1
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ I think you have a typo. It should read “for each $1 \le k \le n$, we have $T(v_k) \in \text{span}(v_1, \dots, v_{\mathbf{k}})$” $\endgroup$ Jun 2, 2021 at 14:53
  • $\begingroup$ Oh, actually that would make sense and then contradict my argument below. Thanks! $\endgroup$
    – ENV
    Jun 2, 2021 at 14:58
  • 1
    $\begingroup$ 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 $[T]_\alpha^\alpha$? $\endgroup$ Jun 2, 2021 at 15:00
  • 1
    $\begingroup$ 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.... $\endgroup$
    – ENV
    Jun 2, 2021 at 15:02
  • 1
    $\begingroup$ Thanks very much, @SamFreedman. This is all clear now, I'll close the question. $\endgroup$
    – ENV
    Jun 2, 2021 at 15:08

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .