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Show that the rank of an upper triangular matrix is at least as large as the number of non-zero main diagonal entries.

What I do not understand with this statement is how can one have a triangular matrix with more linearly independent vectors than non-zero main diagonal entries. If $T = [t_1 \quad t_2 \quad \dots t_n]$ is upper triangular ($t_i$ being the column vectors), cannot $t_j$ always be expressed as linear combination of the set $\{t_1, \dots, t_{j-1} \}$ if the diagonal element $t_{jj} = 0$? In that case the rank should be equal to the number of non-zero diagonal entries. Is there any counter example?

Is it even meaningful to consider non-square matrices?

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"What I do not understand with this statement is how can one have a triangular matrix with more linearly independent vectors than non-zero main diagonal entries."
Take an upper-triangular square matrix where all diagonal entries are zero, i.e., a strictly upper-triangular matrix, different from zero. It's rank will be bigger than zero, the number of non-zero diagonal elements. Explicitly, consider $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$

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  • $\begingroup$ yes of course, thanks for the obvious reminder :) $\endgroup$ – hakanc Apr 18 '16 at 12:56

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