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Lect2-04-web

Lect2-04-web - MATH 304 Fall 2011 Linear Algebra Homework...

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MATH 304, Fall 2011 Linear Algebra

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Homework assignment #5 (due Thursday, October 5) All problems are from Leon’s book (8th edition). Section 3.3: 2b, 2c, 2e, 8c, 9d, 14 Section 3.4: 8c, 10, 14c, 15a
MATH 304 Linear Algebra Lecture 11: Basis and dimension.

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Linear independence Definition. Let V be a vector space. Vectors v 1 , v 2 , . . . , v k V are called linearly dependent if they satisfy a relation r 1 v 1 + r 2 v 2 + · · · + r k v k = 0 , where the coefficients r 1 , . . . , r k R are not all equal to zero. Otherwise the vectors v 1 , v 2 , . . . , v k are called linearly independent . That is, if r 1 v 1 + r 2 v 2 + · · · + r k v k = 0 = r 1 = · · · = r k = 0. An infinite set S V is linearly dependent if there are some linearly dependent vectors v 1 , . . . , v k S . Otherwise S is linearly independent . Remark. If a set S (finite or infinite) is linearly independent then any subset of S is also linearly independent.
Theorem Vectors v 1 , . . ., v k V are linearly dependent if and only if one of them is a linear combination of the other k 1 vectors. Examples of linear independence. Vectors e 1 = (1 , 0 , 0), e 2 = (0 , 1 , 0), and e 3 = (0 , 0 , 1) in R 3 . Matrices E 11 = parenleftbigg 1 0 0 0 parenrightbigg , E 12 = parenleftbigg 0 1 0 0 parenrightbigg , E 21 = parenleftbigg 0 0 1 0 parenrightbigg , and E 22 = parenleftbigg 0 0 0 1 parenrightbigg . Polynomials 1 , x , x 2 , . . . , x n , . . .

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Spanning set Let S be a subset of a vector space V . Definition. The span of the set S is the smallest subspace W V that contains S . If S is not empty then W = Span ( S ) consists of all linear combinations r 1 v 1 + r 2 v 2 + · · · + r k v k such that v 1 , . . ., v k S and r 1 , . . . , r k R . We say that the set S spans the subspace W or that S is a spanning set for W . Remark. If S 1 is a spanning set for a vector space V and S 1 S 2 V , then S 2 is also a spanning set for V .
Basis Definition. Let V be a vector space. A linearly independent spanning set for V is called a basis . Suppose that a set S V is a basis for V . “Spanning set” means that any vector v V can be represented as a linear combination v = r 1 v 1 + r 2 v 2 + · · · + r k v k , where v 1 , . . . , v k are distinct vectors from S and r 1 , . . . , r k R . “Linearly independent” implies that the above representation is unique: v = r 1 v 1 + r 2 v 2 + · · · + r k v k = r 1 v 1 + r 2 v 2 + · · · + r k v k = ( r 1 r 1 ) v 1 + ( r 2 r 2 ) v 2 + · · · + ( r k r k ) v k = 0 = r 1 r 1 = r 2 r 2 = . . . = r k r k = 0

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Examples. Standard basis for R n : e 1 = (1 , 0 , 0 , . . ., 0 , 0), e 2 = (0 , 1 , 0 , . . ., 0 , 0),. . . , e n = (0 , 0 , 0 , . . ., 0 , 1). Indeed, ( x 1 , x 2 , . . . , x n ) = x 1 e 1 + x 2 e 2 + · · · + x n e n . Matrices parenleftbigg 1 0 0 0 parenrightbigg , parenleftbigg 0 1 0 0 parenrightbigg , parenleftbigg 0 0 1 0 parenrightbigg , parenleftbigg 0 0 0 1 parenrightbigg form a basis for M 2 , 2 ( R ).
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