(this exercise is like the previous question I've written today, but now I have a 3-by-3 matrix (with a real parameter $k$). I need to say where the matrix could be diagonalized, if it's possible. (meaning, find the scalar k, in order to diagonalize the matrix)
I have this matrix: $$
\begin{matrix}
2 & 0 & 2 \\
0 & 4 & 0 \\
k & 0 & 2 \\
\end{matrix}
$$
I've computed the characteristic polynomial, and this is the final result:
$$
\begin{matrix}
t-2 & 0 & -2 \\
0 & t-4 & 0 \\
-k & 0 & t-2 \\
\end{matrix}
$$
I've computed the determinant using Laplace expansion using the 2nd column. And this is the final result:
$$ (t-4)*(t-2)^2 -2k = 0$$
now, if k = 0, I have 3 eigenvalues (one eigenvalue is counted twice, because of its multiplicity): t = 4, with algebraic multiplicity = 1 = geometric multiplicity;
t = 2, with algebraic multiplicity = 2.
Now, I need to find the geometric multiplicity of the second eigenvalue. I did this: $2 = n – rk(matrix)$.
where 2 is the algebraic multiplicity, n is the order of the matrix (i.e 3) and rk is the rank of the matrix with t = 2 (the eigenvalue).
what I've found out is that 2 = 3-2; therefore I have 2 = 1, which is false. So, it couldn't be diagonalized as k = 0.
now, I have to study the diagonalization as k not equal to zero, and here is located my problem. Detailed explanation:
this is the characteristic polynomial:
$$ (t-4)*(t-2)^2-2k = 0 $$
the problem is that if k isn't equal to zero, I have a difficult polynomial to handle, it's a 3 degree polynomial but it has a parameter k (which I don't know, so it could be anything ranging from $-\infty$ to $+\infty$). this is the polynomial: $$t^3-6t+(8-2k) = 0$$. I don't even know where to start to solve this polynomial.
other question: in the solution of the exercise I read "k not equal to 2, the matrix couldn't be diagonalized" instead of just k = 0. Why is the matrix not diagonalizable as k = 2?
perhaps, I guess that if k = 2, the original matrix has 2 linearly dependent columns, so the determinant is zero. I still don't get the link between what I said above, and the diagonalization problem. why must I put k not equal to 2?
Best Answer
The characteristic polynomial of your matrix is$$-t^3+8 t^2+2 k t-20 t-8 k+16.\tag1$$But it is clear, looking at the middle collumn of the matrix, that $4$ is an eigenvalue, and therefore a root of $(1)$. Dividing $(1)$ by $t-4$, you get $-\bigl((t-2)^2-2 k\bigr)$, and so the other roots of $(1)$ are $2\pm\sqrt{2k}$. Therefore, if $k>0$ and $k\ne2$, then $(1)$ has three distinct roots, and therefore your matrix is diagonalizable. If $k<0$, then it is not diagonalizable over the reals (although it is diagonalizable over $\Bbb C$). So, now you only have to check what happens when $k=0$ and when $k=2$. In turns out that it is not diagonalizable when $k=0$, whereas it is diagonalizable when $k=2$.