Assume $n \in \mathbb{N}, A=(a_{ij})\in \mathbb{R}^{n \times n}, A$ is upper triangular matrix. Show that $A$ has full rank iff all numbers on the diagonal $\ne 0$.
I'm not sure how to show this as it seems pretty natural to me and feels like it's just definition, but there's more to it, I guess.
Can someone show me what I'm missing?
This can be done by induction. If $n=1$, then we are just asserting that a matrix $[a]$ has rank $1$ if and only if $a\neq0$, which is clear.
Now, assume that the statement holds for $(n-1)\times(n-1)$ matrices and let $A$ be a $n\times n$ matrix. If $\operatorname{rank}A=n$, then $a_{11}\neq0$, because otherwise the first entry of every row of $A$ would be $0$ and therefore $\operatorname{rank}A$ would be, at most, $n-1$. Let $A^\ast$ be the $(n-1)\times(n-1)$ matrix obtaned from $A$ removing from it the first row and the first column. Then, since $\operatorname{rank}A=n$ and since $a_{i1}=0$ is $i>1$, $\operatorname{rank}A^\ast=n-1$ and therefore, by the induction hypothesis, all entrieas from the main diagonal of $A^\ast$ are different from $0$.
And if $\operatorname{rank}A<n$ then either $a_{11}=0$, in which case it is proved that some entry from the main diagonal is equal to $0$, or $a_{11}\neq0$. In this later case, $\operatorname{rank}A^\ast<n-1$ and so, again, the induction hypothesis can be applied to prove that some entry of the main diagonal of $A^\ast$ is equal to $0$.