# Eigenvector of A to given Eigenvalue which requires row swapping to get reduced echelon form

Given the Matrix $$A = \left(\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -3 & -3 \end{matrix}\right)$$

calculate the eigenvalues and the corresponding eigenvectors of it.

$$P(\lambda) = det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynome with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$

So there is only one eigenvalue $-1$

Now, in order to get the corresponding eigenvector, I only need to solve the equation of the form $$Ax=b$$ obviously with the Gauss elimination.

The problem I face though is that implies swapping rows 1 and 3, and then 3 and 2.

1. How to solve this equation, thus getting the eigenvector for the eigenvalue $-1$?
2. Why not not swap them at all? Isn't there any way of solving it without swapping?

# Update

Oh, I think I've missed the fact that I don't use the matrix $A$, but $(A-(-1)I)$ so: $(A+I)=0$

But if this is true, then I have a question:

• is "the last row" always 0 when calculating the eigenvectors?

# Update 2

I've got the final answer $$E_{-1}=\left(\begin{matrix}1 \\ -1 \\ 1 \end{matrix}\right)$$

but answers to the questions left open are still welcome, as well as corrections (I'm not gifted in mathematical formalism).

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If one of the rows in the row reduction process did not become all zeros, then that would mean $(A+I)x=0$ has a unique solution. But $x=0$ obviously is a solution, so this would imply that there is no eigenvector, a contradiction. So, yeah, finding an eigenvector means that you need to find a non-trivial solution to a homogeneous system. So that special case of the algorithm for solving a linear system is a must. –  Jyrki Lahtonen Jan 6 '13 at 22:47

Given:

$$A = \left(\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -3 & -3 \end{matrix}\right)$$

We arrive at:

$$P(\lambda) = det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynomial with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$

Solving for the eigenvectors, we arrive at::

$$\lambda_1 = -1, v_1 = (1, -1, 1)$$

$$\lambda_2 = -1, v_2 = (2, -1, 0)$$

$$\lambda_2 = -1, v_3 = (3,-1, 0)$$

From these eigenvalues and eigenvectors, the Jordan Normal Form can be written as:

$$A = P J P^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 1 & 0 & 0\end{bmatrix} \cdot \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ -1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$

Regards

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Nice work! All this work, and no prior thumbs up! Well, that's history, now. ;-) –  amWhy Apr 26 '13 at 1:28
Your welcome...I feel the same...and have deleted many posts...feeling they must not be/have been helpful. But if this is typical of your work (and your work IS helpful, period), then it has, and will likely help others, even if only to serve as an example for a similar problem. –  amWhy Apr 26 '13 at 1:34
@amWhy: you are correct, and I do think twice about as I have seen old answers get positive feedback at later times. I just hate clutter! :-) So, I am much more careful before I remove answers. –  Amzoti Apr 26 '13 at 1:37