# Show that the mapping T R2 rightarrow P2 defined by T(a, b) = at2

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Show that the mapping T : R2 rightarrow P2 defined by
T(a, b) = at2 + (a + b)t + a - 2b, (a, b) R2 is a linear
transformation. Then, find the kernel of T. Let T : V
rightarrow W be a linear transformation from a vector
space V into a vector space W. Prove that the range of T is
a subspace of W.
Solution
1. To show that T is linear, we need to show that it is
closed under addition and scalar multiplication... that is:
A) T(a,b)+T(c,d) = T(a+c,b+d)
and B) kT(a,b)=T(ka,kb).
A) T(a,b)+T(c,d)
= at2+ (a + b)t + a - 2b + ct2 + (c + d)t + c - 2d (apply T to
(a,b) and (c,d) )
= (a+c)t2 + (a+c+b+d)t + (a + c) - 2(b + d) (collect terms)
= T(a+c, b+d) (Hey look, it worked!)
B) kT(a,b)
=k[at2+ (a + b)t + a - 2b]
=kat2+ k(a + b)t + ka - 2kb
=kat2+ (ka + kb)t + ka - 2kb

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=T(ka, kb)
Since we\'ve proved A and B, T is in fact linear. The kernel
of T is the set of all (a,b) such that T(a,b)=0
so if (a,b) is in the kernel, 0 = at2+ (a + b)t + a - 2b
so a = 0, a + b = 0 and a - 2b = 0 . (all of the coefficients
of t have to be 0).
Solving we get a=0 and b=0, so the kernel is just {(0,0)}
2. A subspace of W only needs to satisfy closure under
For any two vectors X and Y in range of T, there exists x,y
such that T(x)=X and T(y)=Y.
But we know that T(x+y) = T(x) + T(y) = X + Y. (closure
under linear transformations).
Hence X + Y is in the range of T and so range of T is
Multiplication:
For any vector X in the range of T and scalar k, there
exists x such that T(x)=X.
Now we know that T(kx) = kT(x) = kX.
So we know that kX lies in the range of T so the range of T
is closed under scalar multiplication.
Therefore the range of T is a subspace of W.

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Show that the mapping T : R2 rightarrow P2 defined by T(a, b) = at2 + (a + b)t + a - 2b, (a, b) R2 is a linear transformation. Then, find the kernel of T. Let T : V rightarrow W be a linear transformation from a vector space V into a vector space W. Prove that the range of T is a subspace of W. Solution 1. To show that T is linear, we need to show that it is closed under addition and scalar multiplication... that is: A) T(a,b)+T(c,d) = T(a+c,b+d) and B) kT(a,b)=T(ka,kb). A) T(a,b)+T(c,d) = at2+ (a + b)t + a - 2b + ct2 + (c + d)t + c - 2d (apply T to (a,b) and (c,d) ) = (a+c)t2 + (a+c+b+d)t + (a + c) - 2(b + d) (collect terms) = T(a+c, b+d) (Hey look, it worked!) B) kT(a,b) =k[at2+ (a + b)t + a - 2b] =kat2+ k(a + b)t + ka - 2kb =kat2+ (ka + kb)t + ka - 2kb =T(ka, kb) Since we\'ve proved A and B, T is in fact linear. The kernel of T is the set of all (a,b) such that T(a,b)=0 so if (a ...
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