Science:Math Exam Resources/Courses/MATH221/December 2008/Question 09 (a)

MATH221 December 2008

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Question 09 (a)
Let $B=\{b_{1},b_{2}\}$ be a basis for $\mathbb {R} ^{2}$ and let $T:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the linear transformation such that $T(b_{1})=2b_{1}+b_{2}$ and $T(b_{2})=b_{2}$. Find the matrix of T relative to the basis B.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. In general, if $\{e_{1},\ldots ,e_{n}\}$ is a basis for $\mathbb {R} ^{n}$ , and T is a linear transformation mapping $\mathbb {R} ^{n}\to \mathbb {R} ^{n}$, we know that for each j, $T(e_{j})$ is a linear combination of the basis vectors, i.e. $T(e_{j})=\sum _{i=1}^{n}a_{ij}e_{i},\quad a_{ij}\in \mathbb {R}$. Recall T is completely determined by its action on the basis vectors. So the coordinates $a_{ij}$ completely characterize the matrix T, in the sense that if we know all the $a_{ij}$, then we know exactly the linear transformation T. Therefore, the matrix whose entry in the ith row and jth column is $a_{ij}$ is called the matrix of T relative to the basis $\{e_{1},\ldots ,e_{n}\}$. For this problem, we are given that $T(b_{1})=2\cdot b_{1}+1\cdot b_{2}$ and $T(b_{2})=0\cdot b_{1}+1\cdot b_{2}$. Therefore, writing $T(b_{j})=\sum _{i=1}^{2}a_{ij}b_{i}\quad (j=1,2)$ we identify $a_{11}=2,a_{21}=1,a_{12}=0,a_{22}=1$. Hence the matrix of T relative to the given basis B is $[T]_{B}={\begin{pmatrix}2&0\\1&1\end{pmatrix}}.$

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