I am supposed to find the Jordan canonical form of a couple of matrices, but I was absent for a few lectures.
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 2 \\
0 & 0 & 3
\end{bmatrix}
Since this is an upper triangular matrix, its eigenvalues are the diagonal entries. Hence $\lambda_{1,2}=1$ and $\lambda_3 = 3$, with corresponding eigenvectors
$(1,2,2)$ and $(1,0,0)$. Now what? I do not know how to proceed, nor what it means that my matrix is built up by Jordan blocks.
Best Answer
Since this is a very small matrix there's actually no need to explicitly find the Jordan form by going through the routine tedious procedure. You know the characteristic polynomial is $$p(x) = (x-1)^2(x-3)$$ and you can check that the minimal polynomial is the same. This means a few things.
These combined gives you the form (up to order of the blocks) $$J=\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 3\end{pmatrix}$$
In general, for small matrices like these (anything up to $6\times 6$) you can find the Jordan form through similar types of analysis using the facts