A matrix equation $Ax=0$ has infinite solutions, does $A^Tx = 0$ have infinite solutions?

Question

I'm wondering whether a system with a transpose of a matrix has the same type of solution that the original matrix system has. If an equation $Ax=0$ equation has a unique solution, would a system with $A$ transpose instead of $A$ also have a unique solution? And what about with no solution, and infinite solutions?

Answer 1

If you are dealing with square matrices, then the answer is yes. You are essentially asking if the matrix is invertible or not, and $A$ is invertible iff $\det(A)\ne0$ and $\det(A)=\det\!\left(A^T\right)$.

However, $$ \begin{bmatrix} 3&1&2\\ 2&0&1 \end{bmatrix} \begin{bmatrix}1\\1\\-2 \end{bmatrix} =\begin{bmatrix}0\\0 \end{bmatrix} $$ and any real multiple of this solution is a solution. Yet $$ \begin{bmatrix} 3&2\\ 1&0\\ 2&1 \end{bmatrix} x =\begin{bmatrix}0\\0\\0 \end{bmatrix} $$ requires $x=\begin{bmatrix}0\\0\end{bmatrix}$.