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Math Help - Linear Alg.

  1. #1
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    Linear Alg.

    Let A be an m x n matrix. Given A\bold{x}=\bold{0} has nontrivial soln's, in each of the following demonstrate an example if you say yes of A and \bold{b} or if you say no explain why.

    Does \exists\, \bold{b} in \mathbb{R}^m such that A\bold{x}=\bold{b} has:

    a.) no solution

    b.) a unique solution

    c.) an infinite # of solutions
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  2. #2
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    Well, we know m can or cannot equal n.

    For a, there will exist a vector when m < n I believe since we can have a pivot in the augmented column.

    For b, I'm not sure.

    For c, an infinite # of solutions means m < n (more columns than rows, because we'll have a pivot. But how can we answer definitely for a-c if we don't know what m is.
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  3. #3
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    Quote Originally Posted by LinAlg View Post
    Let A be an m x n matrix. Given A\bold{x}=\bold{0} has nontrivial soln's, in each of the following demonstrate an example if you say yes of A and \bold{b} or if you say no explain why.

    Does \exists\, \bold{b} in \mathbb{R}^m such that A\bold{x}=\bold{b} has:

    a.) no solution

    b.) a unique solution

    c.) an infinite # of solutions
    If \bold{x}_0 is a solution to A\bold{x} = \bold{b} and \bold{x}_1 is a non-trivial solution to A\bold{x} = \bold{0} then A(\bold{x}_1+\bold{x}_0) = A\bold{x}_1+A\bold{x_0} = \bold{b}. Furthermore, \bold{x}_1+\bold{x}_0\not =  \bold{x}_0 since \bold{x}_1 is non-trivial. Therefore, it is not possible to have unique solutions to A\bold{x} = \bold{b}. Therefore (b) is always false.

    Define T:\mathbb{R}^n \to \mathbb{R}^m by T(\bold{x}) = A\bold{x}.

    The dimension of the space of all vectors \bold{b} \in \mathbb{R}^m such that A\bold{x}=\bold{b} is the rank of T.

    The dimension of all vectors x\in \mathbb{R}^n such that A\bold{x} = \bold{0} is the nullity of T.

    By rank-nullity theorem we have \text{rank}(T) + \text{nullity}(T) = n.
    ---

    Case m=n:

    Since \text{nullity}(T) \geq 1 by assumption it means \text{rank}(T)<n therefore there is \bold{b} such that A\bold{x} = \bold{b} is not solvable. Therefore (a) is true.

    If we can show that there is a \bold{b} so that A\bold{x} = \bold{b} is solvable (say \bold{x}_0) then \bold{x}_0 + t\bold{x}_1 is a solution (where \bold{x}_1 is non-trivial which solves A\bold{x}=0) where t\in \mathbb{R}. To show that there is such a \bold{b} we need to show that \text{rank}(T) \geq 1 and it is sufficient to prove \text{nullity}(T) < n. But if this true? Well, almost. If A is a matrix with zero entries then \text{nullity}(T) = n, and this is the only exception. Otherwise, it is true.

    Try doing other cases.
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