For each of the following linear systems of differential equations, (i) find the general real solution (ii) show that the solutions are linearly independent (iii) draw the phase portrait
a. $$\dot x=\begin{bmatrix}1&3\\3&1\end{bmatrix}x$$
characteristic equation: $$(1-λ)^2-(3^2)=0$$ $$(λ-4)(λ+2)$$ with roots $λ_1=-2$ and $λ_2=4$
for the eigenvalue, $λ_1=-2$: $$A-λ_1I=A+2I$$ $$=\begin{bmatrix}3&3\\3&3\end{bmatrix}$$ $$=\begin{bmatrix}1&1\\0&0\end{bmatrix}$$
The problem is I forgot how to calculate eigenvectors and I can't see to find a good tutorial, could someone give me a wlakthrough please? the eigenvector $v^1=?$
the first solution is given by: $x^1(t)=e^{-2t}v^1$
for the eigenvalue, $λ_1=4$: $$A-λ_1I=A-4I$$ $$=\begin{bmatrix}-3&3\\3&-3\end{bmatrix}$$ $$=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$$
the eigen vector $v^2=?$
the second solution is given by: $x^2(t)=e^{4t}v^2$
I'll provide an example of computing an eigenvector for a different matrix. Consider the matrix below which has eigenvalues $\lambda_1=2$ and $\lambda_2=0$,
$$A = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right].$$
To find $v_1$ we write the equation $A v_1 = \lambda_1 v_1$. Here we represent $v_1$ as the column vector $(a,b)$,
$$ \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{c} a \\ b \end{array} \right] = 2 \left[ \begin{array}{c} a \\ b \end{array} \right], $$
this gives two equations. The goal is to solve for $a$ and $b$, however you must remember that there is no unique solution to this system since any multiple of an eigenvector is also an eigenvector. To solve this we only need to express $a$ in terms of $b$.
$$ a + b = 2a \qquad \textbf{(1)} $$ $$ a + b = 2b \qquad \textbf{(2)} $$
According to $\textbf{(1)}$ we should have $a=b$ which is consistent with equation $\textbf{(2)}$ (one of the equations is always unnecessary since they are not independent).
So our eigenvectors for the eigenvalue $\lambda_1=2$ are,
$$v_1 = \left[ \begin{array}{c} a \\ a \end{array} \right],$$
it is customary in many situations to choose to work with the normalized eigenvector. In that case the vector is,
$$v_1 = \frac{1}{\sqrt{2}}\left[ \begin{array}{c} 1 \\ 1 \end{array} \right].$$