Show that there is a mapping $ S \ $ such that $ \ S^{-1} AS =\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix} \ $ Let $ A \in \mathbb{R}^{2 \times 2} \ $  and has repeated eigen value $ \lambda \ $. Let $ \vec u \ $ be the only Eigen vector associated with $ \lambda \ $ . 
Show that there is a mapping $ S  \ $ such that $ \ S^{-1} AS =\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix} \ $
Answer:
Since $ \lambda \ $ is repeated Eigen value and $ \vec u \ $ is  the only Eigen vector , we can find another Eigen independent  vector in the following way :
$ |A-\lambda I| \vec v=\vec u \ $ 
i.e., $ A \vec v=\vec u+\lambda \vec v \ $
Let $ S=[ \  \vec u \ | \  \vec v  \ ] \ $ be a matrix consisting of the Eigen vectors .
I think  $ S^{-1} AS=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}  \ $   but I can not prove it .
Is there any help proving this ?
 A: The $S$ you've constructed is a change of basis matrix $M(I, B, B_{\mathrm{st}})$, where $B = (\vec{u}, \vec{v})$ and $B_{\mathrm{st}}$ is the standard basis. Then
$$S^{-1} = M(I^{-1}, B_{\mathrm{st}}, B) = M(I, B_{\mathrm{st}}, B).$$
If we interpret $A = M(T, B_{\mathrm{st}}, B_{\mathrm{st}})$ for some operator $T$, then,
$$S^{-1} A S = M(I, B_{\mathrm{st}}, B)M(T, B_{\mathrm{st}}, B_{\mathrm{st}})M(I, B, B_{\mathrm{st}}) = M(T, B, B).$$
The columns of this matrix are computed by considering $[T\vec{u}]_B$ and $[T\vec{v}]_B$, which yield the columns you want.
A: I am not sure this is exact proof, but I just wanted to play with it. 
Why not insert the Eigen-decomposition into the original form:
$$
\begin{align}
S^{-1}AS &= S^{-1}VUV^TS
\end{align}
$$
U is a diagonal, V is an orthonormal matrix. Then from orthogonality of V:
$$
\begin{align}
S^{-1}AS &= (V^TS)^{-1}UV^TS
\end{align}
$$
If we write $K = (V^TS)^{-1}$ and assume that it is also a real orthogonal matrix, i.e. $(V^TS)^{-1}=(V^TS)^{T}$, then:
$$
\begin{align}
S^{-1}AS &= KUK^T
\end{align}
$$
which means that if we select $S$ to be properly (orthonormal), then A can be reduced to a diagonal form (diagonalizable). 
Would that lead you to some result?
