I understand that Upper triangular matrices must have at least one eigenvector, but why does this mean that the basis of $[T]_B$ must contain an eigenvector for $[T]_B$ to be upper triangular?
2026-04-12 01:39:25.1775957965
Triangularisation of a linear transformation
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If $B=(v_1,v_2,\dots)$ is such that $[T]_B$ is triangular, then $v_1$ is an eigenvector. This follows from the definition of the matrix $[T]_B$.