# Miscalculated Eigen Vectors

### #1

Posted 26 November 2007 - 01:55 AM

68.75 0 0

0 68.75 0

0 0 0

I executed the eigVc command and I obtained

1 0.707 0

0 -0707 0

0 0 1

I almost died when i saw this!!!

How can I fix this???

PS: If u dont realize...the first column is not ortogonal with 2nd column....its suppused that i should get the ijk vector system....

### #2

Posted 28 November 2007 - 02:58 AM

!!!!

### #3

Posted 01 December 2007 - 10:57 AM

0 1 0

1 0 0

0 0 1

which is normal because your matrix is already diagonal.

### #4

Posted 02 December 2007 - 12:39 PM

what can i do?

### #5 Guest_Guest_*

Posted 01 April 2008 - 01:20 AM

### #6

Posted 01 April 2008 - 09:35 AM

### #7

Posted 01 April 2008 - 08:26 PM

I surprised a lot qhe i put the following matrix

68.75 0 0

0 68.75 0

0 0 0

I executed the eigVc command and I obtained

1 0.707 0

0 -0707 0

0 0 1

I almost died when i saw this!!!

How can I fix this???

PS: If u dont realize...the first column is not ortogonal with 2nd column....its suppused that i should get the ijk vector system....

**aacs88:**

Definition: let

**A**be a square matrix and k is the eigenvalue of

**A**if there exist nonzero vector

**P**such that

**AP=**k

**P .**every nonzero vector

**P**is called eigenvector of

**A**associated with the eigenvalue k. this implies that both side are parallel vectors.

According to this definition your answer is correct. Since your eigenvalue are

k1 =68.75

k2=68.75

k3=0

then

|68.75 0 0||1| |1|

|0 68.75 0||0|= k1 |0|

|0 0 0||0| |0|

also

|68.75 0 0||0.707 | |0.707 |

|0 68.75 0||-0.707 |=k2 |-0.707|

|0 0 0| | 0 | | 0 |

And

|68.75 0 0||0| |0|

|0 68.75 0||0|= k3 |0|

|0 0 0||1| |1|

Notice that these eigenvectors are not unique i.e. there are infinitely many possible solutions as long as they satisfy

**AP=**k

**P.**

|0 1 0|

|1 0 0| is also a possible solution

|0 0 1|

The eigenvectors are not necessarily orthogonal, for your case you have a repeated eigenvalues in the diagonal matrix.

On my classpad it gives

0 1 0

1 0 0

0 0 1

which is normal because your matrix is already diagonal.

**Orwell:**

I've tried to solve this problem with my CP it gave me the answer

1 0.707 0

0 -0.707 0

0 0 0

I'm running OS v3.03, what method did you use to calculate the eigenvectors ?

Inform me please.

### #8 Guest_Zwei9_*

Posted 10 April 2008 - 11:41 PM

1 1 0

0 0 0

0 0 1

for the EigenValues

68.75 68.75 0

Just my two cents

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