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For some reason, this particular problem is throwing me off:

Find the kernel of the linear transformation:

$T: P_2 \rightarrow P_1$
$T(a_0+a_1x+a_2x^2)=a_1+2a_2x$

Since the kernel is the set of all vectors in $V$ that satisfy $T(\vec{v})=\vec{0}$, it's obvious that $a_0$ can be any real number. What matters, if I understand correctly, is that $a_1$ and $a_2$ should equal 0 in order to satisfy the zero vector (i.e. $0+2(0)x$).

Granted that what I stated is correct, why would my book say that the $\ker(T)=\{a_0: a_0 \; \text{is real}\}$?

Yes, $a_0$ can be any real number, but what must $a_1$ or $a_2$ equal? I don't see it specified within the set. Perhaps it's implied - I'm not sure.

Let me add some more detail:

Again, if I understand correctly, I could make a system of equations as such:
$a_1 = 0$
$2a_2 = 0$

From that I can translate it into a matrix and find that $a_1$ and $a_2$ really does equal zero.

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1 Answer 1

up vote 3 down vote accepted

Your argument is totally correct. Your book means that $\ker(T)=\{a_0+0\cdot x+0\cdot x^2|a_0\in \mathbb{R}\}$, i.e. $a_1=0$ and $a_2=0$, which is the same as you proved.

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Thanks, I thought there was something I completely missed! For some reason, my book states the answer for any transformation for polynomials in that form. –  Tim Nov 9 '12 at 20:19
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If the vectors in that space were given as tuples $(a_0,a_1,a_2)$, you'd have to say that $\ker T=\{(a_0,0,0)\mid a_0\in\mathbb R\}$. But here we are allowed to write simply $a_0$ instead of $a_0+0\cdot x+0\cdot x^2$. –  Hagen von Eitzen Nov 9 '12 at 21:03
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