# Greatest value a such that the eigenvectors are linearly independent

Question: The maximum value of "a" such that the matrix $$\begin{bmatrix}-3 & 0 &-2\\ 1 &-1 &0 \\ 0& a& -2 \end{bmatrix}$$ has three linearly independent real eigenvectors is?

The approach I was thinking about is, if an eigenvector has linearly independent eigenvectors I need to make sure that the eigenvalues are distinct. So find the characteristic equation ($$\lambda^3 + 6\lambda^2 + 11\lambda + 6 + 2a =0$$) should have distinct roots.

But am I not missing some cases?? because we can even have linearly independent eigenvectors when the eigenvalues are same. We are just sure that eigenvalues will be linearly independent if eigenvalues are distinct but that is not a necessary condition for linear independence of eigenvectors.

How do I approach this problem? How to ensure they are linear independent?

Also, how to approach that they are real? For real eigenvectors does it mean that the eigenvalues should be real?

There is no such $$a$$.
The characteristic polynomial of your matrix is $$P(\lambda)=-\lambda^3-6\lambda^2-11\lambda-6+2a$$. The discriminant of this polynomial is $$4-108a^2$$. The polynomial $$P(\lambda)$$ has only real roots if and only if $$4-108a^2\geqslant0$$, which means that $$a\in\left[-\frac1{3\sqrt3},\frac1{3\sqrt3}\right]$$. If $$a\in\left(-\frac1{3\sqrt3},\frac1{3\sqrt3}\right)$$, then there are three distinct real roots and therefore three linearly independent real eigenvectors. So, either $$a=\frac1{3\sqrt3}$$ is a solution of the problem or there is no solution.
However, a direct computation shows that, for this $$a$$, there are only two (up to multiplication by a non-zero scalar) real eigenvectors: $$\left(3+\sqrt3,-3,1\right)$$ (eigenvalue: $$-\frac13\left(6+\sqrt3\right)$$) and $$\left(-6+4\sqrt3,6,1\right)$$ (eigenvalue: $$\frac23\left(-3+\sqrt3\right)$$).