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Unformatted text preview: MAS 213: Linear Algebra II. Problem list for Week #7. Solutions. Problem 1: (Problem 2.5.2, 2.5.3, various parts, from [FIS].) In the parts below you are given a vector space V and two bases for it, β and β ′ . Find the changeofcoordinates matrix which takes β ′coordinates to βcoordinates. (i) V = R 2 , β = { (1 , 0) , (0 , 1) } , and β ′ = { ( a 1 ,a 2 ) , ( b 1 ,b 2 ) } . (ii) V = P 2 ( R ), β = { 2 x 2 − x, 3 x 2 + 1 ,x 2 } and β ′ = { 1 ,x,x 2 } . (iii) V = P 2 ( R ), β = { x 2 − x,x 2 + 1 ,x − 1 } , β ′ = { 5 x 2 − 2 x − 3 , − 2 x 2 + 5 x + 5 , 2 x 2 − x − 3 } . Solution to part (i). This is the standard procedure. We go through the basis we are moving “from” (in this case it is β ′ ), expressing each vector as a linear combination of the vectors of the basis we are moving “to”. The coeﬃcients we find then form the columns of the changeofcoordinates matrix. • ( a 1 ,a 2 ) = a 1 ∗ (1 , 0) + a 2 ∗ (0 , 1). • ( b 1 ,b 2 ) = b 1 ∗ (1 , 0) + b 2 ∗ (0 , 1). The coeﬃcients are assembled into the “changeofcoeﬃcients” matrix: Q = [ I ] β β ′ = [ a 1 b 1 a 2 b 2 ] . Solution to part (ii). We do the same thing here. This part is going to be different, though, because now we are moving from the standard ordered basis, not to it. That makes the calculations a bit harder. We need to express the three vectors in β ′ as linear combinations of the vectors in β . The equations we get can be solved simultaneously by solving the following augmented matrix: 0 1 0 1 0 0 − 1 0 0 0 1 0 2 3 1 0 0 1 → 1 0 0 − 1 0 0 1 0 1 0 0 2 3 1 0 1 → 1 0 0 − 1 0 0 1 0 1 0 0 0 0 1 − 3 2 1 . Let’s decode exactly what we have just calculated. We have just calculated that: • 1 = (0) ∗ (2 x 2 − x ) + (1) ∗ (3 x 2 + 1) + ( − 3) ∗ ( x 2 ). • x = ( − 1) ∗ (2 x 2 − x ) + (0) ∗ (3 x 2 + 1) + (2) ∗ ( x 2 ). • x 2 = (0) ∗ (2 x 2 − x ) + (0) ∗ (3 x 2 + 1) + (1) ∗ ( x 2 ). 1 2 So the required changeofcoordinates matrix is: − 1 0 1 0 0 − 3 2 1 . We could also have solved this by writing down the changeofcoordinates matrix which changes from βcoordinates to β ′coordinates (this would be easy because the standard ordered basis is in the “to” position) and then taking the inverse via some other matrix algorithm. Solution to part (iii). We’ll be brief here, because it is very similar to part (ii). In this case the calculation which expresses the three vectors of β ′ as combinations of the three vectors of β proceeds: 0 1 − 1 − 3 5 − 3 − 1 0 1 − 2 5 − 1 1 1 5 − 2 2 → 1 1 5 − 2 2 0 1 − 1 − 3 5 − 3 0 1 1 3 3 1 → 1 1 5 − 2 2 0 1 − 1 − 3 5 − 3 0 0 2 6 − 2 4 → 1 1 0 5 − 2 2 0 1 0 4 − 1 0 0 1 3 − 1 2 → 1 0 0 5 − 6 3 0 1 0 4 − 1 0 0 1 3 − 1 2 ....
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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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