Let $B = v_1, . . . , v_n$ be an ordered basis for $V$ and $T : V → V$ be a linear transformation. Prove that $[T]_B$ is upper triangular if and only if $T(v_j ) \in span(v_1, . . . , v_j)$
for all $j = 1, . . . , n$.
My attempt:
Suppose that $[T]_B$ is upper-triangular. Then,
$$[T(v_j)]_B = \begin{pmatrix}
A_{1j}\\
.\\
.\\
.\\
A_{nj}
\end{pmatrix}$$
$$\implies T(v_j) = \sum_{i = 1}^{n} A_{ij}v_i$$
Since $A_{ij} = 0$ for $i>j$, it follows that $T(v_j ) \in span(v_1, . . . , v_j)$.
So, $[T]_B$ is upper-triangular $\implies T(v_j ) \in span(v_1, . . . , v_j)$.
Now, suppose $T(v_j ) \in span(v_1, . . . , v_j)$, then,
$$T(v_j) = c_1v_1 + … + c_jv_j$$ for scalars $c_1,…,c_n \in F$
[FROM HERE, I HAVE NO IDEA WHAT TO DO]
So, can anyone explain or show what is meant to be done. I can't figure out how the matrix is formed. Also, I'm not sure if the former part of the proof is correct. Any help is much appreciated.
Best Answer
If $T(v_j)\in\operatorname{span}\{v_1,v_2,\ldots,v_j\}$, then $T(v_j)$ can be written as $a_{1j}v_1+a_{2j}v_2+\cdots+a_{jj}v_j$. But then$$[T]_B=\begin{bmatrix}a_{11}&a_{12}&a_{13}&\ldots&a_{1n}\\0&a_{22}&a_{23}&\ldots&a_{2n}\\0&0&a_{33}&\ldots&a_{3n}\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a_{nn}\end{bmatrix},$$which is an upper triangular matrix.