An example of a $2\times2$ matrix $A$ without zero entries and with eigenvalues $\lambda_{1}=3,\lambda_{2}=-4$ I am trying to do an exercise that asks :

Find an example of a $2\times2$ matrix $A$ without zero entries and with eigenvalues $\lambda_{1}=3,\lambda_{2}=-4$ 

I am having trouble thinking of a way to find such a matrix, of course $\mathrm{diag}(3,-4)$ have $\lambda_{1}=3,\lambda_{2}=-4$ as eigenvalues but I can't figure if I can use it to get to a matrix without zero entries...
Help is appreciated!
 A: Hint. Start with the diagonal matrix and think of it as corresponding to the basis $\{(1,0),(0,1)\}$. Then conjugate by an invertible matrix (and think of it as a change of basis) in a way that gives a result with no zero components. For instance, what happens if you change to the basis $\{(1,1), (1,-1)\}$? Will the image of either basis vector have a $0$ coordinate?
Alternate method. Find $a$, $b$, and $c$, nonzero, such that $ac-b^2 = \lambda_1\lambda_2$ and $a+c = \lambda_1+\lambda_2$. Then
$$\left(\begin{array}{cc}a&b\\b&c\end{array}\right)$$
will do. (Why?)
A: Here are two ways to solve the problem:
First, if you make $P$ the matrix of eigenvectors and $D$ the diagonal matrix with the eigenvalues, we have
$$ A = PDP^{-1} $$
You have eigenvalues, but you can pick whatever eigenvectors you want.  So, let the eigenvectors be $[1,1]$ and $[3,4]$ so that $P$ is $\begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix}$.  Now see if $A=PDP^{-1}$ has any zero entries.  If it does, try different eigenvectors.
A dirtier way can be had by realizing that the determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues.  Thus, you want your matrix to have determinant -12 and trace -1.
If we write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we want $ad-bc=-12$ and $a+d=-1$.  So, let's take $a = 1$.  Then $d=-2$ from the trace formula.  Plugging these into the determinant formula yields $-2-bc=-12$, so $bc=10$.  Take $b=2$ and $c=5$ for one possible solution.
