Why is this true for matrices? Linearly dependent columns $\implies$ not invertible 
If $A$ is a square matrix with linearly dependent columns, then $A$ is not invertible.

Why is this true for matrices?

Let $A_1, \ldots, A_n$ be the columns of $A$. If they're linearly dependent, then there are constants $c_i$ (not all zero) such that
$$c_1\,A_1 + \cdots + c_n \,A_n=\vec0.$$
The trick is to note that if $\vec c = \left[\begin{array}{c} c_1\\ \vdots\\ c_n\end{array}\right]$, then the above equation says precisely that $A\vec c = \vec 0$, with $\vec c \not=\vec0$. (In general, $x_1\,A_1 + \cdots + x_n \,A_n=A\vec x$, for any vector $\vec x$.)
Now you use the fact that if $A$ is invertible, then the only solution to $A\vec x = \vec0$ is $\vec x = \vec0$. So $A$ can't be invertible. $~\square$
