Use Theorem 2 to diagonalize the matrix A= [latex]\begin{bmatrix} 1&0&0 \\6&-2&0 \\ 7&-4&2 \end{bmatrix}[/latex]

Since A is a triangular matrix, by Proposition 1 of Sec. 5.1, the eigenvalues of the matrix A are the diagonal entries [latex]\lambda _{1} = 1 \quad\lambda _{2} = −2 and \quad\lambda _{3} = 2[/latex] The corresponding eigenvectors, which are linearly independent, are given, respectively, by [latex]V_{1}= \begin{bmatrix} 1 \\2 \\ 1 \end{bmatrix} \quad V_{2}= \begin{bmatrix} 0 \\1 \\ 1 \end{bmatrix} \quad and \quad V_{3}= \begin{bmatrix} 0\\0\\ 1 \end{bmatrix} [/latex] Therefore, by Theorem 2, D = [latex]P^{-1}[/latex]AP, where D= [latex]\begin{bmatrix} 1&0&0 \\0&-2&0 \\ 0&0&2 \end{bmatrix}[/latex] and P= [latex]\begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1\end{bmatrix}[/latex] To verify that D =[latex]P^{-1}[/latex]AP, we can avoid finding [latex]P^{-1}[/latex] by showing that PD = AP In this case, PD = [latex]\begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\0&-2&0 \\ 0&0&2 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\2&-2&0 \\ 1&-2&2 \end{bmatrix}[/latex] = [latex]\begin{bmatrix} 1&0&0 \\6&-2&0 \\ 7&-4&2 \end{bmatrix} \begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1 \end{bmatrix}[/latex] = AP

Question 5.2.2: Use Theorem 2 to diagonalize the matrix A = [ 1 0 0 6 -2 0 7...

Introduction to Linear Algebra with Applications

Use Theorem 2 to diagonalize the matrix

A= $\begin{bmatrix} 1&0&0 \\6&-2&0 \\ 7&-4&2 \end{bmatrix}$

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Since A is a triangular matrix, by Proposition 1 of Sec. 5.1, the eigenvalues of the matrix A are the diagonal entries
$\lambda _{1} = 1 \quad\lambda _{2} = −2 and \quad\lambda _{3} = 2$
The corresponding eigenvectors, which are linearly independent, are given, respectively, by

V_{1}= \begin{bmatrix} 1 \\2 \\ 1 \end{bmatrix}    \quad V_{2}= \begin{bmatrix} 0 \\1 \\ 1 \end{bmatrix}  \quad  and  \quad V_{3}= \begin{bmatrix} 0\\0\\ 1 \end{bmatrix}

Therefore, by Theorem 2, D = $P^{-1}$ AP, where

D= $\begin{bmatrix} 1&0&0 \\0&-2&0 \\ 0&0&2 \end{bmatrix}$ and P= $\begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1\end{bmatrix}$

To verify that D = $P^{-1}$ AP, we can avoid finding $P^{-1}$ by showing that
PD = AP
In this case,

PD = $\begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\0&-2&0 \\ 0&0&2 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\2&-2&0 \\ 1&-2&2 \end{bmatrix}$

= $\begin{bmatrix} 1&0&0 \\6&-2&0 \\ 7&-4&2 \end{bmatrix} \begin{bmatrix} 1&0&0 \\2&1&0 \\ 1&1&1 \end{bmatrix}$

= AP