Can I have O as eigen vector?

[Abhishekdas]

This is my question:
[FONT=&quot]Find Eigen Values and Vectors for the matrix
[/FONT][FONT=&quot] [/FONT]




Is this correct? Can I have O as my eigen vector? This means that my origin does not get rotated of its axis after the transformation of the plane, right?

 

[Steven G]

This is my question:
Find Eigen Values and Vectors for the matrix
View attachment 10674

View attachment 10672

View attachment 10673
Is this correct? Can I have O as my eigen vector? This means that my origin does not get rotated of its axis after the transformation of the plane, right?

No, the O vector can NOT be an eigen vector

 

[HallsofIvy]

If, when solving $\displaystyle Av= \lambda v$, you get only $\displaystyle v= 0$ then $\displaystyle \lambda$ is NOT an eigenvalue!

You have, as characteristic equation, $\displaystyle \lambda^3+ 3\lambda^2+ 3\lambda+ 1= (\lambda+ 1)^3= 0$ so the only eigenvalue is $\displaystyle \lambda= -1$. When the characteristic equation has a "triple root" the space of eigenvector can have dimension, depending on the precise values in the matrix, one, two, or three.

Here, with eigenvalue $\displaystyle \lambda= 3$, you have the three equations
$\displaystyle 7x_1- 6x_2+ 5x_3= 0$
$\displaystyle 14x_1- 12x_2+ 10x_3= 0$
$\displaystyle 7x_1- 6x_2+ 5x_3= 0$

Obviously the first and third equations are the same and the second equation is just twice the other 2. You really have just one equation. And far from having only (0, 0, 0) as solution, we have a two dimensional space of solutions!

We can solve any one of the equations for one of the unknowns in terms of the other two. For example, $\displaystyle x_1= \frac{6}{7}x_2- \frac{5}{7}x_3$. In particular, taking $\displaystyle x_2= 7$, $\displaystyle x_3= 0$ we get the eigenvector $\displaystyle (6, 7, 0)$ and, taking $\displaystyle x_2= 0$,$\displaystyle x_3= 7$ we get the independent eigenvector $\displaystyle (-5, 0 , 7)$. The space of eigenvectors is the span of those two vectors so any eigenvector, corresponding to eigenvalue -1 is of the form $\displaystyle a(6, 7, 0)+ b(-5, 0, 7)= (6a- 5b, 7a, 7b) for any numbers a and b.$

 

[Abhishekdas]

Thanks a lot

SO, we get two eigen vectors from your process. Can we find the third vector somehow? Thanks for the help! I really appreciate it

 

[Steven G]

SO, we get two eigen vectors from your process. Can we find the third vector somehow? Thanks for the help! I really appreciate it

No, you can't find a 3rd (independent) vector