In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:
$\textbf{Conditions for upper-triangular matrix:}$
Suppose $T \in \mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:
(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular
(b) $Tv_j \in span(v_1,...,v_j)$ for each $j=1,...,n$
(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$
In typical Sheldon Axler fashion, his proof begins as follows:
"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"
While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$\Rightarrow$(b) and (c)$\Rightarrow$(b)?
$a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)\in span(v_1,...,v_j)$.
$c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)\in span(v_1,...,v_j)$ since $v_j\in (v_1,...,v_j)$