# Thread: Linear Alg.

1. ## 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

2. 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.

3. Originally Posted by LinAlg
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 $\text{rank}(T) + \text{nullity}(T) = n$.
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Case $m=n$:

Since $\text{nullity}(T) \geq 1$ by assumption it means $\text{rank}(T) 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.