Let $A$ be a square matrix of order $n$ such that $Ax = 0$ only has the trivial solution

Question

(I) $A^Tx = 0$ only has the trivial solution. (II) $A^{-1}x = 0$ only has the trivial solution.

Which of the above statements is/are always true?

Answer 1

HINT

Let consider that $A$ is full rank and then

• the columns and the rows of $A$ are linearly independent vectors
• $A$ is invertible and then $Ax=0\implies A^{-1}Ax=A^{-1}0$

Answer 2

For $(i)$ recall that $A$ is invertible if and only if $A^T$ is invertible, the inverse being $(A^{-1})^T$. \begin{align} Ax = 0 \text{ has only the trivial solution } &\iff A\text{ is invertible} \\ &\iff A^T\text{ is invertible} \\ &\iff A^Tx = 0 \text{ has only the trivial solution } \end{align}

For $(ii)$, recall that $A$ is invertible if and only if $A^{-1}$ is invertible, the inverse being $A$ itself.

\begin{align} Ax = 0 \text{ has only the trivial solution } &\iff A\text{ is invertible} \\ &\iff A^{-1}\text{ is invertible} \\ &\iff A^{-1}x = 0 \text{ has only the trivial solution } \end{align}