Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 21E from Chapter 6.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 21E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We have to find an orthogonal $m \times m$ matrix $\underline { Q } _ { 1 }$ and an invertible $n \times n$ upper triangular matrix $R$ such that $A = Q _ { 1 } \left[ \begin{array} { l } { R } \\ { 0 } \end{array} \right]$ .
Step-1:
Let the columns of A are:\[Q = \left[ \begin{array} { l l l l } { \mathbf { q } _ { 1 } } & { \mathbf { q } _ { 2 } } & { \cdots } & { \mathbf { q } _ { n } } \end{array} \right]\]
Step-2:
Let $v _ { 1 }$ be the first vector in the standard basis of $\mathbb { R } ^ { m }$ that is not in $W _ { m } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } \right\}$
Let $\mathbf { u } _ { 1 } = \mathbf { v } _ { 1 } -$ proj $_ { v _ { r } } \mathbf { v } _ { 1 }$ and $\mathbf { q } _ { n + 1 } = \dfrac { \mathbf { u } _ { 1 } } { \left\| \mathbf { u } _ { 1 } \right\| }$
Then, $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } \right\}$ is the orthonormal basis for $W _ { n + 1 } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } \right\}$
Step-3:
Similarly, find $u_2$\[\mathbf { u } _ { 2 } = \mathbf { v } _ { 2 } - \operatorname { proj } _ { W _ { n + 1 } } \mathbf { v } _ { 2 } \text { and } \mathbf { q } _ { n + 2 } = \dfrac { \mathbf { u } _ { 2 } } { \left\| \mathbf { u } _ { 2 } \right\| }\]Then $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \mathbf { q } _ { n + 2 } \right\}$ is an orthogonal basis for $W _ { n + 1 } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \mathbf { q } _ { n + 2 } \right\}$So m-n such vectors are added, then $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \ldots \mathbf { q } _ { m } \right\}$ is an orthonormal basis for $\mathbb { R } ^ { m }$
Step-4:
Let $Q _ { o } = \left[ \begin{array} { l l l } { \mathbf { q } _ { \text {n+1 } } } & { \mathbf { q } _ { \text {n+2 } } } & { \ldots } & { \mathbf { q } _ { \text {m } } } \end{array} \right]$ and $Q _ { 1 } = \left[ \begin{array} { l l } { Q } & { Q _ { 0 } } \end{array} \right]$. So, by partitioned matrix multiplication,
\[Q _ { 1 } \left[ \begin{array} { l } { R } \\ { 0 } \end{array} \right] = Q R = A\]
We have to find an orthogonal $m \times m$ matrix $\underline { Q } _ { 1 }$ and an invertible $n \times n$ upper triangular matrix $R$ such that $A = Q _ { 1 } \left[ \begin{array} { l } { R } \\ { 0 } \end{array} \right]$ .
Step-1:
Let the columns of A are:\[Q = \left[ \begin{array} { l l l l } { \mathbf { q } _ { 1 } } & { \mathbf { q } _ { 2 } } & { \cdots } & { \mathbf { q } _ { n } } \end{array} \right]\]
Step-2:
Let $v _ { 1 }$ be the first vector in the standard basis of $\mathbb { R } ^ { m }$ that is not in $W _ { m } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } \right\}$
Let $\mathbf { u } _ { 1 } = \mathbf { v } _ { 1 } -$ proj $_ { v _ { r } } \mathbf { v } _ { 1 }$ and $\mathbf { q } _ { n + 1 } = \dfrac { \mathbf { u } _ { 1 } } { \left\| \mathbf { u } _ { 1 } \right\| }$
Then, $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } \right\}$ is the orthonormal basis for $W _ { n + 1 } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } \right\}$
Step-3:
Similarly, find $u_2$\[\mathbf { u } _ { 2 } = \mathbf { v } _ { 2 } - \operatorname { proj } _ { W _ { n + 1 } } \mathbf { v } _ { 2 } \text { and } \mathbf { q } _ { n + 2 } = \dfrac { \mathbf { u } _ { 2 } } { \left\| \mathbf { u } _ { 2 } \right\| }\]Then $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \mathbf { q } _ { n + 2 } \right\}$ is an orthogonal basis for $W _ { n + 1 } = \operatorname { Span } \left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \mathbf { q } _ { n + 2 } \right\}$So m-n such vectors are added, then $\left\{ \mathbf { q } _ { 1 } , \ldots , \mathbf { q } _ { n } , \mathbf { q } _ { n + 1 } , \ldots \mathbf { q } _ { m } \right\}$ is an orthonormal basis for $\mathbb { R } ^ { m }$
Step-4:
Let $Q _ { o } = \left[ \begin{array} { l l l } { \mathbf { q } _ { \text {n+1 } } } & { \mathbf { q } _ { \text {n+2 } } } & { \ldots } & { \mathbf { q } _ { \text {m } } } \end{array} \right]$ and $Q _ { 1 } = \left[ \begin{array} { l l } { Q } & { Q _ { 0 } } \end{array} \right]$. So, by partitioned matrix multiplication,
\[Q _ { 1 } \left[ \begin{array} { l } { R } \\ { 0 } \end{array} \right] = Q R = A\]