Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 5 - Inner Product Spaces - 5.3 The Gram-Schmidt Process - Problems - Page 366: 22

Answer

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Work Step by Step

We are given: $A_1=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ A_2=\begin{bmatrix} 0 & 1\\ 1 & 1 \end{bmatrix}\\ A_3=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}$ According to Gram-Schmidt we have: $v_1=A_1=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ v_2=A_2-\frac{(A_2,v_1)}{||v_1||^2}v_1\\ =\begin{bmatrix} 0 & 1\\ 1 & 1 \end{bmatrix}-\frac{(5.0.0+2.1.1+3.1.1+5.1.0)}{5.0.0+2.1.1+3.1.1+5.0.0}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \\ =\begin{bmatrix} 0 & 1\\ 1 & 1 \end{bmatrix}-\frac{0+2+3+0}{0+2+3+0}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ =A_1=\begin{bmatrix} 0 & 1\\ 1 & 1 \end{bmatrix}-\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ =\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}\\ v_3=A_3-\frac{(A_3,v_1)}{||v_1||^2}v_1-\frac{(A,v_2)}{||v_2||^2}v_2\\ =\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}-\frac{5.1.0+2.1.1+3.1.1+5.0.0}{5.0.0+2.0.0+3.0.0+5.1.1}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}-\frac{5.1.0+2.1.0+3.1.0+5.0.1}{5.0.0+2.0.0+3.0.0+5.1.1}\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}\\ =\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}-\frac{0+2+3+0}{0+2+3+0}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ =\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}-\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\\ =\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}$ Hence, an orthogonal basic spanned by the given factors is: $\{\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix},\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\}$ Determine a subspace of $M_2(R)$ Assume $\alpha, \beta, \gamma,a,b,c,d \in R$ Obtain the system: $\alpha v_1+\beta v_2+\gamma v_3=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ $\alpha\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}+\beta \begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}+\gamma \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ $\begin{bmatrix} \gamma & \alpha\\ \alpha & \beta \end{bmatrix}=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\ \rightarrow a=c$ The subspace spanned by $\{A_1,A_2,A_3\}: \{\begin{bmatrix} a & b\\ c & d \end{bmatrix}: a,b,c \in R\}$
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