Determine if a particular matrix is diagonalizable my teacher gave me this exercise:
Determine if this matrix is diagonalizable
$
\begin{pmatrix}
1 & 1&1&1\\ 
1&2&3&4\\ 
1&-1&2&-2\\ 
0&0&1&-2 
\end{pmatrix}
$
I have tried to calculate the characteristic polynomial, that is $-13 + 10 x + x^2 - 3 x^3 + x^4$, but I don't know how to go on. 
I tried to look at the roots of the characteristic polynomial with Wolfram Alpha and they are horrible! So I think that must exist an alternative way to do it!
Have you got any ideas? Thank you!
 A: Hint: The matrix is diagonalizable (over $\mathbb C$) if, and only if, it has 'four' eigenvalues. In particular, if the matrix has four distinct eigenvalues, then it is diagonalizable. Using the characteristic polynomial, can you prove that the matrix has four distinct eigenvalues?
Further hints: Recall that given a polynomial function $P$ whose coefficients are all real, it holds  that $\forall z\in \mathbb C\left(P(z)=0\implies P(\overline z)=0\right)$. Since the characteristic polynomial is a real fourth (even!) degree polynomial, either all roots are non-real, there exactly 'two' (beware of multiplicities) real roots or all roots are real. Use the intermediate value theorem theorem to prove that there are at least two real roots and check the derivative to prove that there are no more than two real roots.
A: If the roots of the characteristic polynomial are distinct, the matrix is diagonalizable (the Jordan normal form is composed of $1\times1$ Jordan blocks). The roots of the polynomial $P$ are distinct if $\deg(\gcd(P,P'))=0$ (if $(x-a)^n$ divides $P$, then $(x-a)^{n-1}$ divides $P'$).
Using the Polynomial Extended Euclidean Algorithm on $\color{#C00000}{P}$ and $\color{#00A000}{P'}$, we get
$$
\begin{align}
&(1822x^3-1987x^2-2613x+10903)(\color{#00A000}{4x^3-9x^2+2x+10})\\
&-(7288x^2-2482x-3659)(\color{#C00000}{x^4-3x^3+x^2+10x-13})\\
&=61463
\end{align}
$$
Thus, $\deg(\gcd(P,P'))=0$, and therefore, the matrix is diagonalizable.
