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Mathematical Methods for Physicists: A concise introduction - Site Map

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MATRIX ALGEBRA<br />

Since<br />

0 1<br />

1<br />

B<br />

@ 1<br />

0<br />

C<br />

A<br />

and<br />

0 1<br />

0<br />

B C<br />

@ 0 A<br />

1<br />

are linearly independent, they are the eigenvectors corresponding to ˆ 5.<br />

For ˆ 1, we have<br />

0<br />

10<br />

1 0 1 0<br />

1 0 1<br />

2 2 0 x 1 0<br />

2x 1 ‡ 2x 2 ‡ 0x 3 0<br />

B<br />

CB<br />

@ 2 2 0 A@<br />

x<br />

C B C B<br />

2 A ˆ @ 0 A or 2x 1 2x 2 ‡ 0x<br />

C B C<br />

@<br />

3 A ˆ @ 0 A:<br />

0 0 4 x 3 0<br />

0x 1 ‡ 0x 2 4x 3 0<br />

Solving this system yields<br />

x 1 ˆ t; x 2 ˆ t; x 3 ˆ 0;<br />

where t is arbitrary. Thus the eigenvectors corresponding to ˆ 1 are non-zero<br />

vectors of the <strong>for</strong>m<br />

0 1 0 1<br />

t 1<br />

B C B C<br />

~X ˆ @ t A ˆ t@<br />

1 A:<br />

0 0<br />

It is easy to check that the three eigenvectors<br />

0 1 0 1 0 1<br />

1<br />

0<br />

1<br />

B C B C B C<br />

~X 1 ˆ @ 1 A; ~X 2 ˆ @ 0 A; ~X 3 ˆ @ 1 A;<br />

0<br />

1<br />

0<br />

are linearly independent. We now <strong>for</strong>m the matrix ~S that has ~X 1 , ~X 2 ,and ~X 3 as its<br />

column vectors:<br />

0 1<br />

1 0 1<br />

B C<br />

~S ˆ @ 1 0 1A:<br />

0 1 0<br />

The matrix S ~ 1 A ~ S ~ is diagonal:<br />

0<br />

10<br />

10<br />

1 0 1<br />

1=2 1=2 0 3 2 0 1 0 1 5 0 0<br />

~S 1 A ~ S ~ B ˆ @ 0 0 1CB<br />

CB<br />

C B C<br />

A@<br />

2 3 0A@<br />

1 0 1A ˆ @ 0 5 0A:<br />

1=2 1=2 0 0 0 5 0 1 0 0 0 1<br />

There is no preferred order <strong>for</strong> the columns of ~S. If had we written<br />

0<br />

1 1<br />

1<br />

0<br />

B ~S ˆ @ 1 1<br />

C<br />

0A<br />

0 0 1<br />

132

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