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# class6 - MAT1341A LECTURE NOTES FOR FALL 2008 MONICA...

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MAT1341A: LECTURE NOTES FOR FALL 2008 MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP) 5. Subspaces and Spanning Sets Last time, we established the concept of a vector space , which is a set V on which we can perform two operations (addition and scalar multiplication) such that 10 axioms are satisfied. The result is: a vector space is algebraically indistinguishable from things like R n (as far as addition and scalar multiplication go). We met some particularly nice vector spaces: R n , n 1 • E , a space of linear equations in 3 variables F [ a, b ], functions on the interval [ a, b ] F ( R ), functions on the real line M m × n ( R ), m × n real matrices And for each of these, we had to check 10 axioms. But we noticed that sometimes, the last 6 axioms (the ones dealing purely with algebraic properties) were “obvious”. Let’s consider this more carefully. 5.1. Subsets of Vectors Spaces. [N, 5.1.2] Suppose V is a vector space, and W V is a subset of V . Example: Let V = R 2 (we know it’s a vector space), and let W = { ( x, 2 x ) | x R } . Use the same operations of addition and scalar multiplication on W as we use in R 2 . What do we really need to verify to decide if it’s a vector space? closure: (1) closure under addition: if ( x, 2 x ) and ( y, 2 y ) are in W , then ( x, 2 x ) + ( y, 2 y ) = ( x + y, 2( x + y )), which is in W because it has the correct form ( z, 2 z ) with z = x + y . YES. (2) closure under scalar multiplication: if ( x, 2 x ) W and c R then c ( x, 2 x ) = ( cx, 2( cx )) W as well. YES. existence: (3) the zero vector is (0 , 0), which is in W . But we don’t need to bother checking if 0 + u = u for each u W , because we know this is true for all u R 2 already (since V is a vector space and so satisfies axiom (3)). So it’s enough to know that 0 W . Date : September 18, 2008. 1

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2 MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP) (4) if u W , then we know that - u = ( - 1) u W because (2) is true. And again, we know that the negative has the correct property because we checked it when we decided that V was a vector space.
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class6 - MAT1341A LECTURE NOTES FOR FALL 2008 MONICA...

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