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Let $T$ be a triangular matrix where $t_{ii}\ne 0$ for all $i$. Show that the rows and the columns of $T$ are linearly independent. I think it is obvious from the structure of $T$. But I do not know how to prove.

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  • $\begingroup$ Note that $T$ and its transpose are invertible. $\endgroup$ – Julien Jun 14 '13 at 15:14
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The argument is basically the same for upper / lower triangular matrices. For an upper triangular matrix, use the bottom-right entry to clear all the entries above it. Now repeat this process until your matrix is diagonal, and has no zero entries. Then manifestly the rows and columns are linearly independent.

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  • $\begingroup$ can you make it an answer? $\endgroup$ – 81235 Jun 14 '13 at 20:45
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Consider the lower triangular case and let's deal with the row case. If $\sum_{i=1}^{n} \lambda_{i} r_{i} = 0,$ and not all $\lambda_{i}$ are $0,$ then choose $j$ as large as possible so that $\lambda_{j} \neq 0.$ Then $\lambda_{j}r_{j} = -\sum_{i=1}^{j-1} \lambda_{i}r_{i}.$ The right hand expression has $(j,j)$-entry $0,$ but the left hand side doesn't, a contradiction. (Strictly speaking I should have also said that there is more than one non-zero $\lambda_{i}$, which there is).

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