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Linear Algebra

Exercise 6.5.27

Answers

1.

We may write

(\begin{matrix} x & y \end{matrix}) (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

Diagonalize the matrix and get

(\begin{matrix} x & y \end{matrix}) P^{∗} (\begin{matrix} 3 & 0 \\ 0 & - 1 \end{matrix}) P (\begin{matrix} x \\ y \end{matrix}),

where

P = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) .

So we have

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = P (\begin{matrix} x \\ y \end{matrix}) .

2.

Diagonalize

(\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})

and get

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

3.

Diagonalize

(\begin{matrix} 1 & - 6 \\ - 6 & - 4 \end{matrix})

and get

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = \frac{1}{\sqrt{13}} (\begin{matrix} 2 & 3 \\ 3 & - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

4.

Diagonalize

(\begin{matrix} 3 & 1 \\ 1 & 3 \end{matrix})

and get

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

5.

Diagonalize

(\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix})

and get

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

jephianlin

2011-06-27 00:00

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