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Linear Algebra

Exercise 5.2.14

Answers

1.

Let

v = (x, y)

and

v^{'} = (x^{'}, y^{'})

and

A = (\begin{matrix} 1 & 1 \\ 3 & - 1 \end{matrix})

. We may write the system of equation as

Av = v^{'}

. We may also diagonalize the matrix

A

by

Q^{- 1} AQ = D

with

D = (\begin{matrix} - 2 & 0 \\ 0 & 2 \end{matrix})

and

Q = (\begin{matrix} 1 & 1 \\ - 3 & 1 \end{matrix})

. This means

D (Q^{- 1} v) = {(Q^{- 1} v)}^{'}

. So we know that

Q^{- 1} v = (\begin{matrix} c_{1} e^{- 2 t} \\ c_{2} e^{2 t} \end{matrix})

and

v = Q (\begin{matrix} c_{1} e^{- 2 t} \\ c_{2} e^{2 t} \end{matrix}),

where $c_{i}$ is some scalar for all $i$ .

2.

Calculate

D = (\begin{matrix} 3 & 0 \\ 0 & - 2 \end{matrix})

and

Q = (\begin{matrix} 2 & 1 \\ - 1 & - 1 \end{matrix})

. So we have

v = Q (\begin{matrix} c_{1} e^{3 t} \\ c_{2} e^{- 2 t} \end{matrix}),

where $c_{i}$ is some scalar for all $i$ .

3.

Calculate

D = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix})

and

Q = (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix})

. So we have

v = Q (\begin{matrix} c_{1} e^{t} \\ c_{2} e^{t} \\ c_{3} e^{2 t} \end{matrix}),

where $c_{i}$ is some scalar for all $i$ .

jephianlin

2011-06-27 00:00

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