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Lecture 22

# MATH 125 Lecture 22: Lecture 22.PDF

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Department
Mathematics
Course
MATH 125
Professor
Scott
Semester
Fall

Description
\$5 Subspaces, Basis, Dimension and Rank Definition: Let S be a set of vectors in Rn. We say that S is a subspace of R" if the following conditions hold: (S1) The zero vector is in S (S2) If i an i are vectors in S, then is also in S (we say that S is closed under addition) (S3) If i is a vector in S, and c E R is any scalar, then ci is in S (we say that S is closed under scalar multipli- cation) Aside: Subspaces of R provide further examples of the so- called vector spaces alluded to in earlier discussion. As we shall see shortly, we have been implicitly working with the concept of a subspace all along. For example, the non- trivial subspaces of R3 are precisely the lines and planes in R3 which pass through the origin. Example 5.1 Let S (1) Find two elements of s (2) Find a vector UE R4 which is not in S (3) Show that S is a subspace of R4 ES, since Solution (1) 1 o 1 1 to 0 1 1 1 2 2) S Sin Ce (3) me check that conditions CSI S 3) hold e S 0 0 O Since CS 1 (s 2) Let u v' E S. Then here d at and d' a bi c I. Hence a a 1 E S b b C C d d Since Ca+ a 53) Let E S. The b where d at b+ c f ER, then 1 v e S 1 d Since \$5 Subspaces, Basis, Dimension and Rank Definition: Let S be a set of vectors in Rn. We say that S is a subspace of R" if the following conditions hold: (S1) The zero vector is in S (S2) If i an i are vectors in S, then is also in S (we say that S is closed under addition) (S3) If i is a vector in S, and c E R is any scalar, then ci is in S (we say that S is closed under scalar multipli- cation) Aside: Subspaces of R provide further examples of the so- called vector spaces alluded to in earlier discussion. As we shall see shortly, we have been implicitly working with the concept of a subspace all along. For example, the non- trivial subspaces of R3 are precisely the lines and planes in R3 which pass through the origin. Example 5.1 Let S (1) Find two elements of s (2) Find a vector UE R4 which is not in S (3) Show that S is a subspace of R4 ES, since Solution (1) 1 o 1 1 to 0 1 1 1 2 2) S Sin Ce (3) me check that conditions CSI S 3) hold e S 0 0 O Since CS 1 (s 2) Let u v' E S. Then here d at and d' a bi c I. Hence a a 1 E S b b C C d d Since Ca+ a 53) Let E S. The b where d at b+ c f ER, then 1 v e S 1 d Since(a b c Hence 3) all hold so S is a of subspace Example 5.2: Let S a ER2 a ER (1) Find two elements of S (2) Show that S is not a subspace of R2. Solution: (1) (2) We must show that a least one of (St S3) fails to hold. In fact, they all do E IR or any St) fails, sin ce CS2) fails, sine, for example, but E S cos not 1) S3) fails, since, for example, its last E S, but 2 Summary To show that a subset S CR is a subspace of Rn, We check that conditions (S1-S3) hold for S. This demands a general argument (examples do not suffice!) To show that a subset S C Rn is not a subspace of Rn, we must provide a counterexample to one of (S1-S3) Example 5.3: Is the set a subspace of R3? Solution No In this case, S1) and (s3) hold (Chech I), but (52 doesn't. For example since 12 o z 13) E S O2 13 12) since but S in ce (a b c Hence 3) all hold so S is a of subspace Example 5.2: Let S a ER2 a ER (1) Find two elements of S (2) Show that S is not a subspace of R2. Solution: (1) (2) We must show that a least one of (St S3) fails to hold. In fact, they all do E IR or any St) fails, sin ce CS2) fails, sine, for example, but E S cos not 1) S3) fails, since, for example, its last E S, but 2 Summary To show that a subset S CR is a subspace of Rn, We check that conditions (S1-S3) hold for S. This demands a general argument (examples do not suffice!) To show that a subset S C Rn is not a subspace of Rn, we must provide a counterexample to one of (S1-S3) Example 5.3: Is the set a subspace of R3? Solution No In this case, S1) and (s3) hold (Chech I), but (52 doesn't. For example since 12 o z 13) E S O2 13 12) since but
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