How to prove question about upper triangular matrix representation of linear transformation? I have the following linear algebra question: Let $V$ be an $n$-dimensional vector space, and let $\alpha = {v_1, . . . , v_n}$ be a basis of $V$. Suppose $T: V \rightarrow V$ is a linear transformation. Let $[T]_{\alpha}^{\alpha}$ be the matrix representation of $T$ under basis $\alpha$ such that $T(v_1, ..., v_n) = (v_1 ... v_n)[T]_{\alpha}^{\alpha}$
Show that $[T]_{\alpha}^{\alpha}$ is upper-triangular if and only if for each $1 \leq k \leq n$, we have $T(v_k) \in$ span($v_1, ..., v_k$).
I'm not sure I understand how to go about this question. I've started trying to prove if $T(v_k) \in$ span($v_1, ..., v_n$), then $[T]_{\alpha}^{\alpha}$ is upper-triangular, but that gives me the following:
Since $T(v_k) \in$ span($v_1, ..., v_k$), then $T(v_k)= \beta_1v_1 + \beta_2v_2 + ... + \beta_nv_n$, for $\beta_i \in \mathbb{R}$. But suppose that $T(v_1)= v_1 + v_2 + ... + v_n$ (here all $\beta_i = 1$). Then wouldn't the first column of $[T]_{\alpha}^{\alpha}$ contain all $1$s and hence wouldn't be an upper triangular matrix?
Please let me know if that reasoning makes sense and how I should approach this proof, if not this way?

There was originally a typo in my question. The argument should be as follows:
Since $T(v_k) \in$ span($v_1, ..., v_k$), then $T(v_k)= \beta_1v_1 + \beta_2v_2 + ... + \beta_nv_k$, for $\beta_i \in \mathbb{R}$. So consider $T(v_1) = a_1v_1$, $T(v_2) = b_1v_2 + b_2v_2$, $T(v_3) = c_1v_3 + c_2v_3 + cv_3$. Then the columns of   $[T]_{\alpha}^{\alpha}$, would necessarily be $a_1$ followed by all zeros. Then $b_1, b_2$ followed by all zeros, then $c_1v_3 + c_2v_3 + cv_3$ followed by all zeros, effectively creating the upper triangular matrix as required.
