How to find eigenvectors and choosing free variable

Question

I have this matrix: $$ \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ and I have to find the eigenvectors of the upper matrix.

As usual, I found the eigenvalues: x$_{1}$= 2, x$_{2}$= 1, x$_{3}$= 0. The associated eigenvectors are: $$ \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix} $$

Due to an exercise I found, I have to tell if the vector: $$ \begin{bmatrix} 2 \\ -2 \\ 0 \\ \end{bmatrix} $$ is also an eigenvector.

To find the third eigenvector I used y (in x, y and z) as the free variable. Does it change if I use the x as free variable in order to invert the first and the second coordinates of the vector in question? So I could find an eigenvector which is a multiple of the last vector I wrote, which is also an eigenvector. Is it correct or the last vector can't be considered an eigenvector?

Answer 1

The way you find the eigenvectors is irrelevant here. Since $\begin{bmatrix}-1&1&0\end{bmatrix}^T$ is an eigenvector, then automatically $\begin{bmatrix}2&-2&0\end{bmatrix}^T$ is also an eigenvector, since$$\begin{bmatrix}2\\-2\\0\end{bmatrix} =(-2)\times\begin{bmatrix}-1\\1\\0\end{bmatrix}.$$

Answer 2

Every nonzero scalar multiple of an eigenvector is also an eigenvector: If $Av=\lambda v$, then $A(cv)=c(Av)=c(\lambda v)=\lambda(cv).$

However, you can determine whether or not $[2,-2,0]^T$ is an eigenvector of $A$ without doing any of that other work. Simply multiply it by $A$ and see what you get: $$\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}2\\-2\\0\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} = 0\begin{bmatrix}2\\-2\\0\end{bmatrix}$$ so by definition, this is an eigenvector of $A$ with eigenvalue $0$.