Find real number $a$ such that matrix $A$ is NOT diagonalisable Consider the matrix:
$\begin{bmatrix}2 & a & -1\\0 & 2 & 1\\-1 & 8 & -1\end{bmatrix}$
Find all $a \in \mathbb R$ such that $A$ is not diagonalisable.
I've never thought of such a problem before. What is the systematic way to make a matrix non-diagonalisable?
 A: (These Linear Algebra course notes by Michael Stoll, 2007 provide good background to my following assertions. As does the Wikipedia page on characteristic polynomials.)
Use the following facts:
a) A matrix $M \in \mathrm{Mat}(n,F)$ (where $n$ is the size of the matrix and $F$ is some field) is diagonalisable  if and only if its minimal polynomial $m_M(x)$ is a product of distinct monic linear factors.
b) The characteristic polynomial of a matrix $M$ is defined as $p_M(x)=\mathrm{det}(xI-M)$
c) Matrices satisfy their own characteristic equations: $p_M(M)=0$ for all $M$. Hence a matrix's minimal polynomial always divides its characteristic polynomial: $m_M(x) \, \vert \, p_M(x)$.
d) The discriminant of a cubic polynomial $ax^3+bx^2+cx+d$ is given by: 
$$ \Delta_3 = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$
If $\Delta_3>0$ then the equation as three distinct real roots. If $\Delta_3=0$ then the equation has a repeated root and all its roots are real. If $\Delta_3<0$ then the equation has just one real root.
For your matrix,
$$ A:= \left( \begin{matrix}2 & a & -1\\0 & 2 & 1\\-1 & 8 & -1\end{matrix} \right) $$
we have that
$$
\begin{align}
p_A(x) &= \mathrm{det} \left( \begin{matrix} x-2 & -a & 1\\0 & x-2 & -1\\1 & -8 & x+1\end{matrix} \right) 
\\
\\ &= (x-2)((x-2)(x+1)-8) + (a - (x-2))
\\ &= x^3 -3x^2-9x+22+a
\end{align}
$$
With some calculation, the discriminant of $p_A$ is equal to $$\Delta_3 = -27(a-5)(a+27)$$ 
Firstly,
$$\Delta_3>0 \iff -27<a<5$$ 
In this case,  $p_A$ has three distinct real roots then it will factorise into three distinct linear factors. Hence $m_A(x)$ will also factorise into distinct linear factors and $A$ will be diagonalisable. 
So $-27<a<5$ implies that $A$ is diagonalisable.
Secondly,
$$\Delta_3=0 \iff a \in \{-27,5\}$$ 
If $a=-27$ then
$$ A:= \left( \begin{matrix}2 & a & -1\\0 & 2 & 1\\-1 & 8 & -1\end{matrix} \right) $$
and this matrix can be checked to have a maximum of two linearly independent eigenvectors 
$$ v_1:= \left( \begin{matrix}-10\\1\\3\end{matrix} \right) \qquad v_2:= \left( \begin{matrix}-8\\-1\\3\end{matrix} \right)$$
and so $A$ is not diagonalisable. (We could also consider the Jordan Form of $A$ as calculated by @WillJagy in their answer to this question.)
Similarly for $a=5$, $A$ turns out to not be diagonalisable.
Thirdly,
$$\Delta_3<0 \iff a \in (-\infty , -27) \cup (5, \infty )$$
In this case $p_A(x)=(x- \lambda )(x^2+ \beta x + \gamma )$ where $\lambda$ , $\beta$ and $\gamma$ are real and $(x^2+ \beta x + \gamma )$ cannot be factorised over the reals. Since $m_A$ divides $p_A$ we have three options:
$$ \begin{align}
\mathrm{(i)} \qquad & m_A(x)=(x- \lambda ) 
\\\mathrm{(ii)} \qquad & m_A(x)=(x^2+ \beta x + \gamma )
\\\mathrm{(iii)} \qquad & m_A(x)=(x- \lambda )(x^2+ \beta x + \gamma )
\end{align}$$
It is obvious that $(A-\lambda I )$ is not equal to $0$ for any $\lambda$ so we can rule out option $\mathrm{(i)}$. But in both of the options $\mathrm{(ii)}$ and $\mathrm{(iii)}$, $m_A$ is not a  product of distinct monic linear factors and hence $A$ is not diagonalisable.
Therefore:
$A$ is not diagonalisable over $\mathbb{R}$ if and only if $a \in (-\infty , -27] \cup [5, \infty )$
Edit: diagonalisability over $\mathbb{C}$
Even if working in $\mathbb{C}$, the cases $\Delta_3>0$ and $\Delta_3=0$ above remain exactly the same: in these cases the characteristic polynomial has no complex roots.
However, $\Delta_3<0$ now gives us that $p_A$ can be factorised into three distinct linear factors in $\mathbb{C}[x]$. Hence $m_A(x)$ will also factorise into distinct linear factors in $\mathbb{C}[x]$ and $A$ will be diagonalisable.
Hence $A$ is not diagonalisable over $\mathbb{C}$ if and only if $a \in \{-27,5\}$.
In general, for $A \in \mathrm{Mat}(n,\mathbb{C})$, if the discriminant of $p_A$ is not equal to $0$ then $p_A$ has no repeated roots and $A$ will be diagonalisable over $\mathbb{C}$. If the discriminant of $p_A$ is equal to $0$ then check the eigenvectors of $A$ (or form the Jordan Form of $A$) in order to check individual cases of diagonalisability.
A: Let's consider such an $a$. Then $A$ must necessarily have a repeated eigenvalue, otherwise it would be diagonalizable. If $f(t) = \det(tI-A)$ is the characteristic polynomial, then $f$ and $f'$ must have a common root. This severely restricts the values that $a$ can take.
Can you continue from here? Note that above is a necessary condition - after you figure out all potential candidates for $a$, you need to check if it actually makes $A$ non-diagionalizable. 
