problemset5solutions

problemset5solutions - MAS 213 Linear Algebra II Problem...

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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 co-domain. 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 co-domain is the same thing, namely: γ = { (1 , , 0) , (0 , 1 , 0) , (0 , , 1) } . To construct the i-th column of the representation matrix [ T ] γ β , you: 1. Evaluate T on the i-th vector of the basis. 2. Express the result as a linear combination of the basis vectors. 3. Put the co-eﬃcients of that expression (ordered according to the or- dering of the basis) into the entries of the i-th 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 co-domain: (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.

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problemset5solutions - MAS 213 Linear Algebra II Problem...

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