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Q48E

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Found in: Page 393

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Let ${\mathbit{U}}{\mathbf{⩾}}{\mathbf{0}}$ be a real upper triangular ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix with zeros on the diagonal. Show that ${\left({l}_{n}+U\right)}^{{\mathbf{t}}}{\mathbf{⩽}}{{\mathbit{t}}}^{{\mathbf{n}}}\left({l}_{n}+U+{U}^{2}+.....+{U}^{n-1}\right)$for all positive integers t. See Exercises 46 and 47.

${\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}\right)}^{\mathrm{t}}{\mathrm{leqt}}^{\mathrm{n}}\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}+{\mathrm{U}}^{2}+.....+{\mathrm{U}}^{\mathrm{n}-1}\right)\mathrm{for}\mathrm{all}\mathrm{positive}\mathrm{integers}\mathrm{t}$

See the step by step solution

## Step 1: Define Upper Triangular Matrix:

A triangular matrix with all components equal to below the main diagonal is called an upper triangular matrix. It's an element-based square matrix

## Step 2: Upper triangular matrix with zeros on the diagonal:

Consider as $U\le 0$ be a real upper triangular matrix $n×n$ with zeros on the diagonal. Therefore is U a nilpotent

${U}^{n}=0$

Now consider${\left({l}_{n}+U\right)}^{t}$ for $t\le n-1$ as represented below:

${\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}\right)}^{\mathrm{t}}={l}_{n}+\sum _{k=1}^{t}\left(\begin{array}{c}t\\ k\end{array}\right){U}^{k}\phantom{\rule{0ex}{0ex}}{\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}\right)}^{\mathrm{t}}={l}_{n}+\left(\begin{array}{c}t\\ 1\end{array}\right)U+\left(\begin{array}{c}t\\ 2\end{array}\right){U}^{2}+....+\left(\begin{array}{c}t\\ n-1\end{array}\right){U}^{n-1}\phantom{\rule{0ex}{0ex}}\left(\begin{array}{c}l\\ k\end{array}\right)\le {t}^{k},fork=0,1,.....,n-1\phantom{\rule{0ex}{0ex}}{l}_{n}+\left(\begin{array}{c}t\\ 1\end{array}\right)U+\left(\begin{array}{c}t\\ 2\end{array}\right){U}^{2}+....+\left(\begin{array}{c}t\\ n-1\end{array}\right){U}^{n-1}\le {t}^{n}\left({l}_{n}+U+{U}^{2}+....+{U}^{n-1}\right)$

substituting (1) in the above inequality we get as represented below:

${\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}\right)}^{\mathrm{t}}\le {\mathrm{t}}^{\mathrm{n}}\left({\mathrm{l}}_{\mathrm{n}}+\mathrm{U}+{\mathrm{U}}^{2}+.....+{\mathrm{U}}^{\mathrm{n}-1}\right)$