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Unformatted text preview: MAS 213: Linear Algebra II. Problem list for Week #5. Tutorial during the lecture on 4th October. Solution to problem 1: Here we will consider several linear transformations T : R n → R m . We let β denote the standard ordered basis of the domain, and let γ denote the standard ordered basis of the codomain. For each linear transformation below, we’ll determine [ T ] γ β , the matrix representation of the given T with respect to these standard ordered bases. Part (i). In this case the linear transformation T : R 3 → R 3 is defined by T ( a 1 , a 2 , a 3 ) = (2 a 2 + a 3 , − a 1 + 4 a 2 + 5 a 3 , a 1 + a 3 ) . We’ll go through this first part in some detail. In this case the standard ordered basis of the domain is: β = { (1 , , 0) , (0 , 1 , 0) , (0 , , 1) } , and the standard ordered basis of the codomain is the same thing, namely: γ = { (1 , , 0) , (0 , 1 , 0) , (0 , , 1) } . To construct the ith column of the representation matrix [ T ] γ β , you: 1. Evaluate T on the ith vector of the basis. 2. Express the result as a linear combination of the basis vectors. 3. Put the coeﬃcients of that expression (ordered according to the or dering of the basis) into the entries of the ith column. So, in particular, to construct the first column of the representation matrix, we first evaluate: T (1 , , 0) = (0 , − 1 , 1) . Then we express this vector as a linear combination of the basis vectors: (0 , − 1 , 1) = 0 ∗ (1 , , 0) + ( − 1) ∗ (0 , 1 , 0) + (1) ∗ (0 , , 1) . 1 Then the first column of the representation matrix is assembled from the coeﬃcients of this expression, namely: · · − 1 · · 1 · · Then we move on to the second column: T (0 , 1 , 0) = (2 , 4 , 0) = 2 ∗ (1 , , 0) + 4 ∗ (0 , 1 , 0) + 0 ∗ (0 , , 1) . Giving: 0 2 · − 1 4 · 1 0 · And the final column is determined by: T (0 , , 1) = (1 , 5 , 1) = 1 ∗ (1 , , 0) + 5 ∗ (0 , 1 , 0) + 1 ∗ (0 , , 1) . And so the result is: [ T ] γ β = 0 2 1 − 1 4 5 1 0 1 Part (ii). In this case the linear transformation T : R n → R n is defined by T ( a 1 , . . . , a n ) = ( a 1 , . . . , a 1 ) . In this case all of the basis vectors of the domain are mapped to the zero vector except the first vector, which is: (1 , , . . . , 0). So all the columns of the representation matrix, except the first column, will be filled with zeros. To determine the first column, we evaluate T on that first vector, getting: T (1 , , , . . . , 0) = (1 , . . . , 1) . This vector is the following linear combination of the basis vectors of the codomain: (1 , . . . , 1) = 1 ∗ (1 , , , . . . , 0) + 1 ∗ (0 , 1 , , . . . , 0) + . . . + 1 ∗ (0 , , , . . . , 1) ....
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This note was uploaded on 12/08/2010 for the course SPMS MAS213 taught by Professor Andrewkricker during the Fall '10 term at Nanyang Technological University.
 Fall '10
 ANDREWKRICKER

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