Find a non-diagonalizable $3\times 3$ complex matrix $A$ with strictly real eigenvalues, $\det A=2$ and $\text{Tr}\,A=1$.

Question

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

Answer 1

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.

Answer 2

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.

$\lambda_1 = \frac 16(1-\sqrt[3]{107-6\sqrt {318}}+\sqrt[3]{107+6\sqrt {318}})\approx -0.85809\\ \lambda_2 \approx -0.85809\\ \lambda_3\approx 2.71618$