$A=\begin{bmatrix}1&3\\3&1\\\end{bmatrix}\qquad $
To find the fundamental matrix of $y' = Ay$ with $y_0 =\begin{bmatrix}2\\4\end{bmatrix}$
I found the eigenvalues of $A$ as $-2,4$
The eigenvectors correspondingly are as $\begin{bmatrix}-1\\1\\\end{bmatrix}$ and $\begin{bmatrix}1\\1\\\end{bmatrix}\ $
Is this sufficient to conclude that the fundamental matrix $\phi=\begin{bmatrix}-e^{-2t}&e^{4t}\\e^{2t}&e^{4t}\\\end{bmatrix}\\$?
I will appreciate if someone can tell on what if there's more to be done in finding the fundamental matrix.
The fundamental matrix is actually
$$\begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{4t} \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}^{-1}.$$
Putting aside the sign error, what you wrote is the product of the first two matrices. The third matrix is needed to change bases to the basis of eigenvectors of $A$.
Things look like this in general except when $A$ is non-diagonalizable, for example this doesn't work as is when $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. There is a generalization for this case, though.