Use Sage to do your computations for you.
See Appendix B.
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For each of the following matrices, determine a transition matrix \(P\) so that \(P^{-1} A P\) is in Jordan normal form, and determine the form matrix \(P^{-1} A P\) without computing \(P^{-1}\).
- \(A = \begin{bmatrix} -5 & 3 & 1 \\ -4 & 2 & 1 \\ -4 & 3 & 0 \end{bmatrix}\)
- \(A = \begin{bmatrix} -3 & 6 & 3 & 2 \\ -2 & 3 & 2 & 2 \\ -1 & 3 & 0 & 1 \\ -1 & 1 & 2 & 0 \end{bmatrix}\)
NOTE:
These are the same matrices as in
You may wish to use your results from the practise problems for triangular block form
as the starting point for this practise problem,
as in Example 34.4.2 in the textbook.
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Demonstrate that the following two matrices have the same eigenvalues with the same algebraic multiplicities,
but are not similar.
Do this by determining some property of the Jordan normal form matrix for each that differs between the two.
\[\begin{align*}
A & = \left[\begin{array}{rrr} 1 & -2 & -2 \\ -1 & 0 & -2 \\ 1 & 2 & 4 \end{array}\right]
&
B & = \left[\begin{array}{rrr} 3 & 0 & 2 \\ -1 & 0 & -2 \\ 0 & 1 & 2 \end{array}\right] &
\end{align*}\]
-
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Verify that \(P^{-1} J' P = J\), where
\[\begin{align*}
J & = \left[\begin{smallmatrix} \lambda \\ 1 & \lambda \\ & 1 & \lambda \\ & & \ddots & \ddots \\ & & & 1 & \lambda \end{smallmatrix}\right] \text{,}
&
J' & = \left[\begin{smallmatrix} \lambda \\ \epsilon & \lambda \\ & \epsilon & \lambda \\ & & \ddots & \ddots \\ & & & \epsilon & \lambda \end{smallmatrix}\right] \text{,}
&
P & = \left[\begin{smallmatrix} 1 \\ & \epsilon \\ & & \epsilon^{2} \\ & & & \ddots \\ & & & & \epsilon^{(n-1)} \end{smallmatrix}\right] \text{.}
\end{align*}\]
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Use part a to explain what is meant by the following statement.
A matrix in Jordan normal form is similar to a matrix that is "arbitrarily close" to being diagonal.
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Suppose \(\epsilon \approx 0\) but \(\epsilon \ne 0\).
While the matrix \(J\) below is in Jordan normal form,
the matrix \(A\) is not,
even though it might be reasonable to consider \(A \approx J\).
\[\begin{align*}
J &= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
&
A &= \begin{bmatrix} 1 + \epsilon & 0 \\ 1 & 1 \end{bmatrix}
\end{align*}\]
Determine the Jordan normal form of \(A\),
and conclude that \(A\) cannot be similar to \(J\).
Remark:
This exercise illustrates that the Jordan normal form can be sensitive to small perturbations in its entries.
This is particularly of concern when performing matrix calculations using approximate data —
for example, data obtained from a physical experiment.
Answers