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I think this is an interesting question:

In the 2-dimensional real vector space, we are given a linear transformation $f$. Suppose we already know the images of the standard bases, say $f(e_1),f(e_2)$. Of course, in some cases the eigenspace might not exist. But assuming the eigenspaces exist, how can we use ruler and compasses to determine them ?

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2 Answers 2

Ok,thas was normal way but corcret! what is your idea about this: $f(e_1)=(f(e_{1}).(1,0)) (1,0)+(f({e_1}).(0,1) )(0,1)=f_{1}(e_1)(1,o)+f_{2}(e_1)(0,1)$ $f(e_2)=(f(e_{2}).(1,0)) (1,0)+(f({e_2}).(0,1) )(0,1)=f_{1}(e_2)(1,o)+f_{2}(e_2)(0,1)$ so matrix of f at standara basis will be : $A= \left[\begin{matrix} f_1 ({e_1}) & f_1 (e_2)\\ f_2 (e_1) & f_2 (e_2) \end{matrix}\right]$ and so we can find eigen value from this relation: det ($A= \left[\begin{matrix} f_1 ({e_1}) & f_1 (e_2)\\ f_2 (e_1) & f_2 (e_2) \end{matrix}\right]$ -$\lambda I$)=0 for this we must solve this relation : $\lambda^{2}+\lambda(-f_{1}(e_1)-f_2(e_2))+f_{1}(e_1)f_2(e_{2})-f_{2}(e_1)f_{1}(e_2)=0$ now if we can find a good formula for $\lambda$ then we can find a good formula for it's eigen vector too.

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I'll assume for convenience that the given points $f(e_1)=(a,b)$ and $f(e_2)=(c,d)$ lie in the first quadrant, with $a,b,c,d>0$. When two real eigenvalues exist, suitable eigenvectors are $$ v_\pm = \left(2c,\alpha \pm \gamma \right) \quad\mbox{where}\quad \alpha=d-a,\quad \gamma= \sqrt{\alpha^2+\beta^2}, \quad \beta= \sqrt{4bc}. $$ To construct these points using ruler and compass, proceed as follows.

  1. Given $e_1=(1,0)$ and $e_2=(0,1)$, first construct $O=(0,0)$, by constructing the perpendicular bisector of segment $\overline{e_1e_2}$ and drawing the circle with this segment as diameter. Then draw the $x$ and $y$ axes.
  2. Draw circles through $O$ with centers at $f(e_1)$ and $f(e_2)$. These intersect the axes at points $(2a,0)$, $(0,2b)$, $(2c,0)$ and $(0,2d)$. By constructing perpendicular bisectors through the segments joining these points with $O$, construct points $(a,0)$, $(0,b)$, $(c,0)$ and $(0,d)$.

  3. Next, construct $P=(0,d-a)=(0,\alpha)$ at distance $a$ from $(0,d)$. Similarly, construct $(c+b,0)$ and $(c-b,0)$.

  4. Using a perpendicular bisector, draw the vertical line $L$ through $(c-b,0)$. Draw the circle with center $(0,0)$ through $(c+b,0)$ and find the intersection with $L$ at $(c-b,\beta)$, since $(c-b)^2+4bc=(c+b)^2$. This constructs $\beta$. Construct $Q=(\beta,\alpha)$.

  5. Find $\gamma$ as the length of the segment $\overline{OQ}$.

  6. Construct the vertical line through $(2c,0)$ and finish by constructing the desired eigenvectors $v_\pm=(2c,\alpha\pm\gamma)$.

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