Conditions for diagonalizability and tridiagonalizability of a given matrix 
Investigate matrix $A \in \mathrm{Mat}_3 (\mathbb{R})$ for diagonalizability and tridiagonalizability, depending on parameter $\lambda$. $$A = \begin{pmatrix} 0 & 10 & 2 \\ \lambda & 2  & 1 \\ 0 & -3 & 0 \end{pmatrix} $$


I looked at the characteristic polynomial
$$p_A(t)=t^3-2t^2+t(3-10\lambda)+6\lambda$$
and tried to factorize but with no success. So with a function plotter i found out that:

*

*for $\lambda \in[-\infty,\frac38)\ p_A(t)=0$ has only one real
solution.

*for $\lambda=\frac38 \ p_A(t)=0$ has 3 real solutions ($-1,\frac32,\frac32$)

*for $\lambda \in (\frac38,\infty] \ p_A(t)=0$ has 3 UNIQUE Solutions

So how can i proof these three statements?
Are there any ways to find some conditions directly from the equations $p_A(t)=0$ itself and i an missing something?
I really appreciate any help cause im stuck on this problem for 2 days now.
 A: I will write $p_\lambda$ instead of $p_A$.
To find how many roots $p_\lambda$ has, study its variations.
$$p_\lambda'(t)=3t^2-4t+(3-10\lambda)$$
We compute the discriminant : $\Delta=120\lambda-20$.
This shows that for $\lambda\le \frac{1}{6}$, $p_\lambda$ is strictly increasing therefore has only one simple real root : $A$ isn't trigonalisable.
For $\lambda>\frac{1}{6}$, $p_\lambda$ is increasing- decreasing- increasing : therefore is reached a local minimum at $$x_{\lambda,+}=\frac{4+\sqrt{120\lambda-20}}{6}=\frac{2+\sqrt{30\lambda-5}}{3}$$ and a local maximum at
$$x_{\lambda,-}=\frac{4-\sqrt{120\lambda-20}}{6}=\frac{2-\sqrt{30\lambda-5}}{3}$$
Let's study now the sign of these local extrema $p_\lambda(x_{\lambda,\pm})$. Indeed :

*

*$p_\lambda(x_{\lambda,+}) > 0 \Longleftrightarrow p_\lambda$  has only one roots;


*$p_\lambda(x_{\lambda,+}) = 0 \Longleftrightarrow p_\lambda$  has  one simple root and one double;


*$p_\lambda(x_{\lambda,-})p_\lambda(x_{\lambda,+}) < 0 \Longleftrightarrow p_\lambda$  has  three distinct roots.


*Let's start with the local minimum $p_\lambda(x_{\lambda,+})$.
Introduce $u=x_{\lambda,+}$. Conversely, $\lambda=\frac{(3u-2)^2+5}{30}$
$$f(u):=p_\lambda(u)=u^3-2u^2+u\left(3-10\frac{(3u-2)^2+5 }{30}\right)+6\left(\frac{(3u-2)^2+5 }{30}\right)$$
One finds $$f(u)=-2u^3+\frac{19}{5}u^2-\frac{12}{5}u+\frac{9}{5}$$
For $\lambda=\frac{3}{8}$, $u=\frac{3}{2}$ and one checks that $f(3/2)=0$.
So on can find factorize $f$. One finds  $$f(u)=(-2u+3)\left(u^2-\frac{2}{5}u+\frac{19}{5}\right)$$
Now $\Delta=(-2/5)^2-4\times 19/5<0$ : the 2nd factor is always $>0$.
This means $f$ is positive  on $]-\infty;3/2[$, zero at $3/2$ and negative on $]3/2;\infty[$.
--> Going back from $u$ to $\lambda$, the local minimum $p_\lambda(x_{\lambda,+})$ is positive on $]1/6;3/8[$, zero at $3/8$ and negative on $]3/8;\infty[$.

*

*Study of the sign of the local maximum $p_\lambda(x_{\lambda,-})$.

Write $v=x_{\lambda,-}$ this time. Again, $\lambda=\frac{(3v-2)^2+5}{30}$.
As before, $$g(v)=p_\lambda(x_{\lambda,-})=...=(-2v+3)\left(v^2-\frac{2}{5}v+\frac{19}{5}\right)$$
So $g$ is positive on $]-\infty;3/2[$, zero at $3/2$ and negative on $]3/2;+\infty[$.
But this time, for any $\lambda> 1/6$, $v<3/2$ so the local maximum is always positive.
Let's sum everything up now :

*

*for $1/6 < \lambda < 3/8$, we have $p_\lambda(x_{\lambda,+}) >0 $ : $p_\lambda$ has only one simples root and A isn't trigonalisable;


*for $\lambda = 3/8$, we have $p_\lambda(x_{\lambda,-})>0$ and $p_\lambda(x_{\lambda,+}) = 0 $ : $p_\lambda$ has two distinct roots therefore three (one simple, one double) : $A$ is trigonalisable (but not diagonalisable because on checks that $E_{3/2}=Vect(4,1,-2)$ so $dimE_{3/2}<2$).


*for $\lambda > 3/8$, we have $p_\lambda(x_{\lambda,-})p_\lambda(x_{\lambda,+}) < 0 $ : $p_\lambda$ has three distinct roots  and $A$ is diagonalisable.
