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\( P = \begin{bmatrix} 1 & -4 & 3 \\ 0 & -4 & 4 \\ 0 & -4 & 0 \end{bmatrix} \)
\( P^{-1} A P = \begin{bmatrix} -1 & 0 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \)
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\( P = \begin{bmatrix} 1 & 1 & -1 & 10 \\ 0 & 0 & 1 & 4 \\ 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 2 \end{bmatrix} \)
\( P^{-1} A P = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} \)
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Both matrices have \(\lambda = 1\) with \(m_1 = 1\) and \(\lambda = 2\) with \(m_2 = 2\).
However,
by comparing \(\operatorname{rank} (2 I - A)\) and \(\operatorname{rank} (2 I - B)\) we see that they are similar to Jordan normal form matrices
\[\begin{align*}
J_A & = \begin{bmatrix} 1 \\ & 2 \\ & & 2 \end{bmatrix} \text{,}
&
J_B & = \begin{bmatrix} 1 \\ & 2 \\ & 1 & 2 \end{bmatrix} \text{,}
\end{align*}\]
respectively.
Since these form matrices are not similar,
neither can \(A\) and \(B\) be similar.
- [Omitted.]
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Since \(2 \times 2\) matrix \(A\) has two distinct eigenvalues,
it is actually diagonalizable,
and is similar to diagonal matrix \( \left[\begin{smallmatrix} 1 + \epsilon \\ & 1 \end{smallmatrix}\right] \)
via transition matrix \( P = \left[\begin{smallmatrix} \epsilon & 0 \\ 1 & 1 \end{smallmatrix}\right] \).
But \(J\) is already in Jordan normal form and is not diagonal,
hence \(A\) cannot be similar to \(J\).