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The exercise asks me verify if there exists a linear operator $F$ such that:

$$F(1,1,1) = (1,2,3)\\F(1,2,3) = (1,4,9)\\F(2,3,4) = (1,8,27)$$

First I tried to write a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$:

$$(x,y,z) = a(1,1,1) + b(1,4,9) + c(1,8,27)$$

then we have:

$$x = a + b + 2c\\y = a + 2b + 3c\\z = a + 3b + 4c$$

which is a system that has determinant $0$, therefore there isn't a way to represent a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$.

If it were possible to write the vector $(x,y,z)$ I would then apply $F$ to both sides and find the linear operator. But I can't do it since that vectors do not form a basis. But my reasoning does not logically prove that such linear operator does not exsists. By the way, my book also says that this is impossible because $(2,3,4)$ should be equal to $(2,6,12)$. Why?

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  • $\begingroup$ The book is being a little overspecific. You could change either of the other two specified results also. Setting $F(1,1,1)=(0,3,18)$ would fix the specification, as alternatively would $F(1,2,3)= (0,6,24)$. All that's needed is that $F(1,1,1)+F(1,2,3)=F(2,3,4)$ $\endgroup$
    – Joffan
    Jan 15 '15 at 0:32
  • $\begingroup$ @Joffan you're rigth :) $\endgroup$ Jan 15 '15 at 0:32
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Note that $F(1,1,1) + F(1,2,3) = (2,6,12)$ and $F(2,3,4) = (1,8,27)$ and $(1,1,1) + (1,2,3) = (2,3,4) $. Then such $F$ never exist .

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If $F$ were linear, then $$F(1,1,1)+F(1,2,3) = F((1,1,1)+(1,2,3)) = F(2,3,4) = (1,8,27).$$ But $F(1,1,1)+F(1,2,3) = (1,2,3)+(1,4,9) = (2,6,12)$.

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The vectors $v_1=(1,1,1)$, $v_2=(1,2,3)$ and $v_3=(2,3,4)$ are not linearly independent; Gaussian elimination gives \begin{align} \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 2 & 2 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \end{align} which says that $v_3=v_1+v_2$.

Since $(1,2,3)+(1,4,9)=(2,6,12)\ne(1,8,27)$ no linear map exists with the requested property.

Had the problem said $F(2,3,4)=(2,6,12)$, then infinitely many linear maps would exist: complete $\{(1,1,1),(1,2,3)\}$ to a basis and send the third basis vector to any vector.


In the RREF above, the dominant columns (those with a leading $1$) are the first and the second column. The coefficients in the third column give exactly the coefficients to use for getting it as a linear combination of the dominant columns to its left.

Change the third column to represent the vector $(1,0,-1)=2v_1-v_2$; Gaussian elimination is \begin{align} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 3 & -1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 2 & -2 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \end{align} and indeed we find the right coefficients.

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  • $\begingroup$ Did you discover that $v_3 = v_1 + v_2$ by gaussian elimination? $\endgroup$ Jan 15 '15 at 0:29
  • $\begingroup$ @GuerlandoOCs Yes: the last column in the RREF says that. Of course in this case guessing is easier, but I prefer showing general methods. $\endgroup$
    – egreg
    Jan 15 '15 at 0:31
  • $\begingroup$ sorry, I don't get how the third column says $v_3 = v_1 + v_2$ $\endgroup$ Jan 15 '15 at 0:33
  • $\begingroup$ @GuerlandoOCs I added another example, just to show how it works. Gaussian elimination preserves linear relations between columns. $\endgroup$
    – egreg
    Jan 15 '15 at 0:39
  • $\begingroup$ I understood now, but don't know why it works $\endgroup$ Jan 15 '15 at 1:30

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