Let $T$ be set of all matrices $A\in M_{2x2} (\mathbb{R})$ that have $\begin{bmatrix} 1\\ 2 \end{bmatrix}$ as eigenvector. Check if $T$ is subspace of $\ M_{2x2} (\mathbb{R})$, and if it is subspace determine basis of $T$ and it's dimension.
Let matrix $A= \begin{bmatrix} a &b\\c&d \end{bmatrix} \space\wedge\space a,b,c,d \in \mathbb{R} $, and my eigenvector is $x_{1}=\begin{bmatrix} 1\\ 2 \end{bmatrix}$, and from the formula $A\cdot x_{1}= \Lambda \cdot x_{1}$ we have $$\begin{bmatrix} a &b\\c&d \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2 \end{bmatrix}= \Lambda\cdot \begin{bmatrix} 1\\ 2 \end{bmatrix} \Rightarrow\\a+2b=\Lambda \\ c+2d=2\Lambda \\$$ Then, when I do $p_{A}(\Lambda)=det(A-\Lambda I)=0,\space$ I get $\space(a-\Lambda)\cdot(d-\Lambda)-b\cdot c=0 \Leftrightarrow \space \Lambda^{2} -\Lambda\cdot (a+d) -(ad-cb)=0\\$ I know I get here two solutions, but there are too many elements I don't know. What am I doing wrong? I hope there is easier way to solve this kind of a problem..
We have that
$$c+2d=2\Lambda =2(a+2b).$$ Thus $T$ is the set of matrices
$$\begin{bmatrix} a &b\\2a+4b-2d &d \end{bmatrix}.$$