Consider the matrix A = 1 1 P = Let x = 1 √2 1 √2 0 -|-~| 1 √6 1 Compute [x]. √6 2 √6 1/11/11/1 I 2 -1 -1 2 and D = " which can be orthogonally diagonalized A = PDPT, where 300 030 00 0 [] 3 . Let 3 be the basis for R³ given by the columns of P.
Consider the matrix A = 1 1 P = Let x = 1 √2 1 √2 0 -|-~| 1 √6 1 Compute [x]. √6 2 √6 1/11/11/1 I 2 -1 -1 2 and D = " which can be orthogonally diagonalized A = PDPT, where 300 030 00 0 [] 3 . Let 3 be the basis for R³ given by the columns of P.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Consider the matrix \( A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \), which can be orthogonally diagonalized \( A = P D P^T \), where
\[ P = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{3}} \\
0 & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{3}}
\end{bmatrix} \]
and
\[ D = \begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0
\end{bmatrix}. \]
Let \(\mathbf{x} = \begin{bmatrix} -4 \\ 3 \\ -1 \end{bmatrix}\). Let \(\mathcal{B}\) be the basis for \(\mathbb{R}^3\) given by the columns of \(P\).
Compute \([\mathbf{x}]_{\mathcal{B}}\).
\[ \begin{bmatrix} \\ \\ \end{bmatrix} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab289986-80a9-4865-81ee-c0ba3a526505%2Fc5e0e2e4-f806-4f9d-8745-e99db1ddae93%2Fjb7yfa_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the matrix \( A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \), which can be orthogonally diagonalized \( A = P D P^T \), where
\[ P = \begin{bmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{3}} \\
0 & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{3}}
\end{bmatrix} \]
and
\[ D = \begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0
\end{bmatrix}. \]
Let \(\mathbf{x} = \begin{bmatrix} -4 \\ 3 \\ -1 \end{bmatrix}\). Let \(\mathcal{B}\) be the basis for \(\mathbb{R}^3\) given by the columns of \(P\).
Compute \([\mathbf{x}]_{\mathcal{B}}\).
\[ \begin{bmatrix} \\ \\ \end{bmatrix} \]
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