How can I find $t$ and $s$ so that this matrix is diagonalizable?
$$A = \begin{pmatrix} 1 & t & 2 \\ 0 & 1 & s \\ 0 & 0 & 2 \\ \end{pmatrix} $$
I started calculating the eigenvalues, and found $\lambda = 2$ and $\lambda=1$ (the last one with multiplicity of 2). I made a system and found also that $t$ must be zero and one eigenvector for $\lambda=1$ is $v = (1,1,0)^T$. For the other eigenvalue, I found $y=s$, so that $v=(2,1,1)^T$ if I choose $s=1$, for example. But what about the third vector? How can I found it so that $A$ can be written as $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix} ?$$
Thank you very much! I am a very very beginner in linear algebra.