Find a non-diagonalizable $3\times 3$ complex matrix $A$ with strictly real eigenvalues, $\det A=2$ and $\text{Tr}\,A=1$. I know that:
$\lambda_1 , \lambda_2 , \lambda_3 \in \Re$ counting multiplicity
Tr(A) = $\lambda_1 + \lambda_2 + \lambda_3$
and 
Det(A) = $\lambda_1 * \lambda_2 * \lambda_3$
I also know that a matrix is non-diagonalizable when the geometric and algebraic multiplicity of it's eigenvalues don't coincide. But I am unsure what to do with this information, any help would be appreciated
 A: First note that we cannot have three different eigenvalues -- if that was the case, we would have three linearly independent eigenvectors, which would form a basis in which your matrix would diagonalize.
Second, we cannot have only one different eigenvalue: $\det(A) = \lambda^3  = 2$ would force $\lambda = \sqrt[3]{2}$, but then $Tr(a) = 3\lambda \ne 1$.
So, we need to have two different eigenvalues. Try to find out what these would be, and then try creating a matrix that is known to not be diagonalizable with these eigenvalues. Hint: do you know any non-diagonalizable matrix? Hint 2: think about Jordan blocks.
A: You have the following constraints on the eigenvalues.
$\lambda_1 + \lambda_2+\lambda_3 = 1\\
\lambda_1\lambda_2\lambda_3 = 2\\
\lambda_1 = \lambda_2$
Then
$\lambda_1^2\lambda_3 = 2\\
\lambda_3 = \frac {2}{\lambda_1^2}\\
2\lambda_1 + \frac {2}{\lambda_1^2} = 1\\
2\lambda_1^3 - \lambda_1^2 +  2 = 0$
Now you need the cubic formula.
$\lambda_1 = \frac 16(1-\sqrt[3]{107-6\sqrt {318}}+\sqrt[3]{107+6\sqrt {318}})\approx -0.85809\\
\lambda_2 \approx -0.85809\\
\lambda_3\approx 2.71618$
