Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Consider the transformation $T (q (x_{1}, x_{2})) = q (x_{1}, 0)$ from $Q_{2}$ to $P_{2}$ . Is T a linear transformation? If so, find the image, rank, kernel, and nullity of T.

the solution is

Yes,T is a linear transformation.

$\begin{aligned} \ker (T) = \{q (x_{1}, x_{2}) = {bx}_{1} x_{2} + {cx}_{2}^{2} \in Q_{2} : b, c \in R\} \\ Im (T) = \{p (x) = {ax}^{2} \in P_{2} : a \in R\} \\ rank (T) = 1 and nullity (T) = 2 \end{aligned}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: given information

$T (q (x_{1}, x_{2})) = q (x_{1}, 0)$

Step 2: linear transformation

Consider $q_{1} (x_{1}, x_{2}) = a_{1} x_{1}^{2} + b_{1} x_{1} x_{2} + c_{1} x_{2}^{2}$

$q_{2} (x_{1}, x_{2}) = a_{2} x_{1}^{2} + b_{2} x_{1} x_{2} + c_{2} x_{2}^{2} \in Q_{2} and α \in R$ then9

$T (a_{1} (x_{1}, x_{2}) + a_{2} (x_{1}, x_{2})) = T (a (a_{1} x_{1}^{2} + b_{1} x_{1} x_{2} + c_{1} x_{2}^{2}) + a_{2} x_{1}^{2} + b_{2} x_{1} x_{2} + c_{2} x_{2}^{2}) = T ((a a_{1} + a_{2}) x_{1}^{2} + (a b_{1} + b_{2}) x_{1} x_{2} + (a_{1} + c_{2}) x_{2}^{2}) = (a a_{1} + a_{2}) x_{1}^{2} = (a a_{1} + a_{2}) x_{1}^{2} = a q_{1} (x_{1}, 0) + a_{2} (x_{1}, 0) = α T (a_{1} (x_{1}, x_{2})) + T (a_{2} (x_{1}, x_{2}))$

Since T satisfy

$T ({αq}_{1} + q_{2}) = αT (q_{1}) + T (q_{2}) forα \in R and q_{1}, q_{2} \in Q_{2}$

$T : Q_{2} \to P_{2}, T (q (x_{1}, x_{2})) = q (x_{1}, 0)$ is a linear transformation.

Step 3: find kernel of T and image of T

Kernel of T:

$\ker (T) = \{q (x_{1}, x_{2}) \in Q_{2} : T (q (x_{1}, x_{2})) = 0 \in P_{2}\} = \{q (x_{1}, x_{2}) \in Q_{2} : q (x_{1}, 0) = 0 \in P_{2}\} = \{q (x_{1}, x_{2}) = a x_{1}^{2} + b x_{1} x_{2} + c x_{2}^{2} \in Q_{2} : a x_{1}^{2} = 0\} = \{q (x_{1}, x_{2}) = a x_{1}^{2} + b x_{1} x_{2} + c x_{2}^{2} \in Q_{2} : a = 0\} = \{q (x_{1}, x_{2}) = b x_{1} x_{2} + c x_{2}^{2} \in Q_{2} : b, c \in R\}$

Image of T:

$Im (T) = \{T (q (x_{1}, x_{2})) \in P_{2} : q (x_{1}, x_{2}) \in Q_{2}\} = \{q (x_{1}, 0) \in P_{2} : q (x_{1}, x_{2}) = a x_{1}^{2} + b x_{1} x_{2} + c x_{2}^{2} \in Q_{2}\} = \{q (x_{1}, 0) = a x_{1}^{2} \in P_{2} : a \in R\} = \{p (x) = a x^{2} \in P_{2} : a \in R\}$

$rank (T) = \dim (Img (T)) = 1 nullity (T) = \dim (\ker (T)) = 2$

Step 4: conclusion

Yes, is a linear transformation.

$\ker (T) = q (x_{1}, x_{2}) = {bx}_{1} x_{2} + {cx}_{2}^{2} \in Q_{2} : b, c \in R Im (T) = p (x) = {ax}^{2} \in P_{2} : a \in R rank (T) = 1 and nullity (T) = 2$

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Linear Algebra With Applications

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Short Answer

Consider the transformation $T (q (x_{1}, x_{2})) = q (x_{1}, 0)$ from $Q_{2}$ to $P_{2}$ . Is T a linear transformation? If so, find the image, rank, kernel, and nullity of T.

Step by Step Solution

Step 1: given information

Step 2: linear transformation

Step 3: find kernel of T and image of T

Step 4: conclusion

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Step by Step Solution

Step 1: given information

Step 2: linear transformation

Step 3: find kernel of T and image of T

Step 4: conclusion

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Consider the transformation $T (q (x_{1}, x_{2})) = q (x_{1}, 0)$ from $Q_{2}$ to $P_{2}$ . Is T a linear transformation? If so, find the image, rank, kernel, and nullity of T.