Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 5E from Chapter 4.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
In Exercises 5-8, determine if the given set is a subspace of...
Step-by-Step Solution
Step 1
Given that ${\bf{p}}\left( t \right)$ is set of polynomials of the form\[{\bf{p}}\left( t \right) = a{t^2}\]We need to satisfy following conditions to prove that the given set is a subspace of $P_2$.
(a) The zero vector of V is in $H^2$
(b) H is closed under vector addition. That is, for each u and v in H, the sumu + v is in H.
(c) H is closed under multiplication by scalars. That is, for each u in H and eachscalar c, the vector cu is in H.
Step 2: Condition (a): Property of zero vector
Let V be a vector space that is set of polynomials of second degree given by:\[{\bf{q}}\left( t \right) = a{t^2} + bt + c\]The 0 vector belongs to set V, then \[{\bf{p}}\left( 0 \right) = a{\left( 0 \right)^2} = 0\]Hence the 0 vector belongs to H also. The first condition is satisfied.
Step 3: Condition (b): Property of vector addition
Let us take two vectors in H:\[\begin{array}{l}u = a{t^2}\\\\v = b{t^2}\end{array}\]Upon adding the vectors:\[\begin{array}{l}u + v = a{t^2} + b{t^2}\\\\= \left( {a + b} \right){t^2}\\\\ = k{t^2}\end{array}\]Since, the addition of the above vectors also lies in H, the closure property of vector addition is satisfied.
Step 4: Condition (c): Property of scalar multiplication
Let, c be an arbitrary scalar and a vector u be $u = a{t^2}$, that lies in H.Upon scalar multiplication:\[\begin{array}{l}u = a{t^2}\\\\cu = c \times a{t^2}\\\\ = d{t^2}\end{array}\]Since, the scalar multiplication also lies in H, the closure property of scalar multiplication is satisfied. Since all three conditions are satisfied, the set H is a subspace of V.
Given that ${\bf{p}}\left( t \right)$ is set of polynomials of the form\[{\bf{p}}\left( t \right) = a{t^2}\]We need to satisfy following conditions to prove that the given set is a subspace of $P_2$.
(a) The zero vector of V is in $H^2$
(b) H is closed under vector addition. That is, for each u and v in H, the sumu + v is in H.
(c) H is closed under multiplication by scalars. That is, for each u in H and eachscalar c, the vector cu is in H.
Step 2: Condition (a): Property of zero vector
Let V be a vector space that is set of polynomials of second degree given by:\[{\bf{q}}\left( t \right) = a{t^2} + bt + c\]The 0 vector belongs to set V, then \[{\bf{p}}\left( 0 \right) = a{\left( 0 \right)^2} = 0\]Hence the 0 vector belongs to H also. The first condition is satisfied.
Step 3: Condition (b): Property of vector addition
Let us take two vectors in H:\[\begin{array}{l}u = a{t^2}\\\\v = b{t^2}\end{array}\]Upon adding the vectors:\[\begin{array}{l}u + v = a{t^2} + b{t^2}\\\\= \left( {a + b} \right){t^2}\\\\ = k{t^2}\end{array}\]Since, the addition of the above vectors also lies in H, the closure property of vector addition is satisfied.
Step 4: Condition (c): Property of scalar multiplication
Let, c be an arbitrary scalar and a vector u be $u = a{t^2}$, that lies in H.Upon scalar multiplication:\[\begin{array}{l}u = a{t^2}\\\\cu = c \times a{t^2}\\\\ = d{t^2}\end{array}\]Since, the scalar multiplication also lies in H, the closure property of scalar multiplication is satisfied. Since all three conditions are satisfied, the set H is a subspace of V.