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Is there a linear transformation $T: R_3[x] \rightarrow R_3[x] $ such that

$T(x^2-1) = 1$

$T(x^2 +x) = x$

$T(x+1) = 2x-2$

Well, I said that $\operatorname{Im}T = Sp(\{T(x^2-1),T(x^2 +x),T(x+1)\})$

Which means, $\operatorname{Im}T = Sp(\{1,x,2x-2\}) = x(1,0,0) +x(0,1,0) +x(-2,2,0)$

So if I put on matrix $(1,0,0) , (0,1,0) , (-2,2,0)$ and do some row operations I get to:

$\operatorname{Im}T = Sp\{(1,0,0),(0,1,0)\} \rightarrow \operatorname{dim}\operatorname{Im}T = 2 \rightarrow \operatorname{dim}\operatorname{Ker}T = 3-2 = 1$

Now I need somehow to find a base in $R^3$ so I say that it's the standard base of$ R^3 $and then I can take $(0,0,1)$ to make it a base with the span of $ImT$ vecors which are $(1,0,0),(0,1,0). $

Then, can I assume $(0,0,1)$ not inside $ImT$ because it's not linearly independent of $(1,0,0),(0,1,0)$ and also assume that $(0,0,1) \in KerT$ because it's not in $ImT$ and thus there is a linear transformation $T: R^3 \rightarrow R^3$ ? If not how can I prove $T: R^3 \rightarrow R^3$ is a linear transformation? Or actually I can't?

Still trying to solve this question: Kernel of linear transformation in $\Bbb R^3$ Would love to get some help.

Community
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Ilan Aizelman WS
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  • I didn't read your question. Hint: If there is such $T$ what happens to $T(x^2+x)-T(x^2-1)-T(x+1)$? Compute it in two different ways. – Git Gud May 31 '14 at 18:02

1 Answers1

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Hints: assume there is such a linear transformation, then

$$(x^2+x)-(x+1)=x^2-1\implies$$

$$ 1=T(x^2-1)=T\left((x^2+x)-(x+1)\right)\stackrel{\text{linearity}}=T(x^2+x)-T(x+1)$$

So, is the above equality true?

DonAntonio
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