Define T:P
2
\u200b
\u2192P
2
\u200b
by T(p)=p(0)\u2212p(1)t+p(2)t
2
.
a. Show that T is a linear transformation. b. Find T(p) when p(t)=\u22122+t . Is p an eigenvector of T ? c. Find the matrix for T relative to the basis {1,t,t
2
} for P
2
\u200b
.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve the problem, we need to analyze the transformation ( T: P_2 \\to P_2 ) defined by ( T(p) = p(0) – p(1)t + p(2)t^2 ). Here, ( P_2 ) is the vector space of polynomials of degree at most 2. We will break down the solution into parts:<\/p>\n\n\n\n To show that ( T ) is a linear transformation, we need to verify two properties:<\/p>\n\n\n\n Let ( p(t) ) and ( q(t) ) be polynomials in ( P_2 ):<\/p>\n\n\n\n Since both properties are satisfied, ( T ) is a linear transformation.<\/p>\n\n\n\n First, we evaluate ( T(p) ):<\/p>\n\n\n\n Now, check if ( p(t) = -2 + t ) is an eigenvector. A polynomial ( p ) is an eigenvector of ( T ) if ( T(p) = \\lambda p ) for some scalar ( \\lambda ). We have: To find the matrix representation, we compute ( T ) for each basis element:<\/p>\n\n\n\n Combining these results, the matrix representation of ( T ) with respect to the basis ( {1, t, t^2} ) is: In conclusion, we demonstrated that ( T ) is a linear transformation by confirming it satisfies the properties of additivity and homogeneity. Evaluating ( T ) at a specific polynomial showed that ( p(t) = -2 + t ) is an eigenvector with eigenvalue ( 1 ). Finally, we derived the matrix representation of ( T ) in the given basis, providing a complete overview of the transformation’s structure in ( P_2 ).<\/p>\n","protected":false},"excerpt":{"rendered":" Define T:P2\u200b\u2192P2\u200bby T(p)=p(0)\u2212p(1)t+p(2)t2.a. Show that T is a linear transformation. b. Find T(p) when p(t)=\u22122+t . Is p an eigenvector of T ? c. Find the matrix for T relative to the basis {1,t,t2} for P2\u200b. The Correct Answer and Explanation is : To solve the problem, we need to analyze the transformation ( T: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-160629","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/160629","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=160629"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/160629\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=160629"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=160629"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=160629"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}a. Show that T is a linear transformation.<\/h3>\n\n\n\n
\n
\n
[
T(p + q) = (p + q)(0) – (p + q)(1)t + (p + q)(2)t^2
]
Using the linearity of evaluation:
[
= (p(0) + q(0)) – (p(1) + q(1))t + (p(2) + q(2))t^2
]
[
= (p(0) – p(1)t + p(2)t^2) + (q(0) – q(1)t + q(2)t^2) = T(p) + T(q)
]<\/li>\n\n\n\n
[
T(cp) = (cp)(0) – (cp)(1)t + (cp)(2)t^2 = c \\cdot p(0) – c \\cdot p(1)t + c \\cdot p(2)t^2 = cT(p)
]<\/li>\n<\/ul>\n\n\n\nb. Find ( T(p) ) when ( p(t) = -2 + t ) and determine if ( p ) is an eigenvector of ( T ).<\/h3>\n\n\n\n
\n
[
T(p) = -2 – (-1)t + 0 \\cdot t^2 = -2 + t
]<\/li>\n<\/ul>\n\n\n\n
[
T(p) = -2 + t \\quad \\text{and} \\quad p = -2 + t
]
This implies ( T(p) = p ), meaning ( \\lambda = 1 ). Hence, ( p ) is indeed an eigenvector of ( T ) with eigenvalue ( \\lambda = 1 ).<\/p>\n\n\n\nc. Find the matrix for T relative to the basis ( {1, t, t^2} ) for ( P_2 ).<\/h3>\n\n\n\n
\n
[
T(1) = 1(0) – 1(1)t + 1(2)t^2 = 0 – t + 0 = -t \\rightarrow \\begin{pmatrix} 0 \\ -1 \\ 0 \\end{pmatrix}
]<\/li>\n\n\n\n
[
T(t) = t(0) – t(1)t + t(2)t^2 = 0 – t + 0 = -t \\rightarrow \\begin{pmatrix} 0 \\ -1 \\ 0 \\end{pmatrix}
]<\/li>\n\n\n\n
[
T(t^2) = t^2(0) – t^2(1)t + t^2(2)t^2 = 0 – t^3 + 2t^2 \\rightarrow \\text{(Not applicable since this exceeds } P_2)
]
Here, we compute it correctly:
[
T(t^2) = 0 – 0 + 2t^2 = 0 + 0 + 2t^2 \\rightarrow \\begin{pmatrix} 0 \\ 0 \\ 2 \\end{pmatrix}
]<\/li>\n<\/ol>\n\n\n\n
[
[T] = \\begin{pmatrix}
0 & 0 & 0 \\
-1 & -1 & 0 \\
0 & 0 & 2
\\end{pmatrix}
]<\/p>\n\n\n\nSummary<\/h3>\n\n\n\n