Prove that a unique linear map exists with these conditions (or not!)












0












$begingroup$


So...I have to either prove or deny that there exists a unique linear map $f: mathbb{R}^2 rightarrow mathbb{R}^2$ where:



$$ f(1,1)=(2,6)$$
$$ f(-1,1)=(2,1)$$
$$ f(2,7)=(5,3)$$



I know that if, for example, we remove the third condition it can easily be proven that $$ (1,1),(-1,1)$$
are linearly independent and so they form a basis on $mathbb{R}^2$, making such a linear map unique.



But now I have three conditions, where $$ (1,1),(-1,1),(2,7)$$ are, as far as I can see, also linearly independent by checking their determinants in pairs. But how come there can be a basis on $mathbb{R}^2$ with three vectors? Is this possible or is it a "trick" question?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean "linear map" instead of "linear combination"?
    $endgroup$
    – Klaus
    Jan 24 at 16:01










  • $begingroup$
    Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
    $endgroup$
    – Lightsong
    Jan 24 at 16:05
















0












$begingroup$


So...I have to either prove or deny that there exists a unique linear map $f: mathbb{R}^2 rightarrow mathbb{R}^2$ where:



$$ f(1,1)=(2,6)$$
$$ f(-1,1)=(2,1)$$
$$ f(2,7)=(5,3)$$



I know that if, for example, we remove the third condition it can easily be proven that $$ (1,1),(-1,1)$$
are linearly independent and so they form a basis on $mathbb{R}^2$, making such a linear map unique.



But now I have three conditions, where $$ (1,1),(-1,1),(2,7)$$ are, as far as I can see, also linearly independent by checking their determinants in pairs. But how come there can be a basis on $mathbb{R}^2$ with three vectors? Is this possible or is it a "trick" question?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean "linear map" instead of "linear combination"?
    $endgroup$
    – Klaus
    Jan 24 at 16:01










  • $begingroup$
    Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
    $endgroup$
    – Lightsong
    Jan 24 at 16:05














0












0








0





$begingroup$


So...I have to either prove or deny that there exists a unique linear map $f: mathbb{R}^2 rightarrow mathbb{R}^2$ where:



$$ f(1,1)=(2,6)$$
$$ f(-1,1)=(2,1)$$
$$ f(2,7)=(5,3)$$



I know that if, for example, we remove the third condition it can easily be proven that $$ (1,1),(-1,1)$$
are linearly independent and so they form a basis on $mathbb{R}^2$, making such a linear map unique.



But now I have three conditions, where $$ (1,1),(-1,1),(2,7)$$ are, as far as I can see, also linearly independent by checking their determinants in pairs. But how come there can be a basis on $mathbb{R}^2$ with three vectors? Is this possible or is it a "trick" question?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$




So...I have to either prove or deny that there exists a unique linear map $f: mathbb{R}^2 rightarrow mathbb{R}^2$ where:



$$ f(1,1)=(2,6)$$
$$ f(-1,1)=(2,1)$$
$$ f(2,7)=(5,3)$$



I know that if, for example, we remove the third condition it can easily be proven that $$ (1,1),(-1,1)$$
are linearly independent and so they form a basis on $mathbb{R}^2$, making such a linear map unique.



But now I have three conditions, where $$ (1,1),(-1,1),(2,7)$$ are, as far as I can see, also linearly independent by checking their determinants in pairs. But how come there can be a basis on $mathbb{R}^2$ with three vectors? Is this possible or is it a "trick" question?






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 16:04







Lightsong

















asked Jan 24 at 15:58









LightsongLightsong

246




246












  • $begingroup$
    Do you mean "linear map" instead of "linear combination"?
    $endgroup$
    – Klaus
    Jan 24 at 16:01










  • $begingroup$
    Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
    $endgroup$
    – Lightsong
    Jan 24 at 16:05


















  • $begingroup$
    Do you mean "linear map" instead of "linear combination"?
    $endgroup$
    – Klaus
    Jan 24 at 16:01










  • $begingroup$
    Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
    $endgroup$
    – Lightsong
    Jan 24 at 16:05
















