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Thread: Linear Alg.

  1. #1
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    Linear Alg.

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

    Does $\displaystyle \exists\, \bold{b}$ in $\displaystyle \mathbb{R}^m$ such that $\displaystyle A\bold{x}=\bold{b}$ 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 $\displaystyle A$ be an m x n matrix. Given $\displaystyle A\bold{x}=\bold{0}$ has nontrivial soln's, in each of the following demonstrate an example if you say yes of $\displaystyle A$ and $\displaystyle \bold{b}$ or if you say no explain why.

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

    a.) no solution

    b.) a unique solution

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

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

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

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

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

    Case $\displaystyle m=n$:

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

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

    Try doing other cases.
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