# T1: A Simple Question 1

\(\textbf{Lemma}\) Given \(Ax = \textbf{0}\), there is a non-trivial solution, if and only if \(A\) is singular.

\(\textbf{Proof}\). (<=). If \(A\) is singular, there is a non-trivial solution \(x\). The proof is simple. Since \(A\) is singular, we can assume that \(A\sim B\) where \(B\) has a row that consists of all 0s. Then, we can say that there are many solutions (by selecting one variable as a parameter).

(=>). If \(Ax = 0\) has a non-trivial solution, \(A\) is singular. It is not clear how to prove this directly. We can prove that, if \(A\) is non-singular, then, \(Ax = \textbf{0}\) has a trivial solution. Since \(A\) is non-singular, there exists \(A^{-1}\).

\[A^{-1}Ax = A^{-1}\textbf{0}\]Then

\[x = \textbf{0}.\]