Finding out Eigenvalues and Eigenvectors Question:
$A=\begin{pmatrix}1&1&2\\0&1&3\\0&0&2\end{pmatrix}$
I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what  $1[(0)\cdot(λ-2)]$ equals to? 
Mainly the eigenvalues and I'm having issues with.
 A: From here
$$|A-\lambda I|=\begin{vmatrix}
1-\lambda & 1 & 2\\
0 & 1-\lambda & 3\\
0 & 0 & 2-\lambda\end{vmatrix}=(1-\lambda)^2(2-\lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $\lambda$ solve $(A-\lambda I)x=0$ to find the corresponding eigenvectors.
A: $$A~=~\begin{pmatrix}1&1&2\\0&1&3\\0&0&2\end{pmatrix}$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~\begin{vmatrix}1-\lambda&1&2\\0&1-\lambda&3\\0&0&2-\lambda\end{vmatrix}=(1-\lambda)\cdot\begin{vmatrix}1-\lambda&3\\0&2-\lambda\end{vmatrix}+1\cdot\begin{vmatrix}3&0\\2-\lambda&0\end{vmatrix}+2\cdot\begin{vmatrix}0&1-\lambda\\0&0\end{vmatrix}=(1-\lambda)^2(2-\lambda)+1\cdot(0)+2\cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-\lambda)^2(2-\lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $\lambda_{1,2}=1,\lambda_3=2$.
A: Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
A: There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my  matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v \neq 0.$ Then the first column, a genuine eigenvector, is  $u =(A-I) v $
$$
R =
\left(
\begin{array}{ccc}
1 & 0 & 5 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{array}
\right)
$$
The Jordan form is $ R^{-1} A R = J$ and has a special shape.
$$
\left(
\begin{array}{ccc}
1 & 0 & -5 \\
0 & 1 & -3 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 5 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{array}
\right)=
\left(
\begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}
\right)
$$
For applications such as finding $e^{At}$ it is the reverse identity that is important, $RJR^{-1} = A,$
$$
\left(
\begin{array}{ccc}
1 & 0 & 5 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & -5 \\
0 & 1 & -3 \\
0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 2
\end{array}
\right)
$$
