Proof Finishing for Upper Triangular Matrix has to have nonzero diagonals for invertibility

Question

I'm in the process of writing the proof that an upper triangular matrix is only invertible if it's diagonal are all non-zero values.

Given the matrix $X \in R^{n \times n}$ = \begin{bmatrix} x_{11} & x_{12} & \dots & x_{1n} \\ 0 & x_{22} & \dots & x_{2n} \\ \vdots & & \ddots & \vdots \\ 0 & \dotsi & 0 & x_{nn} \end{bmatrix}$

My proof start..

Assume there exists $i \in \{1,\dots,n\}$ such that $x_{ii} = 0$, then we define $i^* = min \{j \mid x_{jj} = 0\}$.

Let $c_1,\dots,c_n$ be the columns of $X$. We will show that $c_{i^*} \in Span(c_{1},\dots,c_{i-1}).$

How do I show that the columns before $c_{i^*}$ are linearly independent but the addition of $c_{i^*}$ makes it linearly dependent and therefore the rank is at most $n-1$. Therefore proving it cannot be invertible if there is a 0 on the diagonal?

Answer 1

Let's consider only the left columns of the matrix before column $i^*$,and show they are linearly independent. Consider the $n \times (i^* -1)$ matrix with these as columns, $$ A = \begin{bmatrix} c_1 & ... & c_{i^* - 1} \end{bmatrix} $$ And note that the reduced row echelon form has $x_{ii}=1$ and zeros everywhere else. namely, these vectors are linearly independent. In particular, these vectors span the space of vectors $$ \left\{ v \in \mathbb{R}^{n \times 1 }:v =\begin{bmatrix} v_1 \\ \vdots \\ v_{i^* - 1} \\ 0 \\ \vdots \\ 0 \end{bmatrix} \right\} $$

Now it easily follows that the vector $$c_{i^*}=\begin{bmatrix} x_{1, i^*} \\ \vdots \\ x_{i^* - 1, i^*} \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$ is a linear combination of the previous ones

Answer 2

Here is a shorter proof, but it makes use of determinants. The matrix $X$ is invertible if and only if its determinant is not equal to $0$. Since $$\text{det}(X) = x_{1,1}x_{2,2} \dotsm x_{n,n}\ ,$$ $\text{det}(X) = 0$ if, and only if, one of the diagonal entries of $X$ is null.