Consider the following homogeneous equation where $A$ and $X$ are matrices.
$$AX = 0$$
I want to know whether there are non trivial solutions for this equations.
Now, if $A^{-1}$ exists, then I can multiply throughout by it and get $X = 0$, so if $A$ is invertible, only the trivial solution exists.
However, I do not understand why $A$ being non-invertible would imply that non-trivial solutions exists, shouldn't it just imply that no solutions exist?
Best Answer
I'm going to assume that your $\mathbf A$ is a square matrix (or else $\mathbf A^{-1}$ doesn't really make sense).
An alternative formula for the matrix multiplication between a $n\times n$ matrix and a $n\times 1$ matrix is a sum of the scalar multiples of $n\times 1$ matrices. That is, let $\mathbf A=\begin{bmatrix} \vec a\ \vec b\ \vec c\end{bmatrix}$, where $\vec a, \vec b, \vec c$ are column vectors, and let $\vec x = \begin{bmatrix} x \\ y \\ z\end{bmatrix}$. Then $\mathbf A\vec x = x\vec a + y\vec b + z \vec c$.
But if $\mathbf A$ is not invertible, then the vectors $\vec a, \vec b, \vec c$ are linearly dependent. So then there will be nontrivial solutions to $x\vec a + y\vec b + z \vec c=\vec 0$ and thus to $\mathbf A\vec x=x\vec a + y\vec b + z \vec c=\vec 0$.