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I have $$T: R^3 \rightarrow R^4 /$$ $$T(1,0,1) = (-1,0,1,2)$$ $$T(0,1,1) = (0,1,1,2)$$ $$T(0,0,1) = (1,1,2,0)$$

I have to find if T exists and is unique, and the formula.

So I did this:

Because $(1,0,1), (0,1,1), (0,0,1)$ are a base of $R^3$ (they are linearly independent, because they are triangulated), there exists a linear transformation and it is unique. (I took this from a theorem that says exactly this).

However, I've tried to solve it, but no luck (from an example I saw in internet):

$$(x,y,z) = x(1,0,1) + y(0,1,1) + z(0,0,1)$$ $$T(x,y,z) = xT(1,0,1) + yT(0,1,1) + zT(0,0,1)$$ $$T(x,y,z) = x(-1,0,1,2) + y(0,1,1,2) + z(1,1,2,0)$$ $$T(x,y,z) = (-x+z, y+z, x+y+2z, 2x+2y)$$

But then evaluating $(1,0,1)$ gives me $(0,1,3,2)$ when I was expecting $(-1,0,1,2)$

Then I took my book and saw an example of how to solve, very similar but in the end it gave me another LT.

$$(x,y,z) = a(1,0,1) + b(0,1,1) + c(0,0,1)$$

$$ \left\{\begin{matrix} & x = a & \\ & y = b & \\ & z = c + b \rightarrow c = z - b \rightarrow c = z - y & \end{matrix}\right. $$

Then

$$T(x,y,z) = aT(1,0,1) + bT(0,1,1) +cT(0,0,1)$$ $$T(x,y,z) = x(-1,0,1,2) + y(0,1,1,2) + (z-y)(1,1,2,0)$$ $$T(x,y,z) = (-x-y+z, z, x+2z-y, 2x+2y)$$

But then again, evaluating in $(1,0,1)$ gives me $(0,1,3,2)$.

So there are 2 problems here:

1) The linear transformation doesn't seem to be unique.

2) Evaluating $(1,0,1)$ does not give me $(-1, 0, 1, 2)$

What am I doing wrong?

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    $x(1,0,1)+y(0,1,1)+z(0,0,1)$ does not equal $(x,y,z)$. Instead it is $(x,y,x+y+z).$ – 2017-02-16
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    Or to put it the other way, $(x,y,z)=x(1,0,1)+y(0,1,1)+(z-x-y)(0,0,1)$. Therefore $T(x,y,x)= x(-1,0,1,2)+y(0,1,1,2)+(z-x-y)(1,1,2,0).$ – 2017-02-16
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    Look at this question: http://math.stackexchange.com/questions/2129477/formula-linear-map-defined-by/2129524#2129524 – 2017-02-16
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    In your second attempt you should have $c=z-x-y$ – 2017-02-16
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    Thanks, so it was an error calculating... The first attemp is a viable one or it will lead me to errors? – 2017-02-16
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    As @ziggurism has already noted, $(x,y,z)\ne x(1,0,1)+y(0,1,1)+z(0,1,1)$, so it’s doomed to failure right from the start. If you make the suggested correction, then it will be essentially identical to your second approach. – 2017-02-16
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    both attempts have an error. But if you make the corrections to both, then yes, they will be identical. – 2017-02-16
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    @ziggurism can you point the error in the first attemp? – 2017-02-16
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    In the first attempt, you write $(x,y,z)=x(1,0,1) + y(0,1,1) + z(0,0,1)$, which is not correct. The only basis for which the components break down like that is the standard basis $(1,0,0),(0,1,0),(0,0,1)$. Instead you should write $(x,y,z)=x(1,0,1)+y(0,1,1)+(z-x-y)(0,0,1).$ In your second attempt, you find the third component of $a(1,0,1)+b(0,1,1)+c(0,0,1)$ to be $c+b$, which is not correct. It should be $a+b+c$. It is essentially the same error in both cases. Add vectors component-wise, and do not forget any. – 2017-02-16

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