Find a $2\times 2$ matrix $A$ such that the eigenvector associated with $\lambda = 1$ equals... 
Find a $2\times 2$ matrix $A$ such that the eigenvector associated
  with $\lambda = 1$ equals $\operatorname{span}\left( \begin{array}{ccc}
 2\\1 \end{array} \right)$ is the only eigenspace.

I believe $A =SBS^{-1}$. So can we use $S$ with first column equal to $\left( \begin{array}{ccc}
2\\1 \end{array} \right)$ and second column equal to anything that makes it invertible, and $B = I_2$, then just compute $SBS^{-1}$ to find $A$?
 A: First of all, your answer cannot be right, for if $B=I$ then $A=SS^{-1}=I$ and the eigenspace will be all of $\Bbb R^2$.

The simplest way to get what you want is to realise that as $A$ has only one independent eigenvector, it will not be diagonalisable and will have a Jordan form instead.  The matrix
$$A=\pmatrix{2&1\cr1&0\cr}\pmatrix{1&1\cr0&1\cr}\pmatrix{2&1\cr1&0\cr}^{-1}
  =\pmatrix{3&-4\cr1&-1\cr}$$
will do what you want.

If you have not yet studied Jordan forms, you can use a bit of trial and error.  The easiest way to make sure that $A$ has $1$ as an eigenvalue, with eigenvector $(2,1)$, is to specify
$$A-I=\pmatrix{1&-2\cr k&-2k\cr}\ ,\tag{$*$}$$
which gives
$$A=\pmatrix{2&-2\cr k&1-2k\cr}\ .$$
Now to make sure that $1$ is the only eigenvalue, we want the trace to be $2$, so $3-2k=2$, so $k=\frac12$ and
$$A=\pmatrix{2&-2\cr {\textstyle\frac12}&0\cr}\ .$$
A: i wonder if it is easier to construct the matrix $A - I$ first. we can use the property that $A$ has eigenvalue $1$ and an associated eigenvector $(2,1)^T,$ then $A-I$ will have eigenvalue $0$ and an associated eigenvector $(2,1)^T.$ 
$(A-I)(2,1)^T = 0$ tells us that the second column of $A-I$ is $-2$ times the first column of $A-I.$ to avoid any further relations between the columns, we can require the first column be not the zero column. i believe any matrix of the form $\pmatrix{a & -2a\cr b & -2b}$ with the constraint $a^2 + b^2 \neq 0$ should do for the matrix $A-I.$
now add $1$ to the diagonals to get the matrix $A.$
