DEFINITION: A vector space is a nonempty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the following axioms (or rules):. The sum of ū and v, denoted by ū + v, is in V. 2. ū + v = v + ū. 3. (ū + v) + w = ū + ( v + w). 4. There is a zero vector in V such that ū + = ū. 5. For each ū in V, there is a vector ū in V such that ū + ( ū) =. 6. The scalar multiple of ū by c, denoted by cū, is in V. 7. c(ū + v) = cū + c v. 8. (c + d)ū = cū + dū. 9. c(dū) = (cd)ū.. ū = ū. These axioms must hold for all vectors ū, v, and w in V and all scalars c and d.
EXAMPLE:. R n = x x 2. x n : x,..., x n R 2. The set P n of polynomials of degree at most n: p(t) = a n t n +... + a 2 t 2 + a t + a where the coefficients a n,..., a and the variable t are real numbers. 3. The set of all real-valued functions defined on R.
DEFINITION: A subspace of a vector space V is a subset H of V that has 3 properties:. The zero vector of V is in H. 2. H is closed under vector addition. That is, for each ū and v in H, the sum ū + v is in H. 3. H is closed under multiplication by scalars. That is, for each ū in H and each scalar c, the vector cū is in H. EXAMPLE: The set consisting of only the zero vector in a vector space V is a subspace of V, called the zero subspace and written as { }.
WARNING: R 2 is not a subspace of R 3, because R 2 is not a subset of R 3. EXAMPLE: The set H = s t is a subspace of R 3. : s and t are real numbers
THEOREM: If v,..., v p are in a vector space V, then Span { v,..., v p } is a subspace of V. EXAMPLE: Let v = 2 v 2 = 3 By the Theorem above Span{ v, v 2 } is a subspace of R 3. 2.
EXAMPLE: Let H be the set of all vectors of the form 4a b 2b a 2b a b where a and b are arbitrary scalars. Show that H is a subspace of R 4.
SOLUTION: We have 4a b 2b a 2b a b We see that = a 4 } {{ } v +b H = Span{ v, v 2 } 2 2 } {{ } v 2 therefore H is a subspace of R 4 by the Theorem above.
EXAMPLE: Let H be the set of all vectors of the form a b b c c a b where a, b and c are arbitrary scalars. Find a set S of vectors that spans H or show that H is not a vector space.
SOLUTION: We have a b b c c a b = a +b } {{ } } {{ } } {{ } v v 2 v 3 and we see that H is a vector space and spans H. { v, v 2, v 3 } +c
EXAMPLE: Let H be the set of all vectors of the form 3a + b 4 a 5b where a and b are arbitrary scalars. Show that H is not a vector space. SOLUTION: H is not a vector space, since H (the second entry is always nonzero).
EXAMPLE: Let H be the set of all vectors of the form: a (a) b (b) a b, where a (c) c a a (d) a b, where a + 2b c = (e) c a b, where 2a b + c =. c Is H a subspace of R 3?
SOLUTION: (a) We have a b = a + b, which is a linear combination and therefore a subspace of R 3. (b) It is not a subspace, since it is not closed under multiplication by a scalar. In fact, ( ) a b c = a b c, and the first entry is negative, which contradicts the initial assumption.
(c) It is not a subspace, since there is no a zero vector (the last entry is always nonzero). (d) We have a a b = b c a + 2b = a + b 2, which is a linear combination and therefore a subspace of R 3.
(e) We have a b = c a b 2a + b, which is not a subspace, since there is no a zero vector. In fact, a b 2a + b = if and only if a = and b =. But in this case the last entry in nonzero, since 2a + b = 2 + =. Contradiction.
DEFINITION: The null space of an m n matrix A, written as Nul A, is the set of all solutions to the homogeneous equation A x =. DEFINITION : The null space of an m n matrix A is the set of all x in R n that are mapped into the zero vector in R m by the linear transformation x A x.
EXAMPLE: Let [ ] 2 A =. 2 3 4 5 Determine if ū = 2 belongs to the null space of A. SOLUTION: Since Aū = ū is in Nul A. [ 2 2 3 4 ] 5 2 = [ ],
EXAMPLE: Let [ ] 3 2 A =. 5 9 5 Determine if ū = 3 belongs to the 2 null space of A. SOLUTION: Since Aū = [ ū is in Nul A. 3 2 5 9 ] 5 3 2 = [ ],
THEOREM: The null space of an m n matrix A is a subspace of R n. Equivalently, the set of all solutions to a system A x = of m homogeneous linear equations in n unknowns is a subspace of R n. EXAMPLE: Find a spanning set for the null space of the matrix 3 6 7 A = 2 2 3. 2 4 5 8 4
SOLUTION: We find the general solution of A x = : 2 3 [A ] 2 2, therefore { x 2x 2 x 4 + 3x 5 = x 3 + 2x 4 2x 5 =, so x x 2 x 3 x 4 x 5 = x 2 = 2 2x 2 + x 4 3x 5 x 2 2x 4 + 2x 5 x 4 x 5 } {{ } } {{ } ū v so Nul A =Span {ū, v, w}. +x 4 2 +x 5 3 2, } {{ } w
DEFINITION: The column space of an m n matrix A, written as Col A, is the set of all linear combinations of the columns of A. REMARK: So, if A = [ā... ā n ], then Col A = Span{ā,..., ā n }.
THEOREM: The column space of an m n matrix is a subspace of R m. EXAMPLE: Let 2 4 2 A = 2 5 7 3. 3 7 8 6 Find a nonzero vector in Col A and a nonzero vector in Nul A.
SOLUTION:. Any column of A is a nonzero vector 2 in Col A. For example, 2 = 3 2 4 2 = 2 + 5 + 7 + 3. 3 7 8 6
2. To find a nonzero vector in Nul A, we row reduce the augmented matrix [A ]: [A ] therefore any vector x x = x 2 x 3 = x 4 9 5 9x 3 5x 3 x 3, is in Nul A. For example, if we put x 3 =, we get 9 ū = 5 is in Nul A.