If $Ax=b$ has infinitely many solutions, prove there exists $c$ such that $Ax=c$ has no solutions

Question

Where $A\in M_n(F)$, and $b,c\in F^n$.

What is the approach? Thanks.

Answer 1

If $Ax = b$ has infinitely many solutions, it means that the nullspace of $A$ is not the zero subspace, but contains at least one nonzero vector, hence has dimension 1 or higher. Thus, by the Rank-Nullity Theorem, the rank of $A$ is less than $n$. It follows that $A$ fails to be onto. This completes the proof.

Answer 2

Let's see $Ax=b$ has infinitely many solutions implies the columns of $A$ are dependent, so it does not give a basis for the entire vector space $F^n$. If the $n$ column vectors of $A$ does not form a basis, this means some vectors $c$ in your vector space is not expressible as $Ax=c$. The linear map $A: F^n\rightarrow F^n$ is not surjective.

Answer 3

As the dimension of the column spaced is less than $n$: what conclusion can you draw?

Answer 4

Hint :

$$Ax=b \text{ has infinitely many solutions } \Longrightarrow A \text{ is not invertible } \Longrightarrow \text{rank}(A) < n$$

Answer 5

If $Ax =b$ has infinitely many solutions, then $A$ has rank less than $n$ and so its column space has at most dimensions $n-1$. So there is the matrix $D$ such that $DAD^{-1}$ is the identity with the zero column at the end. So you can pick $\tilde{c}$ to be $e_n$ (the $nth$ standard basis vector). Now does $DAD^{-1}x = \tilde{c}$ have any solutions?