How to put a matrix in Jordan canonical form, when it has a multiple eigenvalue? 
Put the matrix $$\begin{bmatrix} 3 & -4\\ 1 & -1\end{bmatrix}$$ in Jordan Canonical Form.  Moreover, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form.

I'm having a lot of trouble with this.  I know that the eigenvalue has multiplicity two and is $\lambda = 1$.  I can find the first eigenvector, which is:
\begin{bmatrix}
        2 \\
        1  \\
        \end{bmatrix}
I'm having trouble finding the second since both eigenvalues tell us the same thing.  But I'm not nearly as concerned about the eigenvectors as I am about what to do after.  
If anyone could explain thoroughly the next steps involved (not necessarily the answer but how to obtain it), I would be forever grateful.  This homework is in 2 days and it may determine my grade letter.
 A: To put a matrix in Jordan normal form requires to know three matrices such that $A=PJP^{-1}$ i.e: a matrix $P^{-1}$ that transform to the canonical basis $(\mathbf{i},\mathbf{j},\mathbf{k})$ to a new basis in which  the matrix $J$ represents the transformation of a vector $\mathbf{v}$ such that the transformed vector $\mathbf{v'}$ is the same as we find when we transform $\mathbf{v}$ with $A$ in the canonical basis. Last the matrix $P$ returns this result to the canonical basis.
As noted in OP the matrix $A$ has eigenvalues $\lambda_1=\lambda_2=1$ and a single eigenvector 
$$
\mathbf{u_1}=\left[
\begin{array}{cccc}
2\\
1
\end {array}
\right]
$$
So the main problem is to find another vector that completes the new basis.
To find such a vector notes that all vectors $\mathbf{x}$ such that $(A-\lambda I)\mathbf{x}=0$ are transformed in the eigenspace generated by the eigenvector $\mathbf{u_1}$, so we want a vector $\mathbf{u_2}$ such that $(A-\lambda I)\mathbf{u_2} \ne 0$, and the way to do this is to find a vector such that : $(A-\lambda I)\mathbf{u_2} = \mathbf{u_1}$. (Note that this equation is the same as $(A-\lambda I)^2\mathbf{u_2}= 0$).
Solving in our case we find:
$$
\left[
\begin{array}{cccc}
2&-4\\
1&-2
\end {array}
\right]
\left[
\begin{array}{cccc}
x\\
y
\end {array}
\right]=
\left[
\begin{array}{cccc}
2\\
1
\end {array}
\right]
$$
so that the components $x$ and $y$ of the searched vector must satisfies $x-2y=1$, and we can find the vector 
$$
\mathbf{u_2}=
\left[
\begin{array}{cccc}
1\\
0
\end {array}
\right]
$$ 
So the matrix $P$ we are searching is
$$
P=[\mathbf{u_1},\mathbf{u_2}]=
\left[
\begin{array}{cccc}
2&1\\
1&0
\end {array}
\right]
$$
And the inverse is:
$$
P^{-1}=
\left[
\begin{array}{cccc}
0&1\\
1&-2
\end {array}
\right]
$$
The matrix $J$ is a typical Jordan block, with the eigenvalues as diagonal elements and an entry $1$ up-right them:
$$
J=
\left[
\begin{array}{cccc}
1&1\\
0&1
\end {array}
\right]
$$
and we can easily verify that $A=PJP^{-1}$.
A: The characteristic polynomial $\det (A - \lambda I) = (\lambda -1)^2.$ when the dimension of the null space(1) of an eigenvalue($\lambda = 1$) is less than the algebraic multiplicity(2), you need to find generalized eigenvectors. in this instance, you need to solve $(A - I_2) \left( \begin{array}{l} x \cr y\end{array} \right) = \left( \begin{array}{l} 2 \cr 1 \end{array} \right).$ this gives you $x = 1, y = 0$
with respect to the basis 
$\{ \left( \begin{array}{l} 2 \cr 1 \end{array} \right), \left( \begin{array}{l} 1 \cr 0 \end{array} \right) \}$ you transformation is represented by the Jordan canonical form 
$\left( \begin{array}{ll} 1 & 1 \cr 0 & 1 \end{array} \right).$ 
