I need some kind of hint to get me going for this question as I'm so lost at it. Any sort of help would be appreciated.
Let E be the set of all 2x2 matrices that have $v={(1,-1)}$ as an eigenvector. For example ${(2,1),(2,1)}$ and ${(2,1),(-1,4)}$ are in E but not ${(1,0),(1,1)}$.
1.Prove that $E$ is a subspace of $M(2,2)$
2.Find a basis for $E$ and determine its dimension.
- Extend that basis and find a basis of $M(2,2)$.
In general, to prove that a sub-set $E$ is a sub-space, you have to show (1) that if $v,w\in E$, then $v+w\in E$, and (2) if $v\in E$, then $\alpha v\in E$ for $\alpha \in \mathbb{C}$ (or whatever field you are working over).
Write $v:=(1,-1)$ and let $A$ and $B$ be elements of $E$ with eigenvalues $\lambda$ and $\mu$ respectively. Then, $$ [A+B]v=Av+Bv=\lambda v+\mu v=(\lambda +\mu )v, $$ so that $v$ is an eigen-vector of $A+B$ (with eigen-value $\lambda +\mu$), and hence $A+B\in E$. You should now be able to check for yourself that $\alpha A\in E$. Thus, $E$ is a sub-space.
To determine a basis, try writing down the condition $Av=\lambda v$ explicitly. If $$ A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, $$ then $$ Av=\lambda v $$ iff $$ a-b=\lambda \text{ and }c-d=-\lambda . $$ Hence, $A$ must be of the form $$ \begin{pmatrix}b+\lambda & b \\ d-\lambda & d\end{pmatrix}. $$ From this, it is pretty clear that the dimension is $3$ because there are three in-dependent parameters. Precisely, we try taking the three combinations $b=1,d=0,\lambda =0$; $b=0,d=1,\lambda =0$; and $b=0,d=0,\lambda =1$ and verify that these are actually linearly independent. Plugging these values in, we find the matrices $$ \begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}\qquad \begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}\qquad \begin{pmatrix}1 & 0 \\ -1 & 0\end{pmatrix}. $$
As the dimension of $M(2,2)$ is $4$, to extend this to a basis of all of $M(2,2)$, it suffices to find one more matrix linearly-independent from the above $3$. By inspection, we see that $$ \begin{pmatrix}0 & 1 \\ 0 & 1\end{pmatrix} $$ is such a matrix. In fact, to show that this is linearly independent from the rest, it suffices to show that $v$ is not an eigen-vector for this matrix, which you can readily verify.