Find all $2\times 2$ matrices $A$ such that $E_7 = \mathbb{R}^2$

Question

Find all $2\times 2$ matrices $A$ such that $E_7 = \mathbb{R}^2$, where $E_7$ is the eigenspace associated to the eigenvalue $7$.

Can someone give me a concise way of completing the problem?

I know that $\det(A-\lambda I_2) = \lambda^2-(\operatorname{tr}A)\lambda+\det A$, but the problem seems more difficult that just finding $\det(A-7I_2) = 49-7(\operatorname{tr}A)+\det A$.

Answer 1

(Look the edit for a much simpler proof) Let $B\in \Bbb R^{2\times 2}$ a matrix which has $7$ as eigenvalue and $\Bbb R^2$ as associated eigenspace. Since $\dim(E_7) = 2 = \dim(\Bbb R^2)$, $B$ is diagonalizable and so it can be written in the form $B=PDP^{-1}$, where $D$ is a diagonal matrix with he eigenvalues of $B$ and $P$ is invertible. In particular we must have $D = 7I_2$ where $I_2$ is the identity matrix. But then $D$ commutes with any matrix and so we have $B=PDP^{-1}=PP^{-1}D = D = 7I_2$.

EDIT: Let $A\in \Bbb R^{2\times 2}$ a matrix which has $7$ as eigenvalue and $\Bbb R^2$ as associated eigenspace. Then $\{\vec{e}_1,\vec{e}_2\}=\{(1,0),(0,1)\}$ is a basis of $E_7$ so $A\vec{e}_1=7\vec{e}_1$ and $A\vec{e}_2 = 7\vec{e}_2$. Now let $\vec{v}=(v_1,v_2) \in \Bbb R^2$, then $v = v_1 \vec{e}_1+v_2\vec{e}_2$ and so $$Av = A(v_1 \vec{e}_1+v_2\vec{e}_2) = v_1 A\vec{e}_1+v_2A\vec{e}_2 = 7v_1 \vec{e}_1+7v_2\vec{e}_2 = 7(v_1 \vec{e}_1+v_2\vec{e}_2) = 7v$$ It follows that $Aw = 7w$ for all $w\in \Bbb R^2$, i.e. $A = 7I_2$.

Answer 2

Even simpler: by definition $A x = \lambda x = \lambda I x$ for $x$ in the eigenspace $E_\lambda$, so if that eigenspace is the whole space $A = \lambda I$.