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Find $f_1, f_2$ of $\mathbb{R}^3$ such that for $i = 1, 2$:

  • $f_1 \neq f_2$
  • $f_i(1, 1, 0) = (1, 1, 1)$
  • $f_i(0, 1, 1) = (1, 2, 2)$
  • $f_i(1, 0, -1) = (0, -1, -1)$
  • $f_i(1, 2, 1) = (2, 3, 3)$
  • $\dim(f_1 \circ f_2) = 1$

Im stuck with this problem. I know that there are only $2$ vectors that are linearly independent, so I guess that $\dim(\operatorname{Im}f_1) = \dim(\operatorname{Im}f_2) = 2$ and $\dim(\operatorname{Ker}f_1) = \dim(\operatorname{Ker}f_2) = 1$.

But after that I have no clue how to solve the problem.

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  • $\begingroup$ What do you mean by $dim(f_1\circ f_2)=1$. What is the dimension of a linear operator? $\endgroup$ Sep 1, 2017 at 16:13
  • $\begingroup$ It's the dimension of the resulting composition of $f_1$ and $f_2$. The dimension of image of their composition. Hope it makes sense $\endgroup$
    – fedemengo
    Sep 1, 2017 at 16:15

1 Answer 1

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The fourth and the fifth conditions follow from the second condition. Expand the set $\{(1,1,0),(0,1,1)\}$ to a basis of $\mathbb{R}^3$ by adding the vector $(0,0,1)$. Set $f_1(0,0,1)=\left(\lambda_1,\lambda_2,\lambda_3\right)$ and $f_2(0,0,1)=\left(\mu_1,\mu_2,\mu_3\right)$. Write $(1,1,1)$, $(1,2,2)$, and $\left(\mu_1,\mu_2,\mu_3\right)$ in terms of the three basis vectors and use it to compute $(f_1\circ f_2)$. You will get three vectors which will involve $\lambda_i$s and $\mu_i$s. For example, $$\left(f_1\circ f_2\right)(1,1,0)=f_1((1,1,1))=f_1((1,1,0)+(0,0,1))=(1,1,1)+\left(\lambda_1,\lambda_2,\lambda_3\right)$$ Play around with the values of $\lambda_i$s and $\mu_i$s so that two of the three vectors you get are zero. A brute force solution. There may be a bit more elegant solution.

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  • $\begingroup$ I hoped to find the elegant solution, I guess I really have to do so many calcs $\endgroup$
    – fedemengo
    Sep 1, 2017 at 19:05

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