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 $A$?
Best Answer
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 $$A=\pmatrix{2&-2\cr {\textstyle\frac12}&0\cr}\ .$$