Do exists such linear transformation :
$F:\mathbb{R}^3\rightarrow \mathbb{R}^3 F(1, 2, 1) = (1, 0, 0), F(1, 0, 1) = (0, 1, 0), F(0, 1, 0) = (1, 0, 0)$
How should I do it in general?
Do exists such linear transformation
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linear-algebra
linear-transformations
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1You can define a linear transformation in any way you like on any *basis* and extend it to the rest of the space. Is your set of input vectors a basis? – 2017-01-26
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0Your 3 vectors $(1,2,1)$, $(1,0,1)$, and $(0,1,0)$ do not form a basis of $\mathbb{R}^{3}$, so you probably have a typo here. – 2017-01-26
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0@pjs36 True, but the OP is asking about linearity of $F$. This doesn’t require knowing all of $F$. In fact, the crux of the exercise appears to be noticing that the input vectors *are* linearly dependent. – 2017-01-26
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0@amd Indeed, I was hoping OP would get to discover that crux for themselves, while being nudged in a productive direction. – 2017-01-26
1 Answers
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Let $a = (1,2,1)$, $b=(1,0,1)$ and $c=(0,1,0)$. Then notice that $F(a)$ should equal $F(b+2c)$. But it doesn't. So the answer is "no such linear transformation exists." To do it in general, write $a, b,$ and $c$ as rows of a matrix $A$ and solve each $Ax = v_i$, where the $v_i$ are in vectors you want to hit in the codomain.