# 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?

• I think you have a typo. It should read “for each $1 \le k \le n$, we have $T(v_k) \in \text{span}(v_1, \dots, v_{\mathbf{k}})$” Jun 2, 2021 at 14:53
• Oh, actually that would make sense and then contradict my argument below. Thanks!
– ENV
Jun 2, 2021 at 14:58
• Your argument is very much on the right track. Since $T(v_k) = \sum_{i = 1}^k \beta_i v_i$, what’s the $k$th column of $[T]_\alpha^\alpha$? Jun 2, 2021 at 15:00
• Right, then for the $k^{th}$ column of $[T]^a_a$ we'd only have $k$ rows defined. So the remaining rows would be zero and hence we'd have an upper triangular matrix. It's the typo that messed me up....
– ENV
Jun 2, 2021 at 15:02
• Thanks very much, @SamFreedman. This is all clear now, I'll close the question.
– ENV
Jun 2, 2021 at 15:08

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.