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Q50E

Expert-verifiedFound in: Page 234

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**a. ****Find all n **

**b. ****Show that the QR factorization of an invertible n **

*n*×*n*matrices that are both orthogonal and upper triangular, with positive diagonal entries are

A = ${I}_{n}$

- ${Q}_{1}={Q}_{2}$ and localid="1659498166202" ${R}_{1}={R}_{2}$

If *A* is an orthogonal and upper triangular matrix, then ${A}^{-1}$ is lower and upper triangular because ${A}^{-1}={A}^{T}$ and it is an inverse of an upper triangular matrix. Thus,

${A}^{T}={A}^{-1}$

= *A*

Because A is orthogonal and diagonal with positive entries must be. So, $A={I}_{n}$.

Consider the theorem below.

**Products and inverses of orthogonal matrices.**

**a. ****The product AB of two orthogonal n **

**b. ****The inverse ${{\mathit{A}}}^{\mathbf{-}\mathbf{1}}$ of an orthogonal n **

Thus, according to the above theorem ${{Q}_{2}}^{-1}{Q}_{1}$ is orthogonal and ${R}_{2}{{R}_{1}}^{-1}$ is upper triangular with positive diagonal entries.

From the part (a) it has

${{Q}_{2}}^{-1}{Q}_{2}={R}_{1}={R}_{2}$

$={I}_{m}$

So that, ${Q}_{1}={Q}_{2}\mathrm{and}{R}_{1}={R}_{2}$ as claimed.

Hence,

*n*×*n*matrices that are both orthogonal and upper triangular, with positive diagonal entries will be

$A={I}_{n}$

- ${Q}_{1}={Q}_{2}\mathrm{and}{R}_{1}={R}_{2}$ and both are orthogonal and upper triangular, with positive diagonal entries.

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