$begingroup$
Do you mean "linear map" instead of "linear combination"?
$endgroup$
– Klaus
Jan 24 at 16:01




$begingroup$
Do you mean "linear map" instead of "linear combination"?
$endgroup$
– Klaus
Jan 24 at 16:01












$begingroup$
Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
$endgroup$
– Lightsong
Jan 24 at 16:05




$begingroup$
Sorry - Yes, I edited it now. I sometimes forget about it and do a literal translation from my mother language.
$endgroup$
– Lightsong
Jan 24 at 16:05










1 Answer
1






active

oldest

votes


















1












$begingroup$

You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
    $endgroup$
    – Lightsong
    Jan 24 at 18:57












  • $begingroup$
    I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
    $endgroup$
    – kccu
    Jan 24 at 19:24










  • $begingroup$
    Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
    $endgroup$
    – Lightsong
    Jan 25 at 11:15











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
    $endgroup$
    – Lightsong
    Jan 24 at 18:57












  • $begingroup$
    I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
    $endgroup$
    – kccu
    Jan 24 at 19:24










  • $begingroup$
    Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
    $endgroup$
    – Lightsong
    Jan 25 at 11:15
















1












$begingroup$

You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
    $endgroup$
    – Lightsong
    Jan 24 at 18:57












  • $begingroup$
    I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
    $endgroup$
    – kccu
    Jan 24 at 19:24










  • $begingroup$
    Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
    $endgroup$
    – Lightsong
    Jan 25 at 11:15














1












1








1





$begingroup$

You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.






share|cite|improve this answer









$endgroup$



You cannot check linear independence of a set of three (or more) vectors by checking them pairwise. That is like saying $6$ and $10$ share a factor, and $10$ and $15$ share a factor, and $6$ and $15$ share a factor, so $6$, $10$, and $15$ all share a common factor (they don't). And in fact, no three vectors in a two-dimensional vector space such as $mathbb{R}^2$ can be linearly independent. So it must be that $(1,1)$, $(-1,1)$, and $(2,7)$ are linearly dependent. Now you need to find a way to write one of them as a linear combination of the others, and see if the three conditions you were given are consistent or not.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 16:07









kccukccu

10.6k11229




10.6k11229












  • $begingroup$
    Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
    $endgroup$
    – Lightsong
    Jan 24 at 18:57












  • $begingroup$
    I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
    $endgroup$
    – kccu
    Jan 24 at 19:24










  • $begingroup$
    Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
    $endgroup$
    – Lightsong
    Jan 25 at 11:15


















  • $begingroup$
    Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
    $endgroup$
    – Lightsong
    Jan 24 at 18:57












  • $begingroup$
    I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
    $endgroup$
    – kccu
    Jan 24 at 19:24










  • $begingroup$
    Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
    $endgroup$
    – Lightsong
    Jan 25 at 11:15
















$begingroup$
Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
$endgroup$
– Lightsong
Jan 24 at 18:57






$begingroup$
Thanks! So indeed they are linearly dependent (example: (1,1)=2/9*(-1,1)+1/9*(2,7), does this mean that I can "reduce" the basis to any of these two vectors just like that?, I guess there is a condition that needs to be met for their images?
$endgroup$
– Lightsong
Jan 24 at 18:57














$begingroup$
I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
$endgroup$
– kccu
Jan 24 at 19:24




$begingroup$
I don't think that linear combination works. $frac{2}{9}(-1,1)+frac{1}{9}(2,7)=left(-frac{2}{9},frac{2}{9}right)+left(frac{2}{9},frac{7}{9}right) = (0,1)$. In any case, if you've written $(1,1)=alpha(-1,1)+beta(2,7)$, then by properties of linear maps you should have $f(1,1)=alpha f(-1,1)+beta f(2,7)$. This is the condition you will have to check (after you've written one of these vectors as a linear combination of another).
$endgroup$
– kccu
Jan 24 at 19:24












$begingroup$
Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
$endgroup$
– Lightsong
Jan 25 at 11:15




$begingroup$
Thanks! Now I got it, it was a simple check for the properties...and indeed it was wrong(α=-5/9,β=2/9)
$endgroup$
– Lightsong
Jan 25 at 11:15


















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