Matrix, linearly independent real eigenvectors

Question

The maximum value of $a$ such that the matrix $$ \left[ \begin{array}{ccc} –3 & 0 & –2\\ 1 & –1 & 0\\ 0 & a & –2\\ \end{array} \right] $$ has three linearly independent real eigen vectors is :

1. $\frac{2}{3\sqrt3}$
2. $\frac{1}{3\sqrt3}$
3. $\frac{1+2\sqrt3}{3\sqrt3}$
4. $\frac{1+\sqrt 3}{3\sqrt3}$

Answer 1

One approach is as follows. The characteristic polynomial is given by $$ p(\lambda) = \lambda^3 + 6 \lambda^2 + 11\lambda + (2a + 6). $$ The discriminant of this polynomial is given by $\Delta = 4 - 108a^2$. The matrix will fail to be diagonalizable over $\Bbb R$ whenever the polynomial fails to have $3$ real roots (either distinct or repeated), which occurs if and only if the discriminant is negative.

The greatest value of $a$ for which this occurs is the positive root to $\Delta = 0$.