Why if the columns of a matrix are not linearly independent the matrix is not invertible? Why if the columns of a matrix are not linearly independent the matrix is not invertible?
I have watched this video about eigenvalues and eigenvectors by Sal from Khan Academy, where he says that for $\lambda$ to be an eigenvalue for the matrix $A$, the following must be true
$$A \cdot \vec{v} = \lambda \cdot \vec{v} \\
\vec{0} = \lambda \cdot \vec{v} - A \cdot \vec{v} \\
\vec{0} = (\lambda - A )\cdot \vec{v} \\
\vec{0} = (\lambda \cdot I - A )\cdot \vec{v}$$
and the determinant of $(\lambda \cdot I - A )$ must be $0$, or in other words $(\lambda \cdot I - A )$ is not invertible, or in other words the columns of $(\lambda \cdot I - A )$ are linearly dependent, or the nullspace of $(\lambda \cdot I - A )$ is non trivial.
Could someone explain me better these statements? What's the relation between a statement and the other? 
I understood some stuff, but some other clarifications might help too.
 A: Here's an answer that completely avoids determinants. Determinants are heinously overrated.
Let $\vec v_1,\vec v_2,\dotsc,\vec v_n$ be the columns of a matrix $A$. That is,
$$
A=
\begin{bmatrix}
\vec v_1 & \vec v_2 & \dotsb & \vec v_n
\end{bmatrix}
$$
Now, suppose the columns of $A$ are not linearly independent. Then there exist scalars $\lambda_1,\lambda_2,\dotsc,\lambda_n$ not all zero such that
$$
\lambda_1\vec v_1+\lambda_2\vec v_2+\dotsb+\lambda_n\vec v_n=\vec 0\tag{1}
$$
But (1) may be re-written in matrix form as
$$
\begin{bmatrix}
\vec v_1 & \vec v_2 & \dotsb & \vec v_n
\end{bmatrix}
\begin{bmatrix}
\lambda_1\\ \lambda_2 \\ \vdots\\ \lambda_n
\end{bmatrix}
=\vec 0
$$
Putting 
$$
\vec \lambda=
\begin{bmatrix}
\lambda_1\\ \lambda_2 \\ \vdots\\ \lambda_n
\end{bmatrix}
$$
then gives $A\vec\lambda=\vec 0$ where $\vec \lambda\neq\vec 0$. Hence $A$ has a nontrivial nullspace and is thus not invertible.
A: If a matrix has columns linearly dependent. Then determinant of transpose of the matrix will be zero.Which means determinant of the given matrix is also zero. A matrix is invertible only if its determinant is non-zero.
A: If $A-\lambda \cdot I$ is invertible, then the only one solution to the equation $(A-\lambda  \cdot I)  \cdot \vec{v} = \vec{0}$ is $\vec{v} = \vec{0}$. 
In fact, suppose $(A-\lambda  \cdot I)^{-1}$ exists, then if we multiply both sides by the inverse:
$$(A-\lambda  \cdot I)^{-1} \cdot (A-\lambda  \cdot I)  \cdot \vec{v} = (A-\lambda  \cdot I)^{-1} \cdot \vec{0}$$
$$I  \cdot \vec{v} = (A-\lambda  \cdot I)^{-1} \cdot \vec{0}$$
$$\vec{v} = (A-\lambda  \cdot I)^{-1} \cdot \vec{0}$$
$$\vec{v} = \vec{0}$$
But eigenvectors are non-zero, so one must require: $$\det(A-\lambda  \cdot I) = 0$$ i.e. the matrix $A-\lambda  \cdot  I$ is not invertible.
A: Suppose $A$ is a $n \times n$ invertible real matrix and $x \in \mathbb{R}^n$. The vector $A^{-1}x$ describes the unique representation of $x$ as a linear combination of the columns of $A$. Specifically, $x=\sum_{i=1}^n (A^{-1} x)_i a_i$. If the columns are not linearly independent, then this representation cannot be unique: given $\sum_{i=1}^n b_i a_i = 0$ (where at least one $b_i$ is not zero) and $\sum_{i=1}^n c_i a_i = x$, you will have $\sum_{i=1}^n (b_i+c_i) a_i = x$. So there will always be at least two representations, and so $A$ cannot be invertible.
In other words: 


*

*A $n \times n$ real matrix is invertible if and only if every $x \in \mathbb{R}^n$ has a unique representation as a linear combination of the columns of $A$

*This representation cannot be unique if zero can be nontrivially represented as a linear combination of the columns of $A$, i.e. if the columns of $A$ are not linearly independent.

