Prove a triangular matrix is invertable if and only if all diagonal values are non-zero. Let $A \in K^{n×n}$ be an (upper) triangular matrix,
so $A = (a_{ij})_{i,j = 1,...,n}$ and $a_{ij} = 0$ for $i > j$.
Then A is invertable exactly then, when all $a_{ii} \neq 0; i = 1, . . . , n$.
I have proven that If A is invertable then all the diagonal elements have to be non-zero like this:
Let A be an invertible Matrix and let one $x_{ii}=0$. Because A is invertible, it represents a bijective linear transformation, so the kernel of T is trivial, which implies linear independence of the columns of the matrix. 
Because A is an upper triangular matrix, its diagonal entries represent a linearly independent set of vectors.
But if one $x_{ii} = 0$, then the set of vectors is no longer linearly independent, which is a contradiction.
My problems are
a) i dont know how to prove the other direction
b) I am unsure as to wether this proof works.
Any help is appreciated :)
P.S.: since we didnt introduce the determinant yet, I am not allowed to just say that the diagonals have to be non-zero, so the determinant is non-zero, so it is invertible...
 A: If some $a_{ii}=0$ then the first $i$ columns are all in the subspace spanned by the first $i-1$ basis vector, hence are linearly dependant.

If all $a_{ii}$ are nonzero, write
$$ A=\begin{pmatrix}B&v\\0&a_{nn}\end{pmatrix}$$
where $B$ is an $(n-1)\times (n-1)$ upper triangular matrix with non-zero diagonal entries. For a proof by induction, we may assume that $B$ is invertible, i.e., $BC=CB=I_{n-1}$ for some $C$.
Observe that
$$\begin{pmatrix}B&v\\0&a_{nn}\end{pmatrix}\begin{pmatrix}C&w\\0&x\end{pmatrix} =\begin{pmatrix}I_{n-1}&Bw+xv\\0&xa_{nn}\end{pmatrix}.$$
Can you pick $x\in\Bbb R$ and $w\in\Bbb R^{n-1}$ to make this $I_n$?
A: You have the right idea, I think, but your proof leaves a lot to be desired.  "Its diagonal entries represent a linearly independent set of vectors," doesn't make sense.  The diagonal entries are numbers, not vectors.
You want to show that the columns form a basis if and only if the all the diagonal entries are nonzero.  Suppose all the diagonal entries are nonzero,  Then for any vector $Y$, we can solve the equation $AX=Y$, by back-substitution, as usual.  That is, the image of $A$, considered as a linear transformation is the whole space, so the columns form a basis.
Conversely, let $k$ be the smallest index such that $a_{kk}=0$.  Then then $k$th column is in the span of the first $k-1$ columns, and the columns aren't linearly independent.  (If $k=1$, then the first column is $0$ and we are done.)
