Letting A be an $m$-by-$n$ matrix, if $m>n$, does there exist an $m$ dimensional vector $b$ that would cause the system $Ax=b$ to have no solution?.

Question

Thinking through the question my first thought is when $m >n$ you have an overdetermined system, but all those equations could still have no solution, which would mean that such vector b exists. Am I onto the right track?

Answer 1

The answer is 'yes' However, explanation depends at what stage the question arises during studing linear algebra.

I'm trying to explain it without using dimension or rank argument but with the following tools

1. The reader is at least familier Gaussian elimination process by multiplying elementary matrices.
2. The reader is familier with RRE form of the matrices.

Suppose that $M$ is an $m \times n$ matrix where $m>n$ and consider the equation $Mx=b$. By the Gaussian elimination process applying to the atached matrix $[M|b]$ we have the RRE form $[R|b']$. Note that $R$ has at most $n$ leading $1$ meaning that the last $m-n$ rows are $0$ rows. Let $P$ be the product of all elementary matrices applied during the process. We can manually invert the process $P^{-1}$ (this do not require the knowledge of inverting matrices). Choose tha vector $b'$ to be a vector with zero entries but the last one, $1$. Let $b=P^{-1}b'$. Then the equation $Mx = b$ is equivalent to $Rx = b'$ which is impossible to solve. This completes the proof.