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Let $T: P_2\to P_3$ be a linear transformation defined by $$T(at^{2} +bt+c) = (a-b+c)t^{3} + (-a+3b-2c)t^{2} + (-a-b)t + (2b-c).$$

Find a basis of $\operatorname{range}(T)$.

Would the basis of $\operatorname{range}(T)$ be {$t^{3}$, $t^{2}$, $t$, $1$}? Since they span $T$ ?

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No, that cannot be right, because the image of $T$ cannot be all of $P_3$ (by dimension considerations). – Arturo Magidin Jun 26 '12 at 16:28
up vote 1 down vote accepted

You cannot span a linear transformation, so saying something "spans $T$" is, ipso facto, wrong.

The set you give spans the codomain of $T$; but the codomain is not the same thing as the range. The codomain is a set that contains the range; the range is just the elements that are actually images under $T$.

By dimensional considerations, the image of $P_2$ has dimension at most 3, and so cannot have a basis with four elements. So your answer is definitely wrong.

There are at least two options here:

  1. Hard way. Try to figure out exactly what vectors of $P_3$ can be images of $T$ by looking at the form of $T$; then find a basis for that subspace.

  2. Easy way. The image of a basis for $P_2$ will span the range of $T$. And every spanning set contains a basis. So look at the image under $T$ of your favorite basis for $P_2$, and extract a basis for its span from that set.

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Thank you for your help- I solved it. – Deborah Jun 26 '12 at 17:38
How do you know off the top of your head that the image of $P_2$ has dimension at most $3$? – CodeKingPlusPlus Nov 11 '12 at 20:23

Choose a basis $1,t,t^2$ for $P_2$, and $1,t,t^2,t^3$ for $P_3$. The the matrix for $T$ is given by:

$$ \tau = \left[\begin{array}{rrr} -1 & 2 & 0 \\ 0 & -1 & -1 \\ -2 & 3 & -1 \\ 1 & -1 & 1 \end{array}\right].$$ It is fairly easy to see that $\tau (2,1,-1)^T= 0$ (ie, the second column is equal to the third column minus twice the first column), and that the first and last columns are linearly independent. Thus these two columns span the range space of $T$. Thus $T(1), T(t^2)$ form a basis for the range.

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@Arturo Magidin: Thanks for the edit. My Latex is very elementary (despite using it for close to 25 years). – copper.hat Jun 26 '12 at 17:13
Thank you for your help. – Deborah Jun 26 '12 at 17:39
@copper.hat: What do you mean by $\tau (2,1,-1)^T?$. Also i took a different order of the basis going from $t^3$ to $1$ and then wrote the matrix in row echelon form and found that $T(t)$ and $T(t^2)$ are linearly independent since corresponding to pivot on columns 1 and 2. Is this correct? after all a basis is not unique! – palio Jun 26 '12 at 18:33
I mean multiply the matrix $\tau$ by the vector $(2,1,-1)^T$. Any two of the three columns are linearly independent, so yes, you could take $T(t), T(t^2)$ as a basis. I picked the others because it was obvious from the matrix $\tau$ that the second column could be expressed easily in terms of the other two. – copper.hat Jun 26 '12 at 18:55

HINT: Transform every single element of a basis of the set of polynomials of degree at least two.

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Don't yell, please. – Arturo Magidin Jun 26 '12 at 16:27
@Sebastian: Thank you for your help – Deborah Jun 26 '12 at 17:39

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