Construct a 3x3 matrix given eigenvalues, but no eigenvectors

Question

I have to construct a 3x3 matrix with eigenvalues of 20, -3, and 0. However, we are not given eigenvectors and it can not be upper triangular. I know there are infinite number of matrices but I don't even know where to start.

Answer 1

Choose some invertible $3\times3$ matrix $M$ such that$$N=M\begin{bmatrix}20&0&0\\0&-3&0\\0&0&0\end{bmatrix}M^{-1}$$is not upper triangular. Then $N$ will work.

Answer 2

every three by three orthogonal matrix with positive determinant (therefore $+1$) can be written, with real $a,b,c,d$ not all zero, as

$$ P = \frac{1}{a^2 + b^2 + c^2 + d^2} \left( \begin{array}{ccc} a^2 + b^2 - c^2 - d^2&2(-ad+bc)&2(ac+bd)\\ 2(ad+bc)&a^2 - b^2+c^2- d^2&2(-ab+cd)\\ 2(-ac+bd)&2(ab+cd)&a^2 - b^2 - c^2 + d^2\\ \end{array} \right) $$

Starting with your diagonal matrix $D,$ you can get a symmetric matrix that is probably not diagonal with $$ P^T D P $$

The first example with $a,b,c,d$ integers and absolute values of the nine elements all distinct has $a=1,b=6,c=2,d=4:$

$$ P = \frac{1}{57} \left( \begin{array}{ccc} 17&16&52\\ 32&-47&4\\ 44&28&-23\\ \end{array} \right) $$

We get one example,

$$ P^T DP = \frac{1}{3249} \left( \begin{array}{ccc} 2708&9952&17296\\ 9952&-1507&17204\\ 17296&17204&54032\\ \end{array} \right) $$