Help me find an eigenvector Good evening all! I'm trying to find the eigenvalues and eigenvectors of the following problem
$$
       \begin{bmatrix}
        -10 & 8\\
        -18 & 14\\
         \end{bmatrix}*\begin{bmatrix}
        x_{1}\\
        x_{2}\\
         \end{bmatrix}
$$
I've found that $λ_{1},_{2}=2$ where $λ$ is double eigenvalue of our matrix and now I'm trying to find its eigenvectors. So we have to solve the above system
$$(-10-λ)*x_{1}+8*x_{2}=0    ,    -18*x_{1}+(14-λ)*x_{2}=0$$
So what I do here is to replace $λ=2$ and we have
$$-12*x_{1}+8*x_{2}=0    ,    -18*x_{1}+12*x_{2}=0$$
From here I get that $3*x_{1}=2*x_{2}$ and I think that gives us the eigenvector $\begin{bmatrix}
        2/3\\
        1\\
         \end{bmatrix}$
My book is telling me that the eigenvector for $λ=2$ is $\begin{bmatrix}
        2\\
        3\\
         \end{bmatrix}$ but I can't find the same solution. Can someone help me?
EDIT : I added my own try at solving it.
 A: You and your book are both correct. If you take an eigenvector $v$ and multiply it by a scalar, you get another eigenvector. Multiply your result by $3$ and you have the book's result.
If you are still not sure, you can try multiplying your vector and the book's vector by the matrix. You will find that both are eigenvectors. The book just adjusted the values so that all the entries are integers.
A: Both $\begin{bmatrix}
        2\\
        3\\
         \end{bmatrix}$ and $\begin{bmatrix}
        2/3\\
        1\\
         \end{bmatrix}$ are valid eigenvectors for the eigenvalue $\lambda$. The reason is as follows.
For any eigenvalue $\lambda$ of a square matrix $A$, there exist an infinite number of eigenvectors. The eigenspace of $A$ for $\lambda$ is the set of all eigenvectors of $\lambda$ for $A$ with the addition of the zero vector. It is key to note that the eigenspace is a vector space.
Since vector spaces are closed under scalar multiplication, then the book's solution of $\begin{bmatrix}
        2\\
        3\\
         \end{bmatrix}$ is a valid solution since $\begin{bmatrix}
        2\\
        3\\
         \end{bmatrix} = 3\begin{bmatrix}
        2/3\\
        1\\
         \end{bmatrix}$, hence $\begin{bmatrix}
        2\\
        3\\
         \end{bmatrix}$ is an element of the eigenspace since you have found that $\begin{bmatrix}
        2/3\\
        1\\
         \end{bmatrix}$ is an eigenvector.
Another way to see this is by noting that the system you posted has infinite solutions. After finding that $3x_1 = 2x_2$, the substitution of $x_1$ for $2/3x_2$ into the second equation of the system results in $0=0$, implying infinite solutions for the system.
