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?
 A: 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. $~\square$
A: If the columns of A are linearly dependent, 
then $a_1\vec{c_1}+\cdots+a_n\vec{c_n}=\vec{0}$ for some scalars $a_1,\cdots, a_n$ (not all 0).
Then $Av=\vec{0}$ where $v=\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}\ne\vec{0}$, so A is not invertible
(since otherwise, multiplying by $A^{-1}$ would give a contradiction).
A: You should know that elementary column operations preserve a zero/nonzero determinant. You should also know that if a column is all zeros, then the determinant is zero. All that remains is to convince yourself that if the columns are linearly dependent, then you can make one column all zeros from a sequence of elementary column operators.
A: Let your matrix be $A$ and let the columns of your matrix be $c_1,c_2,\cdots,c_n$. Saying there's a subset of linearly dependent columns means that there are columns $c_{i_1},c_{i_2},...,c_{i_k}$ that are linearly dependent, so that there exist nontrivial $a_k$ such that:
$$a_1c_{i_1}+a_2c_{i_2}+\cdots+a_kc_{i_k}=0.$$
Now define the vector $v$ so that $v_{i_k}=a_k$ and $v_i=0$ otherwise. What do you suppose $Av$ equals?
A: Hint: If $A$ has linearly dependent columns, then $\det(A)=0$.

Another approach without using the determinant: let $A\in \mathbb R^{n\times n}$ (or any other field $\mathbb F$). We write $v_1,\dots v_n$ for the columns of $A$. As the columns are linearly dependent there exists $a_1,\dots,a_n\in \mathbb R$ and $a_i\neq 0$ for at least one $i\in \{1,\dots n\}$ such that $a_1v_1+\dots +a_nv_n=0$. Let $\varphi$ be the linear mapping associated with $A$, then: $$\varphi\left(\begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix}\right)=A\cdot \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} = a_1v_1+\dots +a_nv_n=0.$$
Thus $\varphi$ is not bijective and we conclude that $A$ is not invertible.
