- #1

rugerts

- 153

- 11

- Homework Statement:
- Find eigenvalues & eigenvectors

- Relevant Equations:
- det(A-r*I) = 0

Thanks for your time.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rugerts
- Start date

- #1

rugerts

- 153

- 11

- Homework Statement:
- Find eigenvalues & eigenvectors

- Relevant Equations:
- det(A-r*I) = 0

Thanks for your time.

- #2

RPinPA

Science Advisor

Homework Helper

- 587

- 329

Multiply your ##\vec v## by the constant (1 + i). As you know, any multiple of an eigenvector is still an eigenvector of the same eigenvalue.

$$(1 + i) {2 \choose {2 - 2i}} = {{2 + 2i} \choose {2(1 - i)(1+i)}} = {{2+2i} \choose 4}$$

and similarly multiply ##\vec w## by the constant (1 - i).

$$(1 - i) {2 \choose {2 + 2i}} = {{2 - 2i} \choose {2(1 + i)(1-i)}} = {{2-2i} \choose 4}$$

Since the order is reversed, so are the eigenvalues with which each is associated.

- #3

Mark44

Mentor

- 36,338

- 8,295

- #4

rugerts

- 153

- 11

Ahh, I see. Thank you

Multiply your ##\vec v## by the constant (1 + i). As you know, any multiple of an eigenvector is still an eigenvector of the same eigenvalue.

$$(1 + i) {2 \choose {2 - 2i}} = {{2 + 2i} \choose {2(1 - i)(1+i)}} = {{2+2i} \choose 4}$$

and similarly multiply ##\vec w## by the constant (1 - i).

$$(1 - i) {2 \choose {2 + 2i}} = {{2 - 2i} \choose {2(1 + i)(1-i)}} = {{2-2i} \choose 4}$$

Since the order is reversed, so are the eigenvalues with which each is associated.

Share:

- Replies
- 11

- Views
- 272

- Last Post

- Replies
- 19

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 419

- Replies
- 1

- Views
- 993

- Replies
- 8

- Views
- 819

- Replies
- 41

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 723

- Last Post

- Replies
- 7

- Views
- 897

- Replies
- 20

- Views
- 557