Math 2331 Linear Algebra 4.1 Vector Spaces & Subspaces

Math 2331 – Linear Algebra 4.1 Vector Spaces & Subspaces 4.1 Vector Spaces & Subspaces Vector Spaces: Definition Vector Spaces: Examples 2 × 2 matrices Polynomials Subspaces: Definition Subspaces: Examples Determining Subspaces 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces Many concepts concerning vectors in Rn can be extended to other mathematical systems. We can think of a vector space in general, as a collection of objects that behave as vectors do in Rn . The objects of such a set are called vectors. Vector Space 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 ten axioms below. The axioms must hold for all u, v and w in V and for all scalars c and d. 1. u + v is in V . 2. u + v = v + u. 3 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces (cont.) Vector Space (cont.) 3. (u + v) + w = u + (v + w) 4. There is a vector (called the zero vector) 0 in V such that u + 0 = u. 5. For each u in V , there is vector −u in V satisfying u + (−u) = 0. 6. cu is in V . 7. c(u + v) =cu+cv. 8. (c + d)u = cu + du. 9. (cd)u = c(du). 10. 1u = u. Math 2331, Linear Algebra 4 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces: Examples Example  Let M2×2 = a b c d   : a, b, c, d are real  In this context, note that the 0 vector is Math 2331, Linear Algebra  . 5 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces: Polynomials Example Let n ≥ 0 be an integer and let Pn = the set of all polynomials of degree at most n ≥ 0. Members of Pn have the form p(t) = a0 + a1 t + a2 t 2 + · · · + an t n where a0 , a1 , . . . , an are real numbers and t is a real variable. The set Pn is a vector space. We will just verify 3 out of the 10 axioms here. Let p(t) = a0 + a1 t + · · · + an t n and q(t) = b0 + b1 t + · · · + bn t n . Let c be a scalar. Math 2331, Linear Algebra 6 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces: Polynomials (cont.) Axiom 1: The polynomial p + q is defined as follows: (p + q) (t) = p(t)+q(t). Therefore, (p + q) (t) = p(t)+q(t) =( which is also a )+( )t + ··· + ( of degree at most ) tn . So p + q is in Pn . Math 2331, Linear Algebra 7 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces: Polynomials (cont.) Axiom 4: 0 =0 + 0t + · · · + 0t n (zero vector in Pn ) (p + 0) (t)= p(t)+0 = (a0 + 0) + (a1 + 0)t + · · · + (an + 0)t n = a0 + a1 t + · · · + an t n = p(t) and so p + 0 = p Math 2331, Linear Algebra 8 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Vector Spaces: Polynomials (cont.) Axiom 6: (cp) (t) = cp(t) = ( )+( )t + ··· + ( which is in Pn . ) tn The other 7 axioms also hold, so Pn is a vector space. Math 2331, Linear Algebra 9 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Subspaces Vector spaces may be formed from subsets of other vectors spaces. These are called subspaces. Subspaces A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. For each u and v are in H, u + v is in H. say H is closed under vector addition.) (In this case we c. For each u in H and each scalar c, cu is in H. (In this case we say H is closed under scalar multiplication.) If the subset H satisfies these three properties, then H itself is a vector space. Math 2331, Linear Algebra 10 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Subspaces: Example Example     a  Let H =  0  : a and b are real . Show that H is a   b subspace of R3 . Solution: Verify properties a, b and c of the definition of a subspace. a. The zero vector of R3 is in H (let a = and b = ). b. Adding two vectors in H always produces another vector whose second entry is and therefore the sum of two vectors in H is also in H. (H is closed under addition) c. Multiplying a vector in H by a scalar produces another vector in H (H is closed under scalar multiplication). Since properties a, b, and c hold, V is a subspace of R3 . Math 2331, Linear Algebra 11 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Subspaces: Example (cont.) Note Vectors (a, 0, b) in H look and act like the points (a, b) in R2 . Graphical Depiction of H Math 2331, Linear Algebra 12 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Subspaces: Example Example  Is H =   x : x is real a subspace of x +1 I.e., does H satisfy properties a, b and c? ? Solution: For H to be a subspace of R2 , all three properties must hold Property (a) fails Property (a) is not true because Therefore H is not a subspace of R2 . Math 2331, Linear Algebra . 13 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Subspaces: Example (cont.) Another way to show that H is not a subspace of R2 : Let  u= 0 1   and v =   1 , then u + v = 2  1 and so u + v = , which is 3 and so H is not a subspace of R2 .   in H. So property (b) fails Property (b) fails Math 2331, Linear Algebra 14 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces A Shortcut for Determining Subspaces Theorem (1) If v1 , . . . , vp are in a vector space V , then Span{v1 , . . . , vp } is a subspace of V . Proof: In order to verify this, check properties a, b and c of definition of a subspace. a. 0 is in Span{v1 , . . . , vp } since 0= v1 + v2 + · · · + vp b. To show that Span{v1 , . . . , vp } closed under vector addition, we choose two arbitrary vectors in Span{v1 , . . . , vp } : u =a1 v1 + a2 v2 + · · · + ap vp and v =b1 v1 + b2 v2 + · · · + bp vp . Math 2331, Linear Algebra 15 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces A Shortcut for Determining Subspaces (cont.) Then u + v = (a1 v1 + a2 v2 + · · · + ap vp ) + (b1 v1 + b2 v2 + · · · + bp vp ) =( v1 + v1 ) + ( v2 + v2 ) + · · · + ( vp + vp ) = (a1 + b1 ) v1 + (a2 + b2 ) v2 + · · · + (ap + bp ) vp . So u + v is in Span{v1 , . . . , vp } . c. To show that Span{v1 , . . . , vp } closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span{v1 , . . . , vp } : v =b1 v1 + b2 v2 + · · · + bp vp . Math 2331, Linear Algebra 16 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces A Shortcut for Determining Subspaces (cont.) Then cv =c (b1 v1 + b2 v2 + · · · + bp vp ) = v2 + · · · + v1 + vp So cv is in Span{v1 , . . . , vp } . Since properties a, b and c hold, Span{v1 , . . . , vp } is a subspace of V. Math 2331, Linear Algebra 17 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Determining Subspaces: Recap Recap 1 To show that H is a subspace of a vector space, use Theorem 1. 2 To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. Math 2331, Linear Algebra 18 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Determining Subspaces: Example Example Is V = {(a + 2b, 2a − 3b) : a and b are real} a subspace of R2 ? Why or why not? Solution: Write vectors in V in column form:       a + 2b a 2b = + 2a − 3b 2a −3b  = 1 2   + 2 −3  So V =Span{v1 , v2 } and therefore V is a subspace of Theorem 1. Math 2331, Linear Algebra by 19 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Determining Subspaces: Example Example     a + 2b  Is H =  a + 1  : a and b are real a subspace of R3 ?   a Why or why not? Solution: 0 is not in H since a = b = 0 or any other combination of values for a and b does not produce the zero vector. So property fails to hold and therefore H is not a subspace of R3 . Math 2331, Linear Algebra 20 / 21 4.1 Vector Spaces & Subspaces Vector Spaces Subspaces Determining Subspaces Determining Subspaces: Example Example  Is the set H of all matrices of the form 2a b 3a + b 3b  a subspace of M2×2 ? Explain. Solution: Since       b 2a 0 0 b 2a = + 3a + b 3b 3a 0 b 3b   =a  Therefore H =Span  +b 2 0 3 0  .    0 1 , and so H is a 1 3 subspace of M2×2 . Math 2331, Linear Algebra 21 / 21