntrexpressibleasalinearcombinationxxivixjvjofthebasic

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Unformatted text preview: A_{1} + c2 A_{2} + … + cn A_{n} = 0 is a nontrivial dependence relation among the columns. THUS GAUSSIAN REDUCTION CAN DETERMINE INDEPENDENCE OR DEPENDENCE OF A SET OF COLUMN VECTORS. Moreover, it tells you who to throw out to pare down to an independent spanning set of columns (every spanning set for a space can be shrunk down to a basis for that space): Throw out all the columns A_{k} that correspond to free parameters x_{k}, and the remaining A_{i} (corresponding to the leading one variables) will be independent: the dependence relations x_{1}A_{1} + … + x_{n}A_{n} = AX = 0 have X = [x_{1],..,x_{n}]^{tr} expressible as a linear combination X = x_{i} V_{i} + .. + x_{j}V_{j} of the basic solutions V_{k}, so the linear dependence relations x_{r}A_{r} + … + x_{s}A_{s} = 0 among the leading one columns have x_{k} = 0 for all the free parameter columns, and therefore are expressed in terms of the basic solutions as X = 0 V_{i} + .. + 0 V_{j}, in other words X = 0, so there are no nontrivial combinations of the leading‐one columns. This method of testing independence and spanning will be useful in the Webwork assignments. 1.5 # 24 Remember that the set of vectors orthogonal to W_{1}, .. w_{r} are the vectors v satisfying v.w_{i} = 0 for I = 1,..,r, which is a system of linear equations v_{1} a_{i1} + v_{2} a_{i2} +.. v_{n} a_{in} = 0, so you put the vectors w_{i} in...
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