solving for a matrix A for Ax = 0

Question

just a question

For Ax = 0, I'm supposed to find a 2x3 matrix A that fulfils this when the solution space is the zero subspace

the answer is that A does not exist

My problem is, to me, wouldn't A just be a zero vector? Or perhaps I don't truly understand what a zero subspace is (span{0}?)

Sorry if I'm making a lot of wrong assumptions, I've just started on my journey in linear algebra

Answer 1

For simplicity, assume the field of scalars is $\mathbb{R}$.

Let $A$ be a $2\times3$ matrix, regarded as a function from $\mathbb{R}^3$ to $\mathbb{R}^2$.

By the Rank-Nullity Theorem, the dimension of $\text{im}(A)$ (the image of $A$ in $\mathbb{R}^2$) plus the dimension of $\text{ker}(A)$ (the kernel of $A$ in $\mathbb{R}^3$, also called the solution space) is equal to the dimension of the domain, $\mathbb{R}^3$, which is $3$.

But the dimension of the image is at most $2$, hence the dimension of the kernel is at least $1$.

It follows that the solution space can't be the zero subspace of $\mathbb{R}^3$.

So that explanation works well if you know the Rank-Nullity theorem.

At a more elementary level, consider the task of solving the equation $$Ax = 0$$ using Gaussian elimination (i.e., row reduction).

In row-reduced-echelon form, each row will have at most one pivot, hence, since $A$ has only two rows, there will be at most two pivots. It follows that of the $3$ variables to be solved for, there will be at least one free variable, and that variable can take any value. Thus, the solution space will contain nonzero vectors, so it can't be the zero subspace.

At an intuitive level, the equation $Ax=0$ corresponds to a system of two linear equations in three unknowns, thus, since we have more unknowns than equations, we wouldn't expect the solution to be unique.

Hence we wouldn't expect $x=0$ to be the only solution.

Answer 2

If $A$ is a $2\times 3$ matrix, then it would have been like below:

$$A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ \end{bmatrix} $$ As a result $x$ should be like:

$$x= \begin{bmatrix} x1\\ x2\\ x3\\ \end{bmatrix} $$ (Because when $A$ is $m\times n$, to multiply, $x$ should be of $n\times 1$ size and the resulting vector is $m\times 1$)

The actual equation here is as below:

$$ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$ So, as you can see, this means you have 2 equations with 2 zero constants. It is a zero vector with size of 2.