Why the transpose of a singular matrix is singular?

Assume for contradiction that $A^T$ was invertible, then there would be a matrix $B$ with $BA^T=I$. But that means $I=I^T=(BA^T)^T=AB^T$, so $B^T$ would be an inverse for $A$, which is impossible.

Formally, a singular matrix $A$ is one for which there does not exist another matrix $B$ with $AB=BA=I$.

The statement here can be proven through the contrapositive: if $A$ is not singular, there exists some $B$ with $AB=I$. Transposing this gives $B^TA^T=I$, so $A^T$ is not singular. Thus if $A^T$ is singular, $A$ is singular. Replacing $A$ with $A^T$ in the last sentence gives the other direction, so the original statement is established.