Let p(t) = f be non-zero constant polynomial, let g(t) = dt+e be a linear polynomial (d‡0), and let r(t) = at² + bt+c be a quadratic polynomial (a + 0). Let's show that {p(t), g(t), r(t)} is a basis for P2. Since P₂ is isomorphic to R³ via the coordinate mapping (with respect to the standard basis for P₂), we can convert this into a question about vectors in IR³. In coordinates, the vectors p, q, r are respectively. To determine whether these vectors are linearly independent, we can cook up a matrix A with the above vectors as its columns and then determine whether A is invertible. Indeed, this matrix is invertible because its determinant is det (A) = and this is quantity is nonzero since a, d, f are all assumed to be nonzero. The determinant was easy to compute because the matrix above is choose one

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.4: Linear Transformations
Problem 3EQ: In Exercises 1-12, determine whether T is a linear transformation. T:MnnMnn defines by T(A)=AB,...
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Let p(t) = f be non-zero constant polynomial, let g(t) = dt+e be a linear polynomial (d‡0), and let r(t) = at² + bt+c be a quadratic polynomial (a 0). Let's show that
{p(t), q(t), r(t)} is a basis for P₂.
Since P₂ is isomorphic to R³ via the coordinate mapping (with respect to the standard basis for P₂), we can convert this into a question about vectors in IR³. In coordinates, the vectors p, q, r are
respectively. To determine whether these vectors are linearly independent, we can cook up a matrix A with the above vectors as its columns and then determine whether A is invertible.
A =
Indeed, this matrix is invertible because its determinant is
det (A) =
and this is quantity is nonzero since a, d, f are all assumed to be nonzero. The determinant was easy to compute because the matrix above is choose one
Transcribed Image Text:Let p(t) = f be non-zero constant polynomial, let g(t) = dt+e be a linear polynomial (d‡0), and let r(t) = at² + bt+c be a quadratic polynomial (a 0). Let's show that {p(t), q(t), r(t)} is a basis for P₂. Since P₂ is isomorphic to R³ via the coordinate mapping (with respect to the standard basis for P₂), we can convert this into a question about vectors in IR³. In coordinates, the vectors p, q, r are respectively. To determine whether these vectors are linearly independent, we can cook up a matrix A with the above vectors as its columns and then determine whether A is invertible. A = Indeed, this matrix is invertible because its determinant is det (A) = and this is quantity is nonzero since a, d, f are all assumed to be nonzero. The determinant was easy to compute because the matrix above is choose one
